{"id":3479,"date":"2015-05-06T03:51:00","date_gmt":"2015-05-06T03:51:00","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=3479"},"modified":"2021-05-11T15:18:47","modified_gmt":"2021-05-11T15:18:47","slug":"relative-strengths-of-acids-and-bases-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/chapter\/relative-strengths-of-acids-and-bases-2\/","title":{"raw":"Relative Strengths of Acids and Bases","rendered":"Relative Strengths of Acids and Bases"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Assess the relative strengths of acids and bases according to their ionization constants<\/li>\r\n \t<li>Rationalize trends in acid\u2013base strength in relation to molecular structure<\/li>\r\n \t<li>Carry out equilibrium calculations for weak acid\u2013base systems<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idm193375984\">The relative strength of an acid or base is the extent to which it ionizes when dissolved in water. If the ionization reaction is essentially complete, the acid or base is termed\u00a0<em data-effect=\"italics\">strong<\/em>; if relatively little ionization occurs, the acid or base is weak. As will be evident throughout the remainder of this chapter, there are many more weak acids and bases than strong ones. The most common strong acids and bases are listed in Table 1.<\/p>\r\n\r\n<table>\r\n<thead>\r\n<tr>\r\n<th colspan=\"2\">Table\u00a01. Some Common Strong acids and \u00a0Strong Bases<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Strong Acids<\/th>\r\n<th>Strong Bases<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>HClO<sub>4<\/sub> perchloric acid<\/td>\r\n<td>LiOH lithium hydroxide<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HCl hydrochloric acid<\/td>\r\n<td>NaOH sodium\u00a0hydroxide<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HBr hydrobromic acid<\/td>\r\n<td>KOH potassium\u00a0hydroxide<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HI hydroiodic acid<\/td>\r\n<td>Ca(OH)<sub>2<\/sub> calcium\u00a0hydroxide<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HNO<sub>3<\/sub> nitric acid<\/td>\r\n<td>Sr(OH)<sub>2<\/sub> strontium\u00a0hydroxide<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H<sub>2<\/sub>SO<sub>4<\/sub> sulfuric acid<\/td>\r\n<td>Ba(OH)<sub>2<\/sub> barium hydroxide<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"CNX_Chem_14_03_strong\" class=\"os-figure\">The relative strengths of acids may be quantified by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the\u00a0<strong><span id=\"term586\" data-type=\"term\">acid-ionization constant,\u00a0<em data-effect=\"italics\">K<\/em><sub>a<\/sub><\/span><\/strong>. For the reaction of an acid HA:<\/div>\r\n<div><\/div>\r\n<div class=\"os-figure\" style=\"text-align: center;\">[latex]\\text{HA}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{A}}^{-}\\left(aq\\right)[\/latex]<\/div>\r\n<p style=\"text-align: left;\">the acid ionization constant is written<\/p>\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{A}}^{-}\\right]}{\\text{[HA]}}[\/latex]<\/p>\r\nwhere the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include [H<sub>2<\/sub>O] in the equation. The larger the\u00a0<em data-effect=\"italics\">K<\/em><sub>a<\/sub>\u00a0of an acid, the larger the concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex]and A<sup>\u2212<\/sup>\u00a0relative to the concentration of the nonionized acid, HA, in an equilibrium mixture, and the stronger the acid. An acid is classified as \u201cstrong\u201d when it undergoes complete ionization, in which case the concentration of HA is zero and the acid ionization constant is immeasurably large (<em data-effect=\"italics\">K<\/em><sub>a<\/sub>\u00a0\u2248 \u221e). Acids that are partially ionized are called \u201cweak,\u201d and their acid ionization constants may be experimentally measured. A table of ionization constants for weak acids is provided in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids.<\/a>\r\n\r\nTo illustrate this idea, three acid ionization equations and\u00a0<em data-effect=\"italics\">K<\/em><sub>a<\/sub>\u00a0values are shown below. The ionization constants increase from first to last of the listed equations, indicating the relative acid strength increases in the order [latex]\\text{CH}_{3}\\text{CO}_{2}\\text{H}[\/latex] &lt; [latex]\\text{HNO}_{2}[\/latex]\u00a0&lt; [latex]{\\text{HSO}}_{4}^{-}:[\/latex]\r\n<p style=\"text-align: left; padding-left: 90px;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+\\text{H}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\left(aq\\right)\\qquad{K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 90px;\">[latex]{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{NO}}_{2}^{-}\\left(aq\\right)\\qquad{K}_{\\text{a}}=4.6\\times {10}^{-4}[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 90px;\">[latex]{\\text{HSO}}_{4}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(aq\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{SO}}_{4}^{2-}\\left(aq\\right)\\qquad{K}_{\\text{a}}=1.2\\times {10}^{-2}[\/latex]<\/p>\r\nAnother measure of the strength of an acid is its percent ionization. The <b>percent ionization<\/b> of a weak acid is defined in terms of the compostion of an equilibrium mixture:\r\n<p style=\"text-align: center;\">[latex]\\text{% ionization}=\\dfrac{{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}_{\\text{eq}}}{{\\left[\\text{HA}\\right]}_{0}}\\times 100[\/latex]<\/p>\r\nwhere the numerator is equivalent to the concentration of the acid's conjugate base (per stoichiometry, [A<sup>\u2212<\/sup>] = [H<sub>3<\/sub>O<sup>+<\/sup>]). Unlike the\u00a0<em>K<\/em><sub>a<\/sub>\u00a0value, the percent ionization of a weak acid varies with the initial concentration of acid, typically decreasing as concentration increases. Equilibrium calculations of the sort described later in this chapter can be used to confirm this behavior.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0Calculation of Percent Ionization from pH<\/h3>\r\nCalculate the percent ionization of a 0.125-<em>M<\/em> solution of nitrous acid (a weak acid), with a pH of 2.09.\r\n[reveal-answer q=\"996427\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"996427\"]\r\n\r\nThe percent ionization for an acid is:\r\n<p style=\"text-align: center;\">[latex]\\dfrac{{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}_{\\text{eq}}}{{\\left[{\\text{HNO}}_{2}\\right]}_{0}}\\times 100[\/latex]<\/p>\r\nThe chemical equation for the dissociation of the nitrous acid is: [latex]{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NO}}_{2}^{-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)[\/latex]. Since 10<sup>\u2212pH<\/sup> = [latex]\\left[\\text{H}_{3}\\text{O}^{+}\\right][\/latex], we find that 10<sup>\u22122.09<\/sup> = 8.1 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>, so that percent ionization is:\r\n<p style=\"text-align: center;\">[latex]\\dfrac{8.1\\times {10}^{-3}}{0.125}\\times 100=6.5\\%[\/latex]<\/p>\r\nRemember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.\r\n\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idp51316352\">Check Your Learning<\/h4>\r\nCalculate the percent ionization of a 0.10-<em>M<\/em> solution of acetic acid with a pH of 2.89.\r\n\r\n[reveal-answer q=\"74260\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"74260\"]\r\n\r\n1.3% ionized[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">Watch this\u00a0<a href=\"https:\/\/phet.colorado.edu\/en\/simulation\/acid-base-solutions\" target=\"_blank\" rel=\"noopener\">simulation of strong and weak acids and bases at the molecular level<\/a>.<\/div>\r\nAs we did with acids, we can measure the relative strengths of bases by measuring their <b>base-ionization constant, (<em>K<\/em><sub>b<\/sub>)<\/b> in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B:\r\n<p style=\"text-align: center;\">[latex]\\text{B}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{HB}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex],<\/p>\r\nwe write the equation for the ionization constant as:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{b}}=\\dfrac{\\left[{\\text{HB}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[\\text{B}\\right]}[\/latex]<\/p>\r\nwhere the concentrations are those at equilibrium. Again, we do not include [H<sub>2<\/sub>O] in the equation because water is the solvent. The chemical reactions and ionization constants of the three bases shown are:\r\n<p style=\"text-align: left; padding-left: 90px;\">[latex]{\\text{NO}}_{2}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{HNO}}_{2}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\qquad{K}_{\\text{b}}=2.22\\times {10}^{-11}[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 90px;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\qquad{K}_{\\text{b}}=5.6\\times {10}^{-10}[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 90px;\">[latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\qquad{K}_{\\text{b}}=1.8\\times {10}^{-5}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">A table of ionization constants of weak bases appears in <a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-bases-2\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Bases<\/a>. As for acids, the relative strength of a base is also reflected in its percent ionization, computed as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\%\\text{ ionization}=[\\text{OH}^{-}]_{eq}\/[\\text{B}]_{0}\\times{100}\\%[\/latex]<\/p>\r\nbut will vary depending on the base ionization constant and the initial concentration of the solution.\r\n<h2>Relative Strengths of Conjugate Acid-Base Pairs<\/h2>\r\nBr\u00f8nsted-Lowry acid-base chemistry is the transfer of protons; thus, logic suggests a relation between the relative strengths of conjugate acid-base pairs. The strength of an acid or base is quantified in its ionization constant,\u00a0<em>K<\/em><sub>a<\/sub>\u00a0or\u00a0<em>K<\/em><sub>b<\/sub>, which represents the extent of the acid or base ionization reaction. For the conjugate acid-base pair HA \/ A<sup>\u2212<\/sup>, ionization equilibrium equations and ionization constant expressions are\r\n<p style=\"padding-left: 90px;\">[latex]\\text{HA}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{A}}^{-}\\left(aq\\right)\\qquad{K}_{\\text{a}}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{A}}^{-}\\right]}{\\left[\\text{HA}\\right]}[\/latex]<\/p>\r\n<p style=\"padding-left: 90px;\">[latex]{\\text{A}}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{OH}}^{-}\\left(aq\\right)+\\text{HA}\\left(aq\\right)\\qquad{K}_{\\text{b}}=\\dfrac{\\left[\\text{HA}\\right]\\left[\\text{OH}\\right]}{\\left[{\\text{A}}^{-}\\right]}[\/latex]<\/p>\r\nAdding these two chemical equations yields the equation for the autoionization for water:\r\n<p style=\"text-align: center;\">[latex]\\cancel{\\text{HA}\\left(aq\\right)}+\\text{H}_{2}\\text{O}\\left(l\\right)+\\cancel{\\text{A}^{-}\\left(aq\\right)}+\\text{H}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons\\text{H}_{3}\\text{O}^{+}\\left(aq\\right)+\\cancel{\\text{A}^{-}\\left(aq\\right)}+\\text{OH}^{-}\\left(aq\\right)+\\cancel{\\text{HA}\\left(aq\\right)}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\text{2H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nAs discussed in another chapter on equilibrium, the equilibrium constant for a summed reaction is equal to the mathematical product of the equilibrium constants for the added reactions, and so\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}\\times {K}_{\\text{b}}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{A}}^{-}\\right]}{\\text{[HA]}}\\times \\dfrac{\\text{[HA]}\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{A}}^{-}\\right]}=\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]={K}_{\\text{w}}[\/latex]<\/p>\r\n<p id=\"fs-idm224463760\">This equation states the relation between ionization constants for any conjugate acid-base pair, namely, their mathematical product is equal to the ion product of water,\u00a0<em data-effect=\"italics\">K<\/em><sub>w<\/sub>. By rearranging this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident:<\/p>\r\n<p style=\"text-align: center;\">[latex]K_{\\text{a}}=K_{\\text{w}}\/K_{\\text{b}}\\qquad\\text{ or }\\qquad{K}_{\\text{b}}=K_{\\text{w}}\/K_{\\text{a}}[\/latex]<\/p>\r\n<p id=\"fs-idm225241056\">The inverse proportional relation between\u00a0<em>K<\/em><sub>a<\/sub>\u00a0and\u00a0<em>K<\/em><sub>b<\/sub>\u00a0means\u00a0<em>the stronger the acid or base, the weaker its conjugate partner<\/em>. Figure 1 illustrates this relation for several conjugate acid-base pairs.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"800\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213731\/CNX_Chem_14_03_strengths.jpg\" alt=\"The diagram shows two horizontal bars. The first, labeled, \u201cRelative acid strength,\u201d at the top is red on the left and gradually changes to purple on the right. The red end at the left is labeled, \u201cStronger acids.\u201d The purple end at the right is labeled, \u201cWeaker acids.\u201d Just outside the bar to the lower left is the label, \u201cK subscript a.\u201d The bar is marked off in increments with a specific acid listed above each increment. The first mark is at 1.0 with H subscript 3 O superscript positive sign. The second is ten raised to the negative two with H C l O subscript 2. The third is ten raised to the negative 4 with H F. The fourth is ten raised to the negative 6 with H subscript 2 C O subscript 3. The fifth is ten raised to a negative 8 with C H subscript 3 C O O H. The sixth is ten raised to the negative ten with N H subscript 4 superscript positive sign. The seventh is ten raised to a negative 12 with H P O subscript 4 superscript 2 negative sign. The eighth is ten raised to the negative 14 with H subscript 2 O. Similarly the second bar, which is labeled \u201cRelative conjugate base strength,\u201d is purple at the left end and gradually becomes blue at the right end. Outside the bar to the left is the label, \u201cWeaker bases.\u201d Outside the bar to the right is the label, \u201cStronger bases.\u201d Below and to the left of the bar is the label, \u201cK subscript b.\u201d The bar is similarly marked at increments with bases listed above each increment. The first is at ten raised to the negative 14 with H subscript 2 O above it. The second is ten raised to the negative 12 C l O subscript 2 superscript negative sign. The third is ten raised to the negative ten with F superscript negative sign. The fourth is ten raised to a negative eight with H C O subscript 3 superscript negative sign. The fifth is ten raised to the negative 6 with C H subscript 3 C O O superscript negative sign. The sixth is ten raised to the negative 4 with N H subscript 3. The seventh is ten raised to the negative 2 with P O subscript 4 superscript three negative sign. The eighth is 1.0 with O H superscript negative sign.\" width=\"800\" height=\"277\" \/> Figure\u00a01. This diagram shows the relative strengths of conjugate acid-base pairs, as indicated by their ionization constants in aqueous solution.[\/caption]\r\n\r\nThe listing of conjugate acid\u2013base pairs shown in Figure 2 is arranged to show the relative strength of each species as compared with water, whose entries are highlighted in each of the table\u2019s columns. In the acid column, those species listed below water are weaker acids than water. These species do not undergo acid ionization in water; they are not Bronsted-Lowry acids. All the species listed above water are stronger acids, transferring protons to water to some extent when dissolved in an aqueous solution to generate hydronium ions. Species above water but below hydronium ion are\u00a0<em data-effect=\"italics\">weak acids<\/em>, undergoing partial acid ionization, wheres those above hydronium ion are\u00a0<em data-effect=\"italics\">strong acids<\/em>\u00a0that are completely ionized in aqueous solution.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"800\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213733\/CNX_Chem_14_03_Corresp.jpg\" alt=\"This figure includes a table separated into a left half which is labeled \u201cAcids\u201d and a right half labeled \u201cBases.\u201d A red arrow points up the left side, which is labeled \u201cIncreasing acid strength.\u201d Similarly, a blue arrow points downward along the right side, which is labeled \u201cIncreasing base strength.\u201d Names of acids and bases are listed next to each arrow toward the center of the table, followed by chemical formulas. Acids listed top to bottom are sulfuric acid, H subscript 2 S O subscript 4, hydrogen iodide, H I, hydrogen bromide, H B r, hydrogen chloride, H C l, nitric acid, H N O subscript 3, hydronium ion ( in pink text) H subscript 3 O superscript plus, hydrogen sulfate ion, H S O subscript 4 superscript negative, phosphoric acid, H subscript 3 P O subscript 4, hydrogen fluoride, H F, nitrous acid, H N O subscript 2, acetic acid, C H subscript 3 C O subscript 2 H, carbonic acid H subscript 2 C O subscript 3, hydrogen sulfide, H subscript 2 S, ammonium ion, N H subscript 4 superscript +, hydrogen cyanide, H C N, hydrogen carbonate ion, H C O subscript 3 superscript negative, water (shaded in beige) H subscript 2 O, hydrogen sulfide ion, H S superscript negative, ethanol, C subscript 2 H subscript 5 O H, ammonia, N H subscript 3, hydrogen, H subscript 2, methane, and C H subscript 4. The acids at the top of the listing from sulfuric acid through nitric acid are grouped with a bracket to the right labeled \u201cUndergo complete acid ionization in water.\u201d Similarly, the acids at the bottom from hydrogen sulfide ion through methane are grouped with a bracket and labeled, \u201cDo not undergo acid ionization in water.\u201d The right half of the figure lists bases and formulas. From top to bottom the bases listed are hydrogen sulfate ion, H S O subscript 4 superscript negative, iodide ion, I superscript negative, bromide ion, B r superscript negative, chloride ion, C l superscript negative, nitrate ion, N O subscript 3 superscript negative, water (shaded in beige), H subscript 2 O, sulfate ion, S O subscript 4 superscript 2 negative, dihydrogen phosphate ion, H subscript 2 P O subscript 4 superscript negative, fluoride ion, F superscript negative, nitrite ion, N O subscript 2 superscript negative, acetate ion, C H subscript 3 C O subscript 2 superscript negative, hydrogen carbonate ion, H C O subscript 3 superscript negative, hydrogen sulfide ion, H S superscript negative, ammonia, N H subscript 3, cyanide ion, C N superscript negative, carbonate ion, C O subscript 3 superscript 2 negative, hydroxide ion (in blue), O H superscript negative, sulfide ion, S superscript 2 negative, ethoxide ion, C subscript 2 H subscript 5 O superscript negative, amide ion N H subscript 2 superscript negative, hydride ion, H superscript negative, and methide ion C H subscript 3 superscript negative. The bases at the top, from perchlorate ion through nitrate ion are group with a bracket which is labeled \u201cDo not undergo base ionization in water.\u201d Similarly, the lower 5 in the listing, from sulfide ion through methide ion are grouped and labeled \u201cUndergo complete base ionization in water.\u201d\" width=\"800\" height=\"717\" \/> Figure\u00a02. The chart shows the relative strengths of conjugate acid-base pairs.[\/caption]\r\n\r\nIf all these strong acids are completely ionized in water, why does the column indicate they vary in strength, with nitric acid being the weakest and perchloric acid the strongest? Notice that the sole acid species present in an aqueous solution of any strong acid is H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>), meaning that hydronium ion is the strongest acid that may exist in water; any stronger acid will react completely with water to generate hydronium ions. This limit on the acid strength of solutes in a solution is called a\u00a0<strong>leveling effect<\/strong>. To measure the differences in acid strength for \u201cstrong\u201d acids, the acids must be dissolved in a solvent that is\u00a0<em>less basic<\/em>\u00a0than water. In such solvents, the acids will be \u201cweak,\u201d and so any differences in the extent of their ionization can be determined. For example, the binary hydrogen halides HCl, HBr, and HI are strong acids in water but weak acids in ethanol (strength increasing HCl &lt; HBr &lt; HI).\r\n<p id=\"fs-idm167383056\">The right column of Figure 2 lists a number of substances in order of increasing base strength from top to bottom. Following the same logic as for the left column, species listed above water are weaker bases and so they don\u2019t undergo base ionization when dissolved in water. Species listed between water and its conjugate base, hydroxide ion, are weak bases that partially ionize. Species listed below hydroxide ion are strong bases that completely ionize in water to yield hydroxide ions (i.e., they are\u00a0<em>leveled<\/em>\u00a0to hydroxide). A comparison of the acid and base columns in this table supports the reciprocal relation between the strengths of conjugate acid-base pairs. For example, the conjugate bases of the strong acids (top of table) are all of negligible strength. A strong acid exhibits an immeasurably large\u00a0<em>K<\/em><sub>a<\/sub>, and so its conjugate base will exhibit a\u00a0<em>K<\/em><sub>b<\/sub>\u00a0that is essentially zero:<\/p>\r\n<p style=\"padding-left: 90px;\">strong acid:\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K_a[\/latex] \u2248 [latex] \\infty [\/latex]<\/p>\r\n<p style=\"padding-left: 90px;\">conjugate base:\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K_b = K_w\/K_a = K_w\/\\infty[\/latex] \u2248 0<\/p>\r\n<p id=\"fs-idm225937888\">A similar approach can be used to support the observation that conjugate acids of strong bases (<em>K<\/em><sub>b<\/sub>\u00a0\u2248 \u221e) are of negligible strength (<em>K<\/em><sub>a<\/sub>\u00a0\u2248 0).<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 2: Calculating Ionization Constants for Conjugate Acid-Base Pairs<\/h3>\r\nUse the <em>K<\/em><sub>b<\/sub> for the nitrite ion, [latex]{\\text{NO}}_{2}^{-}[\/latex], to calculate the <em>K<\/em><sub>a<\/sub> for its conjugate acid.\r\n\r\n[reveal-answer q=\"534789\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"534789\"]\r\n\r\n<em>K<\/em><sub>b<\/sub> for [latex]{\\text{NO}}_{2}^{-}[\/latex] is given in this section as 2.22 \u00d7 10<sup>\u221211<\/sup>. The conjugate acid of [latex]{\\text{NO}}_{2}^{-}[\/latex] is HNO<sub>2<\/sub>; <em>K<\/em><sub>a<\/sub> for HNO<sub>2<\/sub> can be calculated using the relationship:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}\\times {K}_{\\text{b}}=1.0\\times {10}^{-14}={K}_{\\text{w}}[\/latex]<\/p>\r\nSolving for <em>K<\/em><sub>a<\/sub>, we get:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\dfrac{{K}_{\\text{w}}}{{K}_{\\text{b}}}=\\dfrac{1.0\\times {10}^{-14}}{2.22\\times {10}^{-11}}=4.5\\times {10}^{-4}[\/latex]<\/p>\r\nThis answer can be verified by finding the <em>K<\/em><sub>a<\/sub> for HNO<sub>2<\/sub> in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids<\/a>.\r\n\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idp28811616\">Check Your Learning<\/h4>\r\nWe can determine the relative acid strengths of [latex]{\\text{NH}}_{4}^{+}[\/latex] and HCN by comparing their ionization constants. The ionization constant of HCN is given in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids<\/a>\u00a0as 4 \u00d7 10<sup>\u221210<\/sup>. The ionization constant of [latex]{\\text{NH}}_{4}^{+}[\/latex] is not listed, but the ionization constant of its conjugate base, NH<sub>3<\/sub>, is listed as 1.8 \u00d7 10<sup>\u22125<\/sup>. Determine the ionization constant of [latex]{\\text{NH}}_{4}^{+}[\/latex], and decide which is the stronger acid, HCN or [latex]{\\text{NH}}_{4}^{+}[\/latex].\r\n\r\n[reveal-answer q=\"218282\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"218282\"]\r\n\r\n[latex]{\\text{NH}}_{4}^{+}[\/latex] is the slightly stronger acid (<em>K<\/em><sub>a<\/sub> for [latex]{\\text{NH}}_{4}^{+}[\/latex] = 5.6 \u00d7 10<sup>\u221210<\/sup>).[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.<\/li>\r\n \t<li>Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.<\/li>\r\n \t<li>Use this list of important industrial compounds (and Figure\u00a02) to answer the following questions regarding: CaO, Ca(OH)<sub>2<\/sub>, CH<sub>3<\/sub>CO<sub>2<\/sub>H, CO<sub>2,<\/sub> HCl, H<sub>2<\/sub>CO<sub>3<\/sub>, HF, HNO<sub>2<\/sub>, HNO<sub>3<\/sub>, H<sub>3<\/sub>PO<sub>4<\/sub>, H<sub>2<\/sub>SO<sub>4<\/sub>, NH<sub>3<\/sub>, NaOH, Na<sub>2<\/sub>CO<sub>3<\/sub>.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Identify the strong Br\u00f8nsted-Lowry acids and strong Br\u00f8nsted-Lowry bases.<\/li>\r\n \t<li>List those compounds in (a) that can behave as Br\u00f8nsted-Lowry acids with strengths lying between those of [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex] and H<sub>2<\/sub>O.<\/li>\r\n \t<li>List those compounds in (a) that can behave as Br\u00f8nsted-Lowry bases with strengths lying between those of H<sub>2<\/sub>O and OH<sup>\u2212<\/sup>.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Explain why the ionization constant, <em>K<\/em><sub>a<\/sub>, for H<sub>2<\/sub>SO<sub>4<\/sub> is larger than the ionization constant for H<sub>2<\/sub>SO<sub>3<\/sub>.<\/li>\r\n \t<li>Explain why the ionization constant, <em>K<\/em><sub>a<\/sub>, for HI is larger than the ionization constant for HF.<\/li>\r\n \t<li>What is the ionization constant at 25 \u00b0C for the weak acid [latex]{\\text{CH}}_{3}{\\text{NH}}_{3}^{+}[\/latex], the conjugate acid of the weak base CH<sub>3<\/sub>NH<sub>2<\/sub>, <em>K<\/em><sub>b<\/sub> = 4.4 \u00d7 10<sup>\u22124<\/sup>.<\/li>\r\n \t<li>What is the ionization constant at 25 \u00b0C for the weak acid [latex]{\\left({\\text{CH}}_{3}\\right)}_{2}{\\text{NH}}_{2}^{+}[\/latex], the conjugate acid of the weak base (CH<sub>3<\/sub>)<sub>2<\/sub>NH, <em>K<\/em><sub>b<\/sub> = 7.4 \u00d7 10<sup>\u22124<\/sup>?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"314309\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"314309\"]\r\n\r\n2.\u00a0The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH<sup>\u2212<\/sup>, which causes the solution to be basic. An example is NaCN. The CN<sup>\u2212<\/sup> reacts with water as follows:\r\n<p style=\"text-align: center;\">[latex]{\\text{CN}}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons \\text{HCN}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\n4.\u00a0The oxidation state of the sulfur in H<sub>2<\/sub>SO<sub>4<\/sub> is greater than the oxidation state of the sulfur in H<sub>2<\/sub>SO<sub>3<\/sub>.\r\n\r\n6. <em>K<\/em><sub>w<\/sub> = <em>K<\/em><sub>a<\/sub> \u00d7 <em>K<\/em><sub>b<\/sub>; thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{K}_{\\text{a}}&amp;=&amp;\\frac{{K}_{\\text{w}}}{{K}_{\\text{b}}}\\\\{K}_{\\text{a}}&amp;=&amp;\\frac{1.0\\times {10}^{-14}}{4.4\\times {10}^{-4}}=2.3\\times {10}^{-11}\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Acid- Base Equilibrium Calculations<\/h2>\r\nThe chapter on chemical equilibria introduced several types of equilibrium calculations and the various mathematical strategies that are helpful in performing them. These strategies are generally useful for equilibrium systems regardless of chemical reaction class, and so they may be effectively applied to acid-base equilibrium problems. This section presents several example exercises involving equilibrium calculations for acid-base systems.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"698\"]<img src=\"https:\/\/openstax.org\/resources\/31169c8dd8328325b22c0ea1eb5ca58a61b9d659\" alt=\"This image shows two bottles containing clear colorless solutions. Each bottle contains a single p H indicator strip. The strip in the bottle on the left is red, and a similar red strip is placed on a filter paper circle in front of the bottle on surface on which the bottles are resting. Similarly, the second bottle on the right contains and orange strip and an orange strip is placed in front of it on a filter paper circle. Between the two bottles is a pack of p Hydrion papers with a p H color scale on its cover.\" width=\"698\" height=\"531\" \/> Figure\u00a03. pH paper indicates that a 0.l-<em>M<\/em> solution of HCl (beaker on left) has a pH of 1. The acid is fully ionized and [H<sub>3<\/sub>O<sup>+<\/sup>] = 0.1 <em>M<\/em>. A 0.1-<em>M<\/em> solution of CH<sub>3<\/sub>CO<sub>2<\/sub>H (beaker on right) is has a pH of 3 ([H<sub>3<\/sub>O<sup>+<\/sup>] = 0.001 <em>M<\/em>) because the weak acid CH<sub>3<\/sub>CO<sub>2<\/sub>H is only partially ionized. In this solution, [H<sub>3<\/sub>O<sup>+<\/sup>] &lt; [CH<sub>3<\/sub>CO<sub>2<\/sub>H]. (credit: modification of work by Sahar Atwa)[\/caption]\r\n<table id=\"fs-idm41095184\" summary=\"This table has two columns and ten rows. The first row is a header row and it labels each column, \u201cIonization Reaction,\u201d and \u201cK subscript a at 25 degrees C.\u201d Under the \u201cIonization Reaction\u201d column are the following equations: \u201cH S O subscript 4 superscript negative sign plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus S O subscript 4 superscript 2 negative sign,\u201d \u201cH F plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus F superscript negative sign,\u201d \u201cH N O subscript 2 plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus N O subscript 2 superscript negative sign,\u201d \u201cH N C O plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sing plus H C O subscript 2 superscript negative sign,\u201d \u201cC H subscript 3 C O subscript 2 H plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus C I O superscript negative sign,\u201d \u201c H B r O plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus B r O superscript negative sign,\u201d and \u201cH C N plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus C N superscript negative sign.\u201d Under the \u201cK subscript a at 25 degrees C\u201d column are the following values: 1.2 times ten to the negative two, 7.2 times ten to the negative 4, 4.5 times ten to the negative 4, 3.46 times ten to the negative 4, 1.8 times ten to the negative 4, 1.8 times ten to the negative 5, 3.5 times ten to the negative 8, 2 times ten to the negative 9, and 4 times ten to the negative ten.\">\r\n<thead>\r\n<tr>\r\n<th colspan=\"2\">Table\u00a02. Ionization Constants of Some Weak Acids<\/th>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<th>Ionization Reaction<\/th>\r\n<th><em>K<\/em><sub>a<\/sub> at 25 \u00b0C<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{HSO}}_{4}^{-}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{SO}}_{4}^{2-}[\/latex]<\/td>\r\n<td>1.2 \u00d7 10<sup>\u22122<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\text{HF}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{F}}^{-}[\/latex]<\/td>\r\n<td>7.2 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{HNO}}_{2}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{NO}}_{2}^{-}[\/latex]<\/td>\r\n<td>4.5 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\text{HNCO}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{NCO}}^{-}[\/latex]<\/td>\r\n<td>3.46 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{HCO}}_{2}\\text{H}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{HCO}}_{2}^{-}[\/latex]<\/td>\r\n<td>1.8 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}[\/latex]<\/td>\r\n<td>1.8 \u00d7 10<sup>\u22125<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\text{HCIO}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{CIO}}^{-}[\/latex]<\/td>\r\n<td>3.5 \u00d7 10<sup>\u22128<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\text{HBrO}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{BrO}}^{-}[\/latex]<\/td>\r\n<td>2 \u00d7 10<sup>\u22129<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\text{HCN}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{CN}}^{-}[\/latex]<\/td>\r\n<td>4 \u00d7 10<sup>\u221210<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nTable\u00a02 gives the ionization constants for several weak acids; additional ionization constants can be found in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids<\/a>.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"641\"]<img src=\"https:\/\/openstax.org\/resources\/cfacb6d6b36c83b23129cf49fc39d1384ff6d850\" alt=\"This photo shows two glass containers filled with a transparent liquid. In between the containers is a p H strip indicator guide. There are p H strips placed in front of each glass container. The liquid in the container on the left appears to have a p H of 10 or 11. The liquid in the container on the right appears to have a p H of about 13 or 14.\" width=\"641\" height=\"368\" \/> Figure\u00a04. pH paper indicates that a 0.1-<em>M<\/em> solution of NH<sub>3<\/sub> (left) is weakly basic. The solution has a pOH of 3 ([OH<sup>\u2212<\/sup>] = 0.001 <em>M<\/em>) because the weak base NH<sub>3<\/sub> only partially reacts with water. A 0.1-<em>M<\/em> solution of NaOH (right) has a pOH of 1 because NaOH is a strong base. (credit: modification of work by Sahar Atwa)[\/caption]The ionization constants of several weak bases are given in Table\u00a03 and in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-bases-2\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Bases<\/a>.\r\n<table id=\"fs-idm84795184\" summary=\"This table has two columns and six rows. The first row is a header row and it labels each column: \u201cIonization Reaction,\u201d and \u201cK subscript b at 25 degrees C.\u201d Under the \u201cIonization Reaction\u201d column are the following reactions: \u201c( C H subscript 3 ) subscript 2 N H plus H subscript 2 O equilibrium arrow ( C H subscript 3 ) subscript 2 N H subscript 2 superscript plus sign plus O H superscript negative sign,\u201d \u201cC H subscript 3 N H subscript 2 plus H subscript 2 O equilibrium arrow C H subscript 3 N H subscript 3 superscript plus sign plus O H superscript negative sign,\u201d \u201c( C H subscript 3 ) subscript 3 N plus H subscript 2 O equilibrium arrow ( C H subscript 3 ) subscript 3 N H superscript plus sign plus O H superscript negative sign,\u201d \u201cN H subscript 3 plus H subscript 2 O equilibrium arrow N H subscript 4 superscript plus sign plus O H superscript negative sign,\u201d and \u201cC subscript 6 H subscript 5 N H subscript 2 plus H subscript 2 O equilibrium arrow C subscript 6 N subscript 5 N H subscript 3 superscript plus sign plus O H superscript negative sign.\u201d Under the \u201cK subscript b at 25 degrees C\u201d column are the following values: \u201c7.4 times ten to the negative 4,\u201d \u201c4.4 times ten to the negative 4,\u201d \u201c7.4 times ten to the negative 5,\u201d \u201c1.8 times ten to the negative 5,\u201d and \u201c4.6 times ten to the negative 10.\u201d\">\r\n<thead>\r\n<tr>\r\n<th colspan=\"2\">Table\u00a03. Ionization Constants of Some Weak Bases<\/th>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<th>Ionization Reaction<\/th>\r\n<th><em>K<\/em><sub>b<\/sub> at 25 \u00b0C<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\left({\\text{CH}}_{3}\\right)}_{2}\\text{NH}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\left({\\text{CH}}_{3}\\right)}_{2}{\\text{NH}}_{2}^{+}+{\\text{OH}}^{-}[\/latex]<\/td>\r\n<td>7.4 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{CH}}_{3}{\\text{NH}}_{2}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{CH}}_{3}{\\text{NH}}_{3}^{+}+{\\text{OH}}^{-}[\/latex]<\/td>\r\n<td>4.4 \u00d7 10<sup>\u22124<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}+{\\text{OH}}^{-}[\/latex]<\/td>\r\n<td>6.3 \u00d7 10<sup>\u22125<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{NH}}_{3}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{NH}}_{4}^{+}+{\\text{OH}}^{-}[\/latex]<\/td>\r\n<td>1.8 \u00d7 10<sup>\u22125<\/sup><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{C}}_{6}{\\text{N}}_{5}{\\text{NH}}_{3}^{+}+{\\text{OH}}^{-}[\/latex]<\/td>\r\n<td>4.6 \u00d7 10<sup>\u221210<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox exercises\">\r\n<h3>Think about It<\/h3>\r\nBoth HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F<sup>\u2212<\/sup> or CN<sup>\u2212<\/sup>, is the stronger base? See Table\u00a03.\r\n\r\n[practice-area rows=\"4\"][\/practice-area]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 <em>M<\/em> in HCO<sub>2<\/sub>H and 0.10 <em>M<\/em> in HClO.<\/li>\r\n \t<li>Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 <em>M<\/em> in HNO<sub>2<\/sub> and 0.120 <em>M<\/em> in HBrO.<\/li>\r\n \t<li>Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 <em>M<\/em> in CH<sub>3<\/sub>NH<sub>2<\/sub> and 0.10 <em>M<\/em> in C<sub>5<\/sub>H<sub>5<\/sub>N (<em>K<\/em><sub>b<\/sub> = 1.7 \u00d7 10<sup>\u22129<\/sup>).<\/li>\r\n \t<li>Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 <em>M<\/em> in NH<sub>3<\/sub> and 0.100 <em>M<\/em> in C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>.<\/li>\r\n \t<li>Using the <em>K<\/em><sub>a<\/sub> values in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids<\/a>, place [latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{3+}[\/latex] in the correct location in Figure\u00a03.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"588268\"]Show Solution to Question 2[\/reveal-answer]\r\n[hidden-answer a=\"588268\"]\r\n<h4>Question 2<\/h4>\r\nThe reactions and equilibrium constants are:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{NO}}_{2}^{-}\\left(aq\\right){K}_{\\text{a}}=4.5\\times {10}^{-4}\\\\ \\text{HBrO}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{BrO}}^{-}\\left(aq\\right){K}_{\\text{a}}=2\\times {10}^{-9}\\end{array}[\/latex]<\/p>\r\nAs <em>K<\/em><sub>a<\/sub> is much larger for HNO<sub>2<\/sub> than for HBrO, the first equilibrium will dominate. The equilibrium expression is [latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{NO}}_{2}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{HNO}}_{2}\\right]}=4.5\\times {10}^{-5}[\/latex]\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[HNO<sub>2<\/sub>]<\/th>\r\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\r\n<th>[NO<sub>2<\/sub><sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.134<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.134 \u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.134 \u2212 <em>x<\/em>) \u2248 0.134 gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{NO}}_{2}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{HNO}}_{2}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.134-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.134}=4.5\\times {10}^{-4}[\/latex]<\/p>\r\nSolving for <em>x<\/em> gives 7.77 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>. Because this value is 5.8% of 0.134, our assumption is incorrect. Therefore, we must use the quadratic formula. Using the data gives the following equation: <em>x<\/em><sup>2<\/sup> + 4.5 \u00d7 10<sup>\u22127<\/sup><em>x<\/em> \u2212 6.03 \u00d7 10<sup>\u22125<\/sup> = 0\r\n\r\nUsing the quadratic formula gives (<em>a<\/em> = 1, <em>b<\/em> = 4.5 \u00d7 10<sup>\u22124<\/sup>, and <em>c<\/em> = \u22126.03 \u00d7 10<sup>\u22125<\/sup>)\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{ }x&amp;=&amp;\\frac{-b\\pm \\sqrt{{b}^{\\text{2+}}-4ac}}{2a}=\\frac{-\\left(4.5\\times {10}^{-4}\\right)\\pm \\sqrt{{\\left(4.5\\times {10}^{-4}\\right)}^{\\text{2+}}-4\\left(1\\right)\\left(-6.03\\times {10}^{-5}\\right)}}{2\\left(1\\right)}\\\\{}&amp;=&amp;\\frac{-\\left(4.5\\times {10}^{-4}\\right)\\pm \\left(1.55\\times {10}^{-2}\\right)}{2}=7.54\\times {10}^{-3}M\\left(\\text{positive root}\\right)\\end{array}[\/latex]<\/p>\r\nThe equilibrium concentrations are therefore:\r\n<ul>\r\n \t<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\left[{\\text{NO}}_{2}^{-}\\right]=7.54\\times {10}^{-3}=7.5\\times {10}^{-3}M[\/latex]<\/li>\r\n \t<li>[HNO<sub>2<\/sub>] = 0.134 \u2212 7.54 \u00d7 10<sup>\u22123<\/sup> = 0.1264 = 0.126<\/li>\r\n<\/ul>\r\n[OH<sup>\u2212<\/sup>] can be calculated using <em>K<\/em><sub>w<\/sub>:\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{OH}}^{-}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{1.0\\times {10}^{-14}}{7.54\\times {10}^{-3}}=1.33\\times {10}^{-12}=1.3\\times {10}^{-12}M[\/latex]<\/p>\r\nFinally, use the other equilibrium to find the other concentrations. Assume for [HBrO] that (0.120 \u2212 <em>x<\/em>) \u2248 0.124 <em>M<\/em>.\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{BrO}}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[\\text{HBrO}\\right]}=2\\times {10}^{-9}\\approx \\frac{\\left[{\\text{BrO}}^{-}\\right]\\left(7.54\\times {10}^{-3}\\right)}{0.120}[\/latex]<\/p>\r\nSolving for [BrO<sup>\u2212<\/sup>] gives:\r\n<ul>\r\n \t<li>[BrO<sup>\u2212<\/sup>] = 3.2 \u00d7 10<sup>\u22128<\/sup> = 3.2 \u00d7 10<sup>\u22128<\/sup><em>M<\/em><\/li>\r\n \t<li>[HBrO] = 0.120 \u2212 3.2 \u00d7 10<sup>\u22128<\/sup> = 0.120 <em>M<\/em><\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n[reveal-answer q=\"399954\"]Show Solution to Question 4[\/reveal-answer]\r\n[hidden-answer a=\"399954\"]\r\n<h4>Question 4<\/h4>\r\nThe reactions and equilibrium constants are:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{}{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{OH}}^{-}\\left(aq\\right)+{\\text{NH}}_{4}^{+}\\left(aq\\right){K}_{\\text{b}}=1.8\\times {10}^{-5}\\\\ {\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{OH}}^{-}\\left(aq\\right)+{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\left(aq\\right){K}_{\\text{b}}=4.6\\times {10}^{-10}\\end{array}[\/latex]<\/p>\r\nAs <em>K<\/em><sub>b<\/sub> is much larger for NH<sub>3<\/sub> than for C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2,<\/sub> the first equilibrium will dominate. The equilibrium expression is [latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=1.8\\times {10}^{-5}[\/latex]\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[NH<sub>3<\/sub>]<\/th>\r\n<th>[OH<sup>\u2212<\/sup>]<\/th>\r\n<th>[NH<sub>4<\/sub><sup>+<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.115<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.115 \u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.115 \u2212 <em>x<\/em>) \u2248 0.115 gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{NH}}_{4}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.115-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.115}=1.8\\times {10}^{-5}[\/latex]<\/p>\r\nSolving for <em>x<\/em> gives 1.44 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>. Because this value is less than 5% of 0.115 <em>M<\/em>, our assumption is correct. The equilibrium concentrations are therefore:\r\n<ul>\r\n \t<li>[OH<sup>\u2212<\/sup>] = [latex]\\left[{\\text{NO}}_{4}^{+}\\right][\/latex] = 1.44 \u00d7 10<sup>\u22123<\/sup> = 0.0014 <em>M<\/em><\/li>\r\n \t<li>[NH<sub>3<\/sub>] = 0.115 \u2212 0.00144 = 0.1136 = 0.144 <em>M<\/em><\/li>\r\n<\/ul>\r\n[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] can be calculated using <em>K<\/em><sub>w<\/sub>:\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{OH}}^{-}\\right]}=\\frac{1.0\\times {10}^{-14}}{0.00144}=6.94\\times {10}^{-12}=6.9\\times {10}^{-12}M[\/latex]<\/p>\r\nFinally, use the other equilibrium to find the other concentrations. Assume for [C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>] that (0.100 \u2212 <em>x<\/em>) \u2248 0.100 <em>M<\/em>:\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{OH}}^{-}\\right]}=\\frac{1.0\\times {10}^{-14}}{0.00144}=6.94\\times {10}^{-12}=6.9\\times {10}^{-12}M[\/latex]<\/p>\r\nSolving for [latex]\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex] gives:\r\n<ul>\r\n \t<li>[latex]\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex] = 3.9 \u00d7 10<sup>\u22128<\/sup><em>M<\/em><\/li>\r\n \t<li>[C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>] = 0.100 \u2212 3.19 \u00d7 10<sup>\u22128<\/sup> = 0.100 <em>M<\/em><\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 3:\u00a0Determination of <em>K<\/em><sub>a<\/sub> from Equilibrium Concentrations<\/h3>\r\nAcetic acid is the principal ingredient in vinegar (Figure\u00a05); that's why it tastes sour. At equilibrium, a solution contains [CH<sub>3<\/sub>CO<sub>2<\/sub>H] = 0.0787 <em>M<\/em> and [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right]=0.00118M[\/latex]. What is the value of <em>K<\/em><sub>a<\/sub> for acetic acid?\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"350\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213737\/CNX_Chem_14_03_Vinegar.jpg\" alt=\"An image shows the label of a bottle of distilled white vinegar. The label states that the contents have been reduced with water to 5 percent acidity.\" width=\"350\" height=\"333\" \/> Figure\u00a05. Vinegar is a solution of acetic acid, a weak acid. (credit: modification of work by \u201cHomeSpot HQ\u201d\/Flickr)[\/caption]\r\n\r\n[reveal-answer q=\"846674\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"846674\"]\r\n<p id=\"fs-idm119280448\">We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\left(aq\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=\\dfrac{\\left(0.00118\\right)\\left(0.00118\\right)}{0.0787}=1.77\\times {10}^{-5}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idm68555680\">Check Your Learning<\/h4>\r\nWhat is the equilibrium constant for the ionization of the [latex]{\\text{HSO}}_{4}^{-}[\/latex] ion, the weak acid used in some household cleansers:\r\n<p style=\"text-align: center;\">[latex]{\\text{HSO}}_{4}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{SO}}_{4}^{2-}\\left(aq\\right)[\/latex]<\/p>\r\nIn one mixture of NaHSO<sub>4<\/sub> and Na<sub>2<\/sub>SO<sub>4<\/sub> at equilibrium, [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 0.027 <em>M<\/em>; [latex]\\left[{\\text{HSO}}_{4}^{-}\\right]=0.29M[\/latex]; and [latex]\\left[{\\text{SO}}_{4}^{2-}\\right]=0.13M[\/latex].\r\n\r\n<em>\r\n<\/em>[reveal-answer q=\"139813\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"139813\"]\r\n\r\n<em>K<\/em><sub>a<\/sub> for [latex]{\\text{HSO}}_{4}^{-}[\/latex] = 1.2 \u00d7 10<sup>\u22122<\/sup>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 4:\u00a0Determination of <em>K<\/em><sub>b<\/sub> from Equilibrium Concentrations<\/h3>\r\nCaffeine, C<sub>8<\/sub>H<sub>10<\/sub>N<sub>4<\/sub>O<sub>2<\/sub> is a weak base. What is the value of <em>K<\/em><sub>b<\/sub> for caffeine if a solution at equilibrium has [C<sub>8<\/sub>H<sub>10<\/sub>N<sub>4<\/sub>O<sub>2<\/sub>] = 0.050 <em>M<\/em>, [latex]\\left[{\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}{\\text{H}}^{+}\\right][\/latex] = 5.0 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>, and [OH<sup>\u2212<\/sup>] = 2.5 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>?\r\n\r\n[reveal-answer q=\"358312\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"358312\"]\r\n\r\nAt equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:\r\n<p style=\"text-align: center;\">[latex]{\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}{\\text{H}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]\r\n[latex]{K}_{\\text{b}}=\\dfrac{\\left[{\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}{\\text{H}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}\\right]}=\\dfrac{\\left(5.0\\times {10}^{-3}\\right)\\left(2.5\\times {10}^{-3}\\right)}{0.050}=2.5\\times {10}^{-4}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idm122074528\">Check Your Learning<\/h4>\r\nWhat is the equilibrium constant for the ionization of the [latex]{\\text{HPO}}_{4}^{2-}[\/latex] ion, a weak base:\r\n<p style=\"text-align: center;\">[latex]{\\text{HPO}}_{4}^{2-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{2}{\\text{PO}}_{4}^{-}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nIn a solution containing a mixture of NaH<sub>2<\/sub>PO<sub>4<\/sub> and Na<sub>2<\/sub>HPO<sub>4<\/sub> at equilibrium, [OH<sup>\u2212<\/sup>] = 1.3 \u00d7 10<sup>\u22126<\/sup><em>M<\/em>; [latex]\\left[{\\text{H}}_{2}{\\text{PO}}_{4}^{-}\\right]=0.042M[\/latex]; and [latex]\\left[{\\text{HPO}}_{4}^{2-}\\right]=0.341M[\/latex].\r\n\r\n[reveal-answer q=\"153773\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"153773\"]\r\n\r\n<em>K<sub>b<\/sub><\/em> for [latex]{\\text{HPO}}_{4}^{2-}=1.6\\times {10}^{-7}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 5:\u00a0Determination of <em>K<\/em><sub>a<\/sub> or <em>K<\/em><sub>b<\/sub> from pH<\/h3>\r\nThe pH of a 0.0516-<em>M<\/em> solution of nitrous acid, HNO<sub>2<\/sub>, is 2.34. What is its <em>K<\/em><sub>a<\/sub>?\r\n<p style=\"text-align: center;\">[latex]{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{NO}}_{2}^{-}\\left(aq\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"978442\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"978442\"]\r\n\r\nThe nitrous acid concentration provided is a\u00a0<em>formal<\/em>\u00a0concentration, one that does not account for any chemical equilibria that may be established in solution. Such concentrations are treated as \u201cinitial\u201d values for equilibrium calculations using the ICE table approach. Notice the initial value of hydronium ion is listed as\u00a0<em>approximately<\/em>\u00a0zero because a small concentration of H<sub>3<\/sub>O<sup>+<\/sup>\u00a0is present (1 \u00d7 10<sup>\u22127<\/sup>\u00a0<em>M<\/em>) due to the autoprotolysis of water. In many cases, such as all the ones presented in this chapter, this concentration is much less than that generated by ionization of the acid (or base) in question and may be neglected.\r\n<p id=\"fs-idm204817712\">The pH provided is a logarithmic measure of the hydronium ion concentration resulting from the acid ionization of the nitrous acid, and so it represents an \u201cequilibrium\u201d value for the ICE table:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]={10}^{-2.34}=0.0046M[\/latex]<\/p>\r\nThe ICE table for this system is then:\r\n\r\n<img class=\"alignnone\" src=\"https:\/\/openstax.org\/resources\/fa4bb024e3500e0f2df46b5d3f916db6e44b1d9f\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of \u201cH N O subscript 2 plus sign H subscript 2 O equilibrium sign H subscript 3 O superscript positive sign plus sign N O subscript 2 superscript negative sign.\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.0516, negative 0.0046, 0.0470. The second column is blank in all three rows. The third column has the following: approximately 0, positive 0.0046, 0.0046. The fourth column has the following: 0, positive 0.0046, 0.0046.\" width=\"1000\" height=\"262\" \/>\r\n\r\nFinally, we calculate the value of the equilibrium constant using the data in the table:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{NO}}_{2}^{-}\\right]}{\\left[{\\text{HNO}}_{2}\\right]}=\\dfrac{\\left(0.0046\\right)\\left(0.0046\\right)}{\\left(0.0470\\right)}=4.5\\times {10}^{-4}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idm166688304\">Check Your Learning<\/h4>\r\nThe pH of a solution of household ammonia, a 0.950-<em>M<\/em> solution of NH<sub>3,<\/sub> is 11.612. What is <em>K<\/em><sub>b<\/sub> for NH<sub>3<\/sub>.\r\n\r\n[reveal-answer q=\"487528\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"487528\"]\r\n\r\n<em>K<sub>b<\/sub><\/em> = 1.8 \u00d7 10<sup>\u22125<\/sup>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 6:\u00a0Equilibrium Concentrations in a Solution of a Weak Acid<\/h3>\r\nFormic acid, HCO<sub>2<\/sub>H, is the irritant that causes the body\u2019s reaction to ant stings (Figure\u00a06).\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"350\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213744\/CNX_Chem_14_03_AntSting.jpg\" alt=\"A photograph is shown of a large black ant on the end of a human finger.\" width=\"350\" height=\"197\" \/> Figure\u00a06. The pain of an ant\u2019s sting is caused by formic acid. (credit: John Tann)[\/caption]\r\n\r\nWhat is the concentration of hydronium ion and the pH in a 0.534-<em>M<\/em> solution of formic acid?\r\n<p style=\"text-align: center;\">[latex]{\\text{HCO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{HCO}}_{2}^{-}\\left(aq\\right){K}_{\\text{a}}=1.8\\times {10}^{-4}[\/latex]<\/p>\r\n[reveal-answer q=\"135863\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"135863\"]\r\n<h4>Step 1: Determine <em>x<\/em> and equilibrium concentrations<\/h4>\r\nThe ICE table for this system is\r\n\r\n<img class=\"alignnone\" src=\"https:\/\/openstax.org\/resources\/cbf133e6eed622c35dfb6c5c537670c8dc581a7b\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of \u201cH C O subscript 2 H plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign.\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.534, blank, 0.534 minus x. The second column is blank in all three rows. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0, positive x, x.\" width=\"1170\" height=\"267\" \/>\r\n\r\nSubstituting the equilibrium concentration terms into the K<sub>a<\/sub> expression gives\r\n<p style=\"text-align: center;\"><span style=\"text-align: center;\">[latex]\\begin{array}{rcl}{K}_{\\text{a}}&amp;=&amp;1.8\\times {10}^{-4}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{HCO}}_{2}^{-}\\right]}{\\left[{\\text{HCO}}_{2}\\text{H}\\right]}\\\\{}&amp;=&amp;\\dfrac{\\left(x\\right)\\left(x\\right)}{0.534-x}=1.8\\times {10}^{-4}\\end{array}[\/latex]<\/span><\/p>\r\nBecause the initial concentration of acid is reasonably large and <em>K<\/em><sub>a<\/sub> is very small, we assume that <em>x<\/em> &lt;&lt; 0.534, which <em>permits<\/em> us to simplify the denominator term as (0.534 \u2212 <em>x<\/em>) = 0.534. This gives:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=1.8\\times {10}^{-4}=\\dfrac{{x}^{\\text{2}}}{0.534}[\/latex]<\/p>\r\nSolve for <em>x<\/em> as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{x}^{\\text{2}}&amp;=&amp;0.534\\times \\left(1.8\\times {10}^{-4}\\right)=9.6\\times {10}^{-5}\\\\x&amp;=&amp;\\sqrt{9.6\\times {10}^{-5}}\\\\&amp;=&amp;9.8\\times {10}^{-3} M\\end{array}[\/latex]<\/p>\r\nTo check the assumption that <em>x<\/em> is small compared to 0.534, we calculate:\r\n<p style=\"text-align: center;\">[latex]\\dfrac{x}{0.534}=\\dfrac{9.8\\times{10}^{-3}}{0.534}=1.8\\times{10}^{-2}\\left(1.8\\%\\text{ of 0.534}\\right)[\/latex]<\/p>\r\n<em>x<\/em> is less than 5% of the initial concentration; the assumption is valid.\r\n<p id=\"fs-idm490302912\">As defined in the ICE table,\u00a0<em>x<\/em>\u00a0is equal to the equilibrium concentration of hydronium ion:<\/p>\r\n<p style=\"text-align: center;\">[latex]x=[\\text{H}_{3}\\text{O}^{+}]=0.0098 M[\/latex]<\/p>\r\n<p style=\"text-align: left;\"><span style=\"font-size: 1rem; text-align: initial;\">Finally, the pH is calculated to be<\/span><\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{pH}=\\text{log}[\\text{H}_{3}\\text{O}^{+}]=-\\text{log}(0.0098)=2.01[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idp80326224\">Check Your Learning<\/h4>\r\nOnly a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-<em>M<\/em> solution of acetic acid, CH<sub>3<\/sub>CO<sub>2<\/sub>H?\r\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\left(aq\\right)\\qquad{K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]<\/p>\r\n(Hint: Determine [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right][\/latex] at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized \u00d7 100, or [latex]\\dfrac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right]}{{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}_{\\text{initial}}}\\times 100[\/latex].\r\n[reveal-answer q=\"26536\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"26536\"]\r\n\r\npercent ionization = 1.3%[\/hidden-answer]\r\n\r\n<\/div>\r\nExample 7 shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid.\r\n<div class=\"textbox examples\">\r\n<h3>Example 7:\u00a0Equilibrium Concentrations in a Solution of a Weak Base<\/h3>\r\nFind the concentration of hydroxide ion in a 0.25-<em>M<\/em> solution of trimethylamine, a weak base:\r\n<p style=\"text-align: center;\">[latex]{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\qquad{K}_{\\text{b}}=6.3\\times {10}^{-5}[\/latex]<\/p>\r\n[reveal-answer q=\"636372\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"636372\"]\r\n\r\nThe ICE table for this system is\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213748\/CNX_Chem_14_03_ICETable4_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of \u201c( C H subscript 3 ) subscript 3 N plus sign H subscript 2 O equilibrium arrow ( C H subscript 3 ) subscript 3 N H superscript positive sign plus sign O H superscript positive sign.\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.25, negative x, 0.25 plus sign negative x. The second column is blank in all three rows. The third column has the following: 0, x, 0 plus x. The fourth column has the following: approximately 0, x, and approximately 0 plus x.\" width=\"697\" height=\"183\" \/>\r\n<h4><\/h4>\r\nAt equilibrium:\r\n<p style=\"text-align: center;\"><span style=\"text-align: center;\">[latex]{K}_{\\text{b}}=\\dfrac{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\right]}=\\dfrac{\\left(x\\right)\\left(x\\right)}{0.25-x}=6.3\\times {10}^{-5}[\/latex]<\/span><\/p>\r\nIf we assume that <em>x<\/em> is small relative to 0.25, then we can replace (0.25 \u2212 <em>x<\/em>) in the preceding equation with 0.25. Solving the simplified equation gives\u00a0[latex]x=4.0\\times {10}^{-3}[\/latex].\r\n\r\nThis change is less than 5% of the initial concentration (0.25), so the assumption is justified. Recall that, for this computation, <em>x<\/em> is equal to the equilibrium concentration of <em>hydroxide ion<\/em> in the solution (see earlier tabulation):\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{OH}}^{-}\\right]=\\text{~}0+x=x=4.0\\times {10}^{-3}M[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]=4.0\\times {10}^{-3}M[\/latex]<\/p>\r\nThen calculate pOH as follows:\r\n<p style=\"text-align: center;\">[latex]\\text{pOH}=-\\text{log}\\left(4.0\\times {10}^{-3}\\right)=2.40[\/latex]<\/p>\r\nUsing the relation introduced in the previous section of this chapter, we get\r\n<p style=\"text-align: center;\">[latex]\\text{pH}+\\text{pOH}=\\text{p}{K}_{\\text{w}}=14.00[\/latex]<\/p>\r\nwhich\u00a0permits the computation of pH:\r\n<p style=\"text-align: center;\">[latex]\\text{pH}=14.00-\\text{pOH}=14.00 - 2.40=11.60[\/latex]<\/p>\r\n\r\n<h4>Step 3: Check the work<\/h4>\r\nA check of our arithmetic shows that <em>K<\/em><sub>b<\/sub> = 6.3 \u00d7 10<sup>\u22125<\/sup>.\r\n\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idm83886080\">Check Your Learning<\/h4>\r\nCalculate the hydroxide ion concentration and the percent ionization of a 0.0325-<em data-effect=\"italics\">M<\/em>\u00a0solution of ammonia, a weak base with a\u00a0<em data-effect=\"italics\">K<\/em><sub>b<\/sub>\u00a0of 1.76 \u00d7 10<sup>\u22125<\/sup>.\r\n\r\n[reveal-answer q=\"145683\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"145683\"]\r\n<ol>\r\n \t<li>7.56 \u00d7 10<sup>\u22124<\/sup><em>M<\/em><\/li>\r\n \t<li>2.33%<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn some cases, the strength of the weak acid or base and its formal (initial) concentration result in an appreciable ionization. Though the ICE strategy remains effective for these systems, the algebra is a bit more involved because the simplifying assumption that\u00a0<em>x<\/em>\u00a0is negligible can not be made. Calculations of this sort are demonstrated in Example 8 below.\r\n<div class=\"textbox examples\">\r\n<h3>Example 8: Calculating Equilibrium Concentrations without simplifying assumptions<\/h3>\r\nSodium bisulfate, NaHSO<sub>4<\/sub>, is used in some household cleansers because it contains the [latex]{\\text{HSO}}_{4}^{-}[\/latex] ion, a weak acid. What is the pH of a 0.50-<em>M<\/em> solution of [latex]{\\text{HSO}}_{4}^{-}?[\/latex]\r\n<p style=\"text-align: center;\">[latex]{\\text{HSO}}_{4}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{SO}}_{4}^{2-}\\left(aq\\right)\\qquad{K}_{\\text{a}}=1.2\\times {10}^{-2}[\/latex]<\/p>\r\n[reveal-answer q=\"23029\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"23029\"]\r\n\r\nThe ICE table for this system is:\r\n\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213751\/CNX_Chem_14_03_ICETable5_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium ( M ). The second column has the header of \u201cH S O subscript 4 superscript negative sign plus sign H subscript 2 O equilibrium sign H subscript 3 O superscript positive sign plus sign S O subscript 4 superscript 2 superscript negative sign.\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.50, negative x, 0.50 plus sign negative x equals 0.50 minus x. The second column is blank for all three rows. The third column has the following: approximately 0, x, 0 plus sign x equals x. The fourth column has the following: 0, x, 0 plus sign x equals x.\" width=\"700\" height=\"224\" \/>\r\n\r\nAs we begin solving for <em>x<\/em>, we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of <em>x<\/em>. At equilibrium:\r\n<p style=\"text-align: center;\"><span style=\"text-align: center;\">[latex]{K}_{\\text{a}}=1.2\\times {10}^{-2}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{SO}}_{4}^{2-}\\right]}{\\left[{\\text{HSO}}_{4}^{-}\\right]}=\\dfrac{\\left(x\\right)\\left(x\\right)}{0.50-x}[\/latex]<\/span><\/p>\r\nIf we assume that <em>x<\/em> is small and approximate (0.50 \u2212 <em>x<\/em>) as 0.50, we find [latex]x=7.7\\times {10}^{-2}M[\/latex].\u00a0When we check the assumption, we calculate:\r\n<p style=\"text-align: center;\">[latex]\\dfrac{x}{{\\left[{\\text{HSO}}_{4}^{-}\\right]}_{\\text{i}}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\dfrac{x}{0.50}=\\dfrac{7.7\\times {10}^{-2}}{0.50}=0.15\\left(15\\%\\right)[\/latex]<\/p>\r\nThe value of <em>x<\/em> is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find <em>x<\/em>.\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=1.2\\times {10}^{-2}=\\dfrac{\\left(x\\right)\\left(x\\right)}{0.50-x}[\/latex]<\/p>\r\nRearranging this equation yields\r\n<p style=\"text-align: center;\">[latex]6.0\\times {10}^{-3}-1.2\\times {10}^{-2}x={x}^{2}[\/latex]<\/p>\r\nWriting the equation in quadratic form gives\r\n<p style=\"text-align: center;\">[latex]{x}^{2}+1.2\\times {10}^{-2}x - 6.0\\times {10}^{-3}=0[\/latex]<\/p>\r\n&nbsp;\r\n\r\nSolving for the two roots of this quadratic equation results in a negative value that may be discarded as physically irrelevant and a positive value equal to\u00a0<em data-effect=\"italics\">x<\/em>. As defined in the ICE table,\u00a0<em data-effect=\"italics\">x<\/em>\u00a0is equal to the hydronium concentration.\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=~0+x=0+7.2\\times {10}^{-2}M=7.2\\times {10}^{-2}M[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\text{pH}=-\\text{log}\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=-\\text{log}(7.2\\times {10}^{-2})=1.14[\/latex]<\/p>\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idm108532480\">Check Your Learning<\/h4>\r\n<ol>\r\n \t<li>Calculate the pH in a 0.010-<em>M<\/em> solution of caffeine, a weak base: [latex]{\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}{\\text{H}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right){K}_{\\text{b}}=2.5\\times {10}^{-4}[\/latex]<\/li>\r\n<\/ol>\r\n(Hint: It will be necessary to convert [OH<sup>\u2212<\/sup>] to [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] or pOH to pH toward the end of the calculation.)\r\n[reveal-answer q=\"774358\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"774358\"]\r\n\r\n[latex]\\text{pH }11.16[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\nCalculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids<\/a>\u00a0and\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-bases-2\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Bases<\/a>.\r\n<h4>Solution 1<\/h4>\r\n0.0092 <em>M<\/em> HClO, a weak acid\r\n\r\n[reveal-answer q=\"936215\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"936215\"]\r\n\r\nThe reaction is: [latex]\\text{HClO}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{ClO}}^{-}\\left(aq\\right)[\/latex]\r\n\r\nThe equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{ClO}}^{-}\\right]}{\\left[\\text{HClO}\\right]}=3.5\\times {10}^{-8}[\/latex]\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[HClO]<\/th>\r\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\r\n<th>[ClO<sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.0092<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.0092\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0092 \u2212 <em>x<\/em>) \u2248 0.0092 gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{ClO}}^{-}\\right]}{\\left[\\text{HClO}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.0092-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.0092}=3.5\\times {10}^{-8}[\/latex]<\/p>\r\nSolving for <em>x<\/em> gives 1.79 \u00d7 10<sup>\u22125<\/sup><em>M<\/em>. This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:\r\n<ul>\r\n \t<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = [ClO] = 1.8 \u00d7 10<sup>\u22125<\/sup><em>M<\/em><\/li>\r\n \t<li>[HClO] = 0.0092 \u2212 1.79 \u00d7 10<sup>\u22125<\/sup> = 0.00918 = 0.00092 <em>M<\/em><\/li>\r\n \t<li>[latex]\\left[{\\text{OH}}^{-}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{1.0\\times {10}^{-14}}{1.79\\times {10}^{-5}}=5.6\\times {10}^{-10}M[\/latex]<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n<h4>Solution 2<\/h4>\r\n0.0784 <em>M<\/em> C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>, a weak base\r\n\r\n[reveal-answer q=\"996941\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"996941\"]\r\n\r\nThe reaction is\u00a0[latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]\r\n\r\nThe equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\right]}=4.6\\times {10}^{-10}[\/latex]\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>]<\/th>\r\n<th>[C<sub>5<\/sub>H<sub>5<\/sub>NH<sub>3<\/sub><sup>+<\/sup>]<\/th>\r\n<th>[OH<sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.0784<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.0784\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0784 \u2212 <em>x<\/em>) \u2248 0.0784 gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.0784-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.0784}=4.6\\times {10}^{-10}[\/latex]<\/p>\r\nSolving for <em>x<\/em> gives 6.01 \u00d7 10<sup>\u22126<\/sup><em>M<\/em>. This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:\r\n<ul>\r\n \t<li>[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right][\/latex] = [OH<sup>\u2212<\/sup>] = 6.0\u00a0\u00d7 10<sup>\u22126<\/sup><em>M<\/em><\/li>\r\n \t<li>[C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>] = 0.0784 \u2212 6.01 \u00d7 10<sup>\u22126<\/sup> = 0.007839 = 0.00784<\/li>\r\n \t<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{OH}}^{-}\\right]}=\\frac{1.0\\times {10}^{-14}}{6.01\\times {10}^{-6}}=1.7\\times {10}^{-9}M[\/latex]<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n<h4>Solution 3<\/h4>\r\n0.0810 <em>M<\/em> HCN, a weak acid\r\n\r\n[reveal-answer q=\"342499\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"342499\"]\r\n\r\nThe reaction is [latex]\\text{HCN}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CN}}^{-}\\left(aq\\right)[\/latex].\r\n\r\nThe equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CN}}^{-}\\right]}{\\left[\\text{HCN}\\right]}=4\\times {10}^{-10}[\/latex]\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[HClO]<\/th>\r\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\r\n<th>[CN<sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.0810<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.0810\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0810 \u2212 <em>x<\/em>) \u2248 0.0810 gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CN}}^{-}\\right]}{\\left[\\text{HCN}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.0810-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.0810}=4\\times {10}^{-10}[\/latex]<\/p>\r\nSolving for <em>x<\/em> gives 5.69 \u00d7 10<sup>\u22126<\/sup><em>M<\/em>. This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:\r\n<ul>\r\n \t<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = [CN<sup>\u2212<\/sup>] = 5.7 \u00d7 10<sup>\u22126<\/sup><em>M<\/em><\/li>\r\n \t<li>[HCN] = 0.0810 \u2212 5.69 \u00d7 10<sup>\u22126<\/sup> = 0.08099 = 0.0810 <em>M<\/em><\/li>\r\n \t<li>[latex]\\left[{\\text{OH}}^{-}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{1.0\\times {10}^{-14}}{5.69\\times {10}^{-6}}=1.8\\times {10}^{-9}M[\/latex]<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n<h4>Solution 4<\/h4>\r\n0.11 <em>M<\/em> (CH<sub>3<\/sub>)<sub>3<\/sub>N, a weak base\r\n\r\n[reveal-answer q=\"210337\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"210337\"]\r\n\r\nThe reaction is\u00a0[latex]{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]\r\n\r\nThe equilibrium expression is\u00a0[latex]{K}_{\\text{b}}=\\frac{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\right]}=7.4\\times {10}^{-5}[\/latex]\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[(CH<sub>3<\/sub>)<sub>3<\/sub>N]<\/th>\r\n<th>[(CH<sub>3<\/sub>)<sub>3<\/sub>NH<sup>+<\/sup>]<\/th>\r\n<th>[OH<sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.11<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.11\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.11 \u2212 <em>x<\/em>) \u2248 0.11 gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.11-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.11}=7.4\\times {10}^{-5}[\/latex]<\/p>\r\nSolving for <em>x<\/em> gives 2.85 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>. This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:\r\n<ul>\r\n \t<li>[latex]\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\right][\/latex] = [OH<sup>\u2212<\/sup>] = 2.9 \u00d7 10<sup>\u22123<\/sup><em>M<\/em><\/li>\r\n \t<li>[(CH<sub>3<\/sub>)<sub>3<\/sub>N] = 0.11 \u2212 2.85 \u00d7 10<sup>\u22123<\/sup> = 0.107 = 0.11 <em>M<\/em><\/li>\r\n \t<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{OH}}^{-}\\right]}=\\frac{1.0\\times {10}^{-14}}{2.85\\times {10}^{-3}}=3.5\\times {10}^{-12}M[\/latex]<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n<h4>Solution 5<\/h4>\r\n0.120 <em>M<\/em> [latex]\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{2+}[\/latex] a weak acid, <em>K<\/em><sub>a<\/sub> = 1.6 \u00d7 10<sup>\u22127<\/sup>\r\n\r\n[reveal-answer q=\"874526\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"874526\"]\r\n\r\nThe reaction is\u00a0[latex]\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{2+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons \\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{+}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)[\/latex]\r\n\r\nThe equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{+}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{2+}\\right]}=1.6\\times {10}^{-7}[\/latex]\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[Fe(H<sub>2<\/sub>O)<sub>6<\/sub><sup>2+<\/sup>]<\/th>\r\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\r\n<th>[Fe(H<sub>2<\/sub>O)<sub>5<\/sub>(OH)<sup>+<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.120<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.120\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.120 \u2212 <em>x<\/em>) \u2248 0.120 gives:\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{+}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{2+}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.120-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.120}=1.6\\times {10}^{-7}[\/latex]<\/p>\r\nSolving for <em>x<\/em> gives 1.39 \u00d7 10<sup>\u22124<\/sup><em>M<\/em>. This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:\r\n<ul>\r\n \t<li style=\"text-align: left;\">[latex]\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{+}\\right][\/latex] = [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 1.4 \u00d7 10<sup>\u22124<\/sup><em>M<\/em><\/li>\r\n \t<li style=\"text-align: left;\">[latex]\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{2+}\\right][\/latex] = 0.120 \u2212 1.39 \u00d7 10<sup>\u22124<\/sup> = 0.1199 = 0.120 <em>M<\/em><\/li>\r\n \t<li style=\"text-align: left;\">[latex]\\left[{\\text{OH}}^{-}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{1.0\\times {10}^{-14}}{1.39\\times {10}^{-4}}=7.2\\times {10}^{-11}M[\/latex]<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Effect of Molecular Structure on Acid-Base Strength<\/h2>\r\nIn the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 7A, the order of increasing acidity is HF &lt; HCl &lt; HBr &lt; HI. Likewise, for group 6A, the order of increasing acid strength is H<sub>2<\/sub>O &lt; H<sub>2<\/sub>S &lt; H<sub>2<\/sub>Se &lt; H<sub>2<\/sub>Te.\r\n\r\nAcross a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is CH<sub>4<\/sub> &lt; NH<sub>3<\/sub> &lt; H<sub>2<\/sub>O &lt; HF; across the third row, it is SiH<sub>4<\/sub> &lt; PH<sub>3<\/sub> &lt; H<sub>2<\/sub>S &lt; HCl (see Figure\u00a07).\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"700\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213752\/CNX_Chem_14_03_AcidpH.jpg\" alt=\"This diagram has two rows and four columns. Red arrows point left across the bottom of the figure and down at the right side and are labeled \u201cIncreasing acid strength.\u201d Blue arrows point left across the bottom and up at the right side of the figure and are labeled \u201cIncreasing base strength.\u201d The first column is labeled 14 at the top and two white squares are beneath it. The first has the number 6 in the upper left corner and the formula C H subscript 4 in the center along with designation Neither acid nor base. The second square contains the number 14 in the upper left corner, the formula C H subscript 4 at the center and the designation Neither acid nor base. The second column is labeled 15 at the top and two blue squares are beneath it. The first has the number 7 in the upper left corner and the formula N H subscript 3 in the center along with the designation Weak base and K subscript b equals 1.8 times 10 superscript negative 5. The second square contains the number 15 in the upper left corner, the formula P H subscript 3 at the center and the designation Very weak base and K subscript b equals 4 times 10 superscript negative 28. The third column is labeled 16 at the top and two squares are beneath it. The first is shaded tan and has the number 8 in the upper left corner and the formula H subscript 2 O in the center along with the designation neutral. The second square is shaded pink, contains the number 16 in the upper left corner, the formula H subscript 2 S at the center and the designation Weak acid and K subscript a equals 9.5 times 10 superscript negative 8. The fourth column is labeled 17 at the top and two squares are beneath it. The first is shaded pink, has the number 9 in the upper left corner and the formula H F in the center along with the designation Weak acid and K subscript a equals 6.8 times 10 superscript negative 4. The second square is shaded a deeper pink, contains the number 17 in the upper left corner, the formula H C l at the center, and the designation Strong acid.\" width=\"700\" height=\"444\" \/> Figure\u00a07. As you move from left to right and down the periodic table, the acid strength increases. As you move from right to left and up, the base strength increases.[\/caption]\r\n<h2>Ternary Acids and Bases<\/h2>\r\nCompounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric, depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group. Such compounds have the general formula O<sub>n<\/sub>E(OH)<sub>m<\/sub>, and include sulfuric acid, O<sub>2<\/sub>S(OH)<sub>2<\/sub>, sulfurous acid, OS(OH)<sub>2<\/sub>, nitric acid, O<sub>2<\/sub>NOH, perchloric acid, O<sub>3<\/sub>ClOH, aluminum hydroxide, Al(OH)<sub>3<\/sub>, calcium hydroxide, Ca(OH)<sub>2<\/sub>, and potassium hydroxide, KOH:\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213754\/CNX_Chem_14_03_OHbonds_img.jpg\" alt=\"A diagram is shown that includes a central atom designated with the letter E. Single bonds extend above, below, left, and right of the E. An O atom is bonded to the right of the E, and an arrow points to the bond labeling it, \u201cBond a.\u201d An H atom is single bonded to the right of the O atom. An arrow pointing to this bond connects it to the label, \u201cBond b.\u201d\" \/>\r\n\r\nIf the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond <em>a<\/em> between the element and oxygen is more readily broken than bond <em>b<\/em> between oxygen and hydrogen. Hence bond <em>a<\/em> is ionic, hydroxide ions are released to the solution, and the material behaves as a base\u2014this is the case with Ca(OH)<sub>2<\/sub> and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds.\r\n\r\nIf, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond <em>a<\/em> relatively strongly covalent. The oxygen-hydrogen bond, bond <em>b<\/em>, is thereby weakened because electrons are displaced toward E. Bond <em>b<\/em> is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic \u2212OH groups that are called <strong>oxyacids<\/strong>.\r\n\r\nIncreasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H<sub>2<\/sub>SO<sub>4<\/sub>, or O<sub>2<\/sub>S(OH)<sub>2<\/sub> (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H<sub>2<\/sub>SO<sub>3<\/sub>, or OS(OH)<sub>2<\/sub> (with a sulfur oxidation number of +4). Likewise nitric acid, HNO<sub>3<\/sub>, or O<sub>2<\/sub>NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO<sub>2<\/sub>, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure\u00a08).\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"880\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213755\/CNX_Chem_14_03_Oxyacid.jpg\" alt=\"A diagram is shown that includes four structural formulas for acids. A red, right pointing arrow is placed beneath the structures which is labeled \u201cIncreasing acid strength.\u201d At the top left, the structure of Nitrous acid is provided. It includes an H atom to which an O atom with two unshared electron pairs is connected with a single bond to the right. A single bond extends to the right and slightly below to a N atom with one unshared electron pair. A double bond extends up and to the right from this N atom to an O atom which has two unshared electron pairs. To the upper right is a structure for Nitric acid. This structure differs from the previous structure in that the N atom is directly to the right of the first O atom and a second O atom with three unshared electron pairs is connected with a single bond below and to the right of the N atom which has no unshared electron pairs. At the lower left, an O atom with two unshared electron pairs is double bonded to its right to an S atom with a single unshared electron pair. An O atom with two unshared electron pairs is bonded above and an H atom is single bonded to this O atom. To the right of the S atom is a single bond to another O atom with two unshared electron pairs to which an H atom is single bonded. This structure is labeled \u201cSulfurous acid.\u201d A similar structure which is labeled \u201cSulfuric acid\u201d is placed in the lower right region of the figure. This structure differs in that an H atom is single bonded to the left of the first O atom, leaving it with two unshared electron pairs and a fourth O atom with two unshared electron pairs is double bonded beneath the S atom, leaving it with no unshared electron pairs.\" width=\"880\" height=\"500\" \/> Figure\u00a08. As the oxidation number of the central atom E increases, the acidity also increases.[\/caption]\r\n\r\nHydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate Al(H<sub>2<\/sub>O)<sub>3<\/sub>(OH)<sub>3<\/sub>, is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, Al(H<sub>2<\/sub>O)<sub>3<\/sub>(OH)<sub>3<\/sub>, is converted into the soluble ion, [latex]{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{2}{\\left(\\text{OH}\\right)}_{4}\\right]}^{-}[\/latex], by reaction with hydroxide ion:\r\n<p style=\"text-align: center;\">[latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{3}{\\left(\\text{OH}\\right)}_{3}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\rightleftharpoons {\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{2}{\\left(\\text{OH}\\right)}_{4}\\right]}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nIn this reaction, a proton is transferred from one of the aluminum-bound H<sub>2<\/sub>O molecules to a hydroxide ion in solution. The Al(H<sub>2<\/sub>O)<sub>3<\/sub>(OH)<sub>3<\/sub> compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion [latex]{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}\\right]}^{3+}[\/latex] by reaction with hydronium ion:\r\n<p style=\"text-align: center;\">[latex]{\\text{3H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{3}{\\left(\\text{OH}\\right)}_{3}\\left(aq\\right)\\rightleftharpoons \\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{3+}\\left(aq\\right)+{\\text{3H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\r\nIn this case, protons are transferred from hydronium ions in solution to Al(H<sub>2<\/sub>O)<sub>3<\/sub>(OH)<sub>3<\/sub>, and the compound functions as a base.\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>Propionic acid, C<sub>2<\/sub>H<sub>5<\/sub>CO<sub>2<\/sub>H (<em>K<\/em><sub>a<\/sub> = 1.34 \u00d7 10<sup>\u22125<\/sup>), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a 0.698-<em>M<\/em> solution of C<sub>2<\/sub>H<sub>5<\/sub>CO<sub>2<\/sub>H?<\/li>\r\n \t<li>White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g\/cm<sup>3<\/sup>, what is the pH?<\/li>\r\n \t<li>The ionization constant of lactic acid, CH<sub>3<\/sub>CH(OH)CO<sub>2<\/sub>H, an acid found in the blood after strenuous exercise, is 1.36 \u00d7\u00a010<sup>\u22124<\/sup>. If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?<\/li>\r\n \t<li>Nicotine, C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub>, is a base that will accept two protons (<em>K<\/em><sub>1<\/sub> = 7 \u00d7\u00a010<sup>\u22127<\/sup>, <em>K<\/em><sub>2<\/sub> = 1.4 \u00d7 10<sup>\u221211<\/sup>). What is the concentration of each species present in a 0.050-<em>M<\/em> solution of nicotine?<\/li>\r\n \t<li>The pH of a 0.20-<em>M<\/em> solution of HF is 1.92. Determine <em>K<\/em><sub>a<\/sub> for HF from these data.<\/li>\r\n \t<li>The pH of a 0.15-<em>M<\/em> solution of [latex]{\\text{HSO}}_{4}^{-}[\/latex] is 1.43. Determine <em>K<\/em><sub>a<\/sub> for [latex]{\\text{HSO}}_{4}^{-}[\/latex] from these data.<\/li>\r\n \t<li>The pH of a 0.10-<em>M<\/em> solution of caffeine is 11.16. Determine <em>K<\/em><sub>b<\/sub> for caffeine from these data:<\/li>\r\n \t<li>The pH of a solution of household ammonia, a 0.950 M solution of NH<sub>3,<\/sub> is 11.612. Determine <em>K<\/em><sub>b<\/sub> for NH<sub>3<\/sub> from these data.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"864039\"]Show Solution to Question 2[\/reveal-answer]\r\n[hidden-answer a=\"864039\"]\r\n\r\nFirst, find the mass of acetic acid. <em>d<\/em> = 1.007 g\/cm<sup>3<\/sup>. Take 1.0 L of solution to have the quantities on a mole basis. Then, since 1000 cm<sup>3<\/sup> = 1.0 L, 1000 cm<sup>3<\/sup>\u00a0\u00d7 1.007 g\/cm<sup>3<\/sup> = 1007 g in 1.0 L. Then, 5.00% of this is the mass of acetic acid:\r\n<p style=\"text-align: center;\">[latex]\\text{Mass (acetic acid)}=\\text{1007 g}\\times \\frac{5.0%}{100%}=\\text{50.35 g}[\/latex]<\/p>\r\nNow calculate the number of moles of acetic acid present. The molar mass of acetic acid is 60.053 g\/mol:\r\n<p style=\"text-align: center;\">[latex]\\text{mol acetic acid}=\\frac{50.35\\cancel{\\text{g}}}{60.053\\cancel{\\text{g}}{\\text{mol}}^{-1}}=\\text{0.838 mol}[\/latex]<\/p>\r\nFrom the moles of acetic acid and <em>K<\/em><sub>a<\/sub>, calculate [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]:[\/latex]\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=1.8\\times {10}^{-5}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}[\/latex]<\/p>\r\n\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>H]<\/th>\r\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\r\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub><sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.838<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.838 \u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstitution gives:\u00a0[latex]{K}_{\\text{a}}=1.8\\times {10}^{-5}=\\frac{{x}^{\\text{2}}}{0.838-x}[\/latex]\r\n\r\nDrop <em>x<\/em> because it is small in comparison with 0.838 <em>M<\/em>.\r\n<ul>\r\n \t<li><em>x<\/em><sup>2<\/sup> = 0.838(1.8 \u00d7 10<sup>\u22125<\/sup>) = 1.508 \u00d7 10<sup>\u22125<\/sup> = 3.88 \u00d7 10<sup>\u22123<\/sup> = 2.41<\/li>\r\n \t<li>pH = \u2212log(3.88 \u00d7 10<sup>\u22123<\/sup>) = 2.41<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n[reveal-answer q=\"287147\"]Show Solution to Question 4[\/reveal-answer]\r\n[hidden-answer a=\"287147\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{ }{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\left({K}_{\\text{b1}}=7\\times {10}^{-7}\\right)\\\\ {\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\left({K}_{\\text{b2}}=1.4\\times {10}^{-11}\\right)\\end{array}[\/latex]<\/p>\r\nFirst set up a concentration table:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub>]<\/th>\r\n<th>[C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub>H<sup>+<\/sup>]<\/th>\r\n<th>[OH<sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.050<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Change (<em>M<\/em>)<\/th>\r\n<td>\u2212<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.050 \u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSubstituting the equilibrium concentrations into the equilibrium equation and making the assumption that (0.050 \u2212 <em>x<\/em>) = 0.050, we get:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{K}_{\\text{b1}}&amp;=&amp;\\frac{\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}\\right]}=7\\times {10}^{-7}\\\\{}&amp;=&amp;\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.050-x\\right)}=\\frac{{x}^{\\text{2+}}}{0.050}=7\\times {10}^{-7}\\end{array}[\/latex]<\/p>\r\nSolving for <em>x<\/em> gives 1.87 \u00d7 10<sup>\u22124<\/sup> = 2 \u00d7 10<sup>\u22124<\/sup><em>M<\/em> = [OH<sup>\u2212<\/sup>]\r\n\r\nBecause <em>x<\/em> is less than 5% of 0.050 and [OH<sup>\u2212<\/sup>] is greater than 4.5 \u00d7 10<sup>\u22127\u00a0<\/sup><em>M<\/em>, our customary assumptions are justified. We can calculate\r\n<ul>\r\n \t<li>[C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub>] = 0.050 \u2212 <em>x<\/em> = 0.050 \u2212 2 \u00d7 10<sup>\u22124<\/sup> = 0.048 <em>M<\/em><\/li>\r\n \t<li>[OH<sup>\u2212<\/sup>] = [latex]\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\right][\/latex] = <em>x<\/em> = 2 \u00d7 10<sup>\u22124<\/sup><em>M<\/em><\/li>\r\n<\/ul>\r\nNow calculate the concentration of [latex]{\\text{C}}_{10}{\\text{H}}_{2}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}[\/latex] in a solution with [OH<sup>\u2212<\/sup>] and [latex]\\left[{\\text{C}}_{10}{\\text{H}}_{2}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}\\right][\/latex] equal to 2 \u00d7 10<sup>\u22124<\/sup><em>M<\/em>. The equilibrium between these species is [latex]{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}_{2}^{\\text{2+}}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]. We know [C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub>H<sup>+<\/sup>] and [OH<sup>\u2212<\/sup>], so we can calculate the concentration of [latex]{\\text{C}}_{10}{\\text{H}}_{2}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}[\/latex] from the equilibrium expression:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{}{K}_{\\text{b2}}&amp;=&amp;\\frac{\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\right]}=1.4\\times {10}^{-11}\\\\{}&amp;=&amp;\\frac{\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}\\right]\\left[2\\times {10}^{-4}\\right]}{\\left[2\\times {10}^{-4}\\right]}\\\\ \\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}\\right]&amp;=&amp;1.4\\times {10}^{-11}M\\end{array}[\/latex]<\/p>\r\nThe concentration of OH<sup>\u2212<\/sup> produced in this ionization is equal to the concentration of [latex]{\\text{C}}_{10}{\\text{H}}_{2}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}[\/latex], 1.4 \u00d7 10<sup>\u221211<\/sup><em>M<\/em>, which is much smaller than the 2 \u00d710<sup>\u22124<\/sup><em>M<\/em> produced in the first ionization; therefore, we are justified in neglecting the OH<sup>\u2212<\/sup> formed from [latex]{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}[\/latex].\r\n\r\nWe can now calculate the concentration of H<sub>2<\/sub>O<sup>+<\/sup> present from the ionization of water:\r\n<p style=\"text-align: center;\"><em>K<\/em><sub>a<\/sub> = 1 \u00d7 10<sup>\u221214<\/sup> = [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] [OH<sup>\u2212<\/sup>]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{1\\times {10}^{-14}}{\\left[{\\text{OH}}^{-}\\right]}=\\frac{1\\times {10}^{-14}}{1.9\\times {10}^{-4}}=5.3\\times {10}^{-11}M[\/latex]<\/p>\r\nWe can now summarize the concentrations of all species in solution as follows:\r\n<ul>\r\n \t<li>[C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub>] = 0.049 <em>M<\/em><\/li>\r\n \t<li>[latex]\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\right][\/latex] = 1.9 \u00d7 10<sup>\u22124\u00a0<\/sup><em>M<\/em><\/li>\r\n \t<li>[latex]\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}\\right]=1.4\\times {10}^{-11}M[\/latex]<\/li>\r\n \t<li>[OH<sup>\u2212<\/sup>] = 1.9 \u00d7 10<sup>\u22124<\/sup><em>M<\/em><\/li>\r\n \t<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 5.3 \u00d7 10<sup>\u221211<\/sup><em>M<\/em><\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n[reveal-answer q=\"969367\"]Show Solution to Question 6[\/reveal-answer]\r\n[hidden-answer a=\"969367\"]\r\n\r\nThe reaction is [latex]{\\text{HSO}}_{4}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{SO}}_{2}^{2-}\\left(aq\\right)[\/latex].\r\n\r\nThe concentrations at equilibrium are [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 10<sup>\u2212pH<\/sup> = 10<sup>\u22121.43<\/sup> = 0.0372 <em>M<\/em>\r\n<ul>\r\n \t<li>[HF] = 0.15 \u2212 0.0372 <em>M<\/em> = 0.113 <em>M<\/em><\/li>\r\n \t<li>[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{SO}}_{4}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{HSO}}_{4}^{-}\\right]}=\\frac{\\left(0.0372\\right)\\left(0.0372\\right)}{\\left(0.113\\right)}=1.2\\times {10}^{-2}[\/latex]<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n[reveal-answer q=\"864327\"]Show Solution to Question 8[\/reveal-answer]\r\n[hidden-answer a=\"864327\"]\r\n\r\nThe reaction is [latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex].\r\n\r\nThe pOH can be determined from pOH = 14.000 \u2212 pH = 14.000 \u2212 11.612 = 2.388. Therefore, the concentrations at equilibrium are [latex]\\left[{\\text{NH}}_{4}^{+}\\right][\/latex] = [OH<sup>\u2212<\/sup>] = 10<sup>\u2212pOH<\/sup> = 10<sup>\u22122.388<\/sup> = 0.004093 <em>M<\/em>\r\n<ul>\r\n \t<li>[NH<sub>3<\/sub>] = 0.950 \u2212 0.004093 = 0.9459 <em>M<\/em><\/li>\r\n \t<li>[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(0.004093\\right)\\left(0.004093\\right)}{\\left(0.9459\\right)}=1.77\\times {10}^{-5}[\/latex]<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=5557241&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=ZNo6gfCAgWE&amp;video_target=tpm-plugin-965fa6qw-ZNo6gfCAgWE\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/AcidsAndBasesChemistry_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Acids and Bases Chemistry - Basic Introduction\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nThe strengths of Br\u00f8nsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion. The strengths of the binary acids increase from left to right across a period of the periodic table (CH<sub>4<\/sub> &lt; NH<sub>3<\/sub> &lt; H<sub>2<\/sub>O &lt; HF), and they increase down a group (HF &lt; HCl &lt; HBr &lt; HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H<sub>2<\/sub>SO<sub>3<\/sub> &lt; H<sub>2<\/sub>SO<sub>4<\/sub>). The strengths of oxyacids also increase as the electronegativity of the central element increases [H<sub>2<\/sub>SeO<sub>4<\/sub> &lt; H<sub>2<\/sub>SO<sub>4<\/sub>].\r\n<h4>Key Equations<\/h4>\r\n<ul>\r\n \t<li>[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{A}}^{-}\\right]}{\\left[\\text{HA}\\right]}[\/latex]<\/li>\r\n \t<li>[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{HB}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[\\text{B}\\right]}[\/latex]<\/li>\r\n \t<li><em>K<\/em><sub>a<\/sub> \u00d7 <em>K<\/em><sub>b<\/sub> = 1.0 \u00d7 10<sup>\u221214<\/sup> = <em>K<\/em><sub>w<\/sub><\/li>\r\n \t<li>[latex]\\text{Percent ionization}=\\frac{{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}_{\\text{eq}}}{{\\left[\\text{HA}\\right]}_{0}}\\times 100[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>The odor of vinegar is due to the presence of acetic acid, CH<sub>3<\/sub>CO<sub>2<\/sub>H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-<em>M<\/em> aqueous solution of this acid.<\/li>\r\n \t<li>Household ammonia is a solution of the weak base NH<sub>3<\/sub> in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1-<em>M<\/em> aqueous solution of this base.<\/li>\r\n \t<li>Which base, CH<sub>3<\/sub>NH<sub>2<\/sub> or (CH<sub>3<\/sub>)<sub>2<\/sub>NH, is the strongest base? Which conjugate acid, [latex]{\\left({\\text{CH}}_{3}\\right)}_{2}{\\text{NH}}_{2}^{+}[\/latex] or (CH<sub>3<\/sub>)<sub>2<\/sub>NH, is the strongest acid?<\/li>\r\n \t<li>Which is the stronger acid, [latex]{\\text{NH}}_{4}^{+}[\/latex] or HBrO?<\/li>\r\n \t<li>Which is the stronger base, (CH<sub>3<\/sub>)<sub>3<\/sub>N or [latex]{\\text{H}}_{2}{\\text{BO}}_{3}^{-}?[\/latex]<\/li>\r\n \t<li>Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>H<sub>2<\/sub>O or HF<\/li>\r\n \t<li>B(OH)<sub>3<\/sub> or Al(OH)<sub>3<\/sub><\/li>\r\n \t<li>[latex]{\\text{HSO}}_{3}^{-}[\/latex] or [latex]{\\text{HSO}}_{4}^{-}[\/latex]<\/li>\r\n \t<li>NH<sub>3<\/sub> or H<sub>2<\/sub>S<\/li>\r\n \t<li>H<sub>2<\/sub>O or H<sub>2<\/sub>Te<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\text{HSO}}_{4}^{-}[\/latex] or [latex]{\\text{HSeO}}_{4}^{-}[\/latex]<\/li>\r\n \t<li>NH<sub>3<\/sub> or H<sub>2<\/sub>O<\/li>\r\n \t<li>PH<sub>3<\/sub> or HI<\/li>\r\n \t<li>NH<sub>3<\/sub> or PH<sub>3<\/sub><\/li>\r\n \t<li>H<sub>2<\/sub>S or HBr<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>acidity: HCl, HBr, HI<\/li>\r\n \t<li>basicity: H<sub>2<\/sub>O, OH<sup>\u2212<\/sup>, H<sup>\u2212<\/sup>, Cl<sup>\u2212<\/sup><\/li>\r\n \t<li>basicity: Mg(OH)<sub>2<\/sub>, Si(OH)<sub>4<\/sub>, ClO<sub>3<\/sub>(OH) (Hint: Formula could also be written as HClO<sub>4<\/sub>).<\/li>\r\n \t<li>acidity: HF, H<sub>2<\/sub>O, NH<sub>3<\/sub>, CH<sub>4<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>acidity: NaHSO<sub>3<\/sub>, NaHSeO<sub>3<\/sub>, NaHSO<sub>4<\/sub><\/li>\r\n \t<li>basicity: [latex]{\\text{BrO}}_{2}^{-}[\/latex], [latex]{\\text{ClO}}_{2}^{-}[\/latex], [latex]{\\text{IO}}_{2}^{-}[\/latex]<\/li>\r\n \t<li>acidity: HOCl, HOBr, HOI<\/li>\r\n \t<li>acidity: HOCl, HOClO, HOClO<sub>2<\/sub>, HOClO<sub>3<\/sub><\/li>\r\n \t<li>basicity: [latex]{\\text{NH}}_{2}^{-}[\/latex], HS<sup>\u2212<\/sup>, HTe<sup>\u2212<\/sup>, [latex]{\\text{PH}}_{2}^{-}[\/latex]<\/li>\r\n \t<li>basicity: BrO<sup>\u2212<\/sup>, [latex]{\\text{BrO}}_{2}^{-}[\/latex], [latex]{\\text{BrO}}_{3}^{-}[\/latex], [latex]{\\text{BrO}}_{4}^{-}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The active ingredient formed by aspirin in the body is salicylic acid, C<sub>6<\/sub>H<sub>4<\/sub>OH(CO<sub>2<\/sub>H). The carboxyl group (\u2212CO<sub>2<\/sub>H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001-<em>M<\/em> aqueous solution of C<sub>6<\/sub>H<sub>4<\/sub>OH(CO<sub>2<\/sub>H).<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"929665\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"929665\"]\r\n\r\n1.\u00a0[H<sub>2<\/sub>O] &gt; [CH<sub>3<\/sub>CO<sub>2<\/sub>H] &gt; [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] \u2248 [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right][\/latex] &gt; [OH<sup>\u2212<\/sup>]\r\n\r\n3.\u00a0The strongest base or strongest acid is the one with the larger <em>K<\/em><sub>b<\/sub> or <em>K<\/em><sub>a<\/sub>, respectively. In these two examples, they are (CH<sub>3<\/sub>)<sub>2<\/sub>NH and [latex]{\\text{CH}}_{3}{\\text{NH}}_{3}^{+}[\/latex].\r\n\r\n5.\u00a0Look up <a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-bases-2\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Bases<\/a> the value of <em>K<\/em><sub>b<\/sub> for (CH<sub>3<\/sub>)<sub>3<\/sub>N and the value of <em>K<\/em><sub>a<\/sub> for H<sub>3<\/sub>BO<sub>3<\/sub>. From the latter, calculate the value of <em>K<\/em><sub>b<\/sub> for [latex]{\\text{H}}_{2}{\\text{BO}}_{3}^{-}[\/latex]. Then compare values:\r\n<ul>\r\n \t<li><em>K<\/em><sub>b<\/sub>(CH<sub>3<\/sub>)<sub>3<\/sub>N = 7.4 \u00d7 10<sup>\u22125<\/sup><\/li>\r\n \t<li>[latex]{K}_{\\text{a}}\\left({\\text{H}}_{3}{\\text{BO}}_{3}\\right)=5.8\\times {10}^{-10}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{b}}}[\/latex], [latex]{K}_{\\text{b}}=\\frac{1.0\\times {10}^{-14}}{5.8\\times {10}^{-10}}=1.7\\times {10}^{-5}[\/latex]<\/li>\r\n<\/ul>\r\nA comparison shows that the larger <em>K<\/em><sub>b<\/sub> is that of triethylamine.\r\n\r\n7.\u00a0The more acidic compounds are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\text{HSO}}_{4}^{-}[\/latex]; higher electronegativity of the central ion.<\/li>\r\n \t<li>H<sub>2<\/sub>O; NH<sub>3<\/sub> is a base and water is neutral, or decide on the basis of <em>K<\/em><sub>a<\/sub> values.<\/li>\r\n \t<li>HI; PH<sub>3<\/sub> is weaker than HCl; HCl is weaker than HI. Thus, PH<sub>3<\/sub> is weaker than HI.<\/li>\r\n \t<li>PH<sub>3<\/sub>; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group.<\/li>\r\n \t<li>HBr; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S, so HBr is the stronger acid.<\/li>\r\n<\/ol>\r\n9. The correct ordered lists are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>NaHSeO<sub>3<\/sub> &lt; NaHSO<sub>3<\/sub> &lt; NaHSO<sub>4<\/sub>; in polyoxy acids, the more electronegative central element\u2014S, in this case\u2014forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner.<\/li>\r\n \t<li>[latex]{\\text{ClO}}_{2}^{-}&lt;{\\text{BrO}}_{2}^{-}&lt;{\\text{IO}}_{2}^{-}[\/latex]; the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.<\/li>\r\n \t<li>HOI &lt; HOBr &lt; HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.<\/li>\r\n \t<li>HOCl &lt; HOClO &lt; HOClO<sub>2<\/sub> &lt; HOClO<sub>3<\/sub>; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases).<\/li>\r\n \t<li>[latex]{\\text{HTe}}^{-}&lt;{\\text{HS}}^{-}&lt;&lt;{\\text{PH}}_{2}^{-}&lt;{\\text{NH}}_{2}^{-}[\/latex]; [latex]{\\text{PH}}_{2}^{-}[\/latex] and [latex]{\\text{NH}}_{2}^{-}[\/latex] are anions of weak bases, so they act as strong bases toward H<sup>+<\/sup>. [latex]{\\text{HTe}}^{-}[\/latex] and HS<sup>\u2212<\/sup> are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion.<\/li>\r\n \t<li>[latex]{\\text{BrO}}_{4}^{-}&lt;{\\text{BrO}}_{3}^{-}&lt;{\\text{BrO}}_{2}^{-}&lt;{\\text{BrO}}^{-}[\/latex]; with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic.<\/li>\r\n<\/ol>\r\n10.\u00a0[latex]\\left[\\text{H}_{2}\\text{O}\\right]\\gt\\left[\\text{C}_{6}\\text{H}_{4}\\text{OH}\\left(\\text{CO}_{2}\\text{H}\\right)\\right]\\gt\\left[\\text{H}^{+}\\right]\\text{O}\\gt\\left[\\text{C}_{6}\\text{H}_{4}\\text{OH}\\left(\\text{CO}_{2}\\right)^{-}\\right]\\gt\\left[\\text{C}_{6}\\text{H}_{4}\\text{O}\\left(\\text{CO}_{2}\\text{H}\\right)^{-}\\right]\\gt\\left[\\text{OH}^{-}\\right][\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li id=\"fs-idm94403824\">Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO<sub>3<\/sub>)<sub>2<\/sub>, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO<sub>3<\/sub> with CuO.<\/li>\r\n \t<li>Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)<sub>2<\/sub> in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs<\/li>\r\n \t<li>What do we represent when we write: [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\left(aq\\right)?[\/latex]<\/li>\r\n \t<li>Explain why equilibrium calculations are not necessary to determine ionic concentrations in solutions of certain strong electrolytes such as NaOH and HCl. Under what conditions are equilibrium calculations necessary as part of the determination of the concentrations of all ions of some other strong electrolytes in solution?<\/li>\r\n \t<li>Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.<\/li>\r\n \t<li>What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid?<\/li>\r\n \t<li>What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak base?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"405128\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"405128\"]\r\n\r\n2.\u00a0[latex]\\begin{array}{cccccc}\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)+&amp; \\text{HCl}\\left(aq\\right)&amp; \\longrightarrow &amp; {\\text{Mg}}^{2+}\\left(aq\\right)+&amp; 2{\\text{Cl}}^{-}\\left(aq\\right)+&amp; {\\text{2H}}_{2}\\text{O}\\left(l\\right)\\\\ \\text{BB}&amp; \\text{BA}&amp; &amp; \\text{CB}&amp; \\text{CA}&amp; \\end{array}[\/latex]\r\n\r\n4.\u00a0Strong electrolytes are 100% ionized, and, as long as the component ions are neither weak acids nor weak bases, the ionic species present result from the dissociation of the strong electrolyte. Equilibrium calculations are necessary when one (or more) of the ions is a weak acid or a weak base.\r\n\r\n6. The two assumptions are\r\n<ol>\r\n \t<li>Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration.<\/li>\r\n \t<li>Assume we can neglect the contribution of water to the equilibrium concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li>Which of the following will increase the percent of NH<sub>3<\/sub> that is converted to the ammonium ion in water (Hint: Use LeCh\u00e2telier\u2019s principle.)?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>addition of NaOH<\/li>\r\n \t<li>addition of HCl<\/li>\r\n \t<li>addition of NH<sub>4<\/sub>Cl<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Which of the following will increase the percent of HF that is converted to the fluoride ion in water?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>addition of NaOH<\/li>\r\n \t<li>addition of HCl<\/li>\r\n \t<li>addition of NaF<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>What is the effect on the concentrations of [latex]{\\text{NO}}_{2}^{-}[\/latex], HNO<sub>2<\/sub>, and OH<sup>\u2212<\/sup> when the following are added to a solution of KNO<sub>2<\/sub> in water:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>HCl<\/li>\r\n \t<li>HNO<sub>2<\/sub><\/li>\r\n \t<li>NaOH<\/li>\r\n \t<li>NaCl<\/li>\r\n \t<li>KNO<\/li>\r\n<\/ol>\r\nThe equation for the equilibrium is: [latex]{\\text{NO}}_{2}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{HNO}}_{2}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/li>\r\n \t<li>What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>HCl<\/li>\r\n \t<li>KF<\/li>\r\n \t<li>NaCl<\/li>\r\n \t<li>KOH<\/li>\r\n \t<li>HF<\/li>\r\n<\/ol>\r\nThe equation for the equilibrium is: [latex]\\text{HF}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{F}}^{-}\\left(aq\\right)[\/latex]<\/li>\r\n \t<li>Why is the hydronium ion concentration in a solution that is 0.10 <em>M<\/em> in HCl and 0.10 <em>M<\/em> in HCOOH determined by the concentration of HCl?<\/li>\r\n \t<li>From the equilibrium concentrations given, calculate <em>K<\/em><sub>a<\/sub> for each of the weak acids and <em>K<\/em><sub>b<\/sub> for each of the weak bases.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>CH<sub>3<\/sub>CO<sub>2<\/sub>H: [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 1.34 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>; [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right][\/latex] = 1.34 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>; [CH<sub>3<\/sub>CO<sub>2<\/sub>H] = 9.866 \u00d7 10<sup>\u22122<\/sup><em>M<\/em><\/li>\r\n \t<li>ClO<sup>\u2212<\/sup>: [OH<sup>\u2212<\/sup>] = 4.0 \u00d7 10<sup>\u22124<\/sup><em>M<\/em>;\u00a0[HClO] = 2.38 \u00d7 10<sup>\u22125<\/sup><em>M<\/em>;\u00a0[ClO<sup>\u2212<\/sup>] = 0.273 <em>M<\/em><\/li>\r\n \t<li>HCO<sub>2<\/sub>H: [HCO<sub>2<\/sub>H] = 0.524 <em>M<\/em>; [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 9.8 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>; [latex]\\left[{\\text{HCO}}_{2}^{-}\\right][\/latex] = 9.8 \u00d7 10<sup>\u22123<\/sup><em>M<\/em><\/li>\r\n \t<li>[latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}:[\/latex] [latex]\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex] = 0.233 <em>M<\/em>;\u00a0[C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>] = 2.3 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>;\u00a0[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 2.3 \u00d7 10<sup>\u22123<\/sup><em>M<\/em><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>From the equilibrium concentrations given, calculate <em>K<\/em><sub>a<\/sub> for each of the weak acids and <em>K<\/em><sub>b<\/sub> for each of the weak bases.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>NH<sub>3<\/sub>: [OH<sup>\u2212<\/sup>] = 3.1 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>;\u00a0[latex]\\left[{\\text{NH}}_{4}^{+}\\right][\/latex] = 3.1 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>;\u00a0[NH<sub>3<\/sub>] = 0.533 <em>M<\/em><\/li>\r\n \t<li>HNO<sub>2<\/sub>: [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 0.011 <em>M<\/em>; [latex]\\left[{\\text{NO}}_{2}^{-}\\right][\/latex] = 0.0438 <em>M<\/em>;\u00a0[HNO<sub>2<\/sub>] = 1.07 <em>M<\/em><\/li>\r\n \t<li>(CH<sub>3<\/sub>)<sub>3<\/sub>N: [(CH<sub>3<\/sub>)<sub>3<\/sub>N] = 0.25 <em>M<\/em>;\u00a0[latex]\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\right][\/latex] = 4.3 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>;\u00a0[OH<sup>\u2212<\/sup>] = 4.3 \u00d7 10<sup>\u22123<\/sup><em>M<\/em><\/li>\r\n \t<li>[latex]{\\text{NH}}_{4}^{+}:[\/latex] [latex]\\left[{\\text{NH}}_{4}^{+}\\right][\/latex] = 0.100 <em>M<\/em>; [NH<sub>3<\/sub>] = 7.5 \u00d7 10<sup>\u22126<\/sup><em>M<\/em>;\u00a0[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 7.5 \u00d7 10<sup>\u22126<\/sup><em>M<\/em><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Determine <em>K<\/em><sub>b<\/sub> for the nitrite ion, [latex]{\\text{NO}}_{2}^{-}[\/latex]. In a 0.10-<em>M<\/em> solution this base is 0.0015% ionized.<\/li>\r\n \t<li>Determine <em>K<\/em><sub>a<\/sub> for hydrogen sulfate ion, [latex]{\\text{HSO}}_{4}^{-}[\/latex]. In a 0.10-<em>M<\/em> solution the acid is 29% ionized.<\/li>\r\n \t<li>Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>F<sup>\u2212<\/sup><\/li>\r\n \t<li>[latex]{\\text{NH}}_{4}^{+}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{AsO}}_{4}^{3-}[\/latex]<\/li>\r\n \t<li>[latex]{\\left({\\text{CH}}_{3}\\right)}_{2}{\\text{NH}}_{2}^{+}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{NO}}_{2}^{-}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{HC}}_{2}{\\text{O}}_{4}^{-}[\/latex] (as a base)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>HTe<sup>\u2212<\/sup> (as a base)<\/li>\r\n \t<li>[latex]{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{HAsO}}_{4}^{3-}[\/latex] (as a base)<\/li>\r\n \t<li>[latex]{\\text{HO}}_{2}^{-}[\/latex] (as a base)<\/li>\r\n \t<li>[latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{HSO}}_{3}^{-}[\/latex] (as a base)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>For which of the following solutions must we consider the ionization of water when calculating the pH or pOH?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>3 \u00d7 10<sup>\u22128<\/sup><em>M<\/em> HNO<sub>3<\/sub><\/li>\r\n \t<li>0.10 g HCl in 1.0 L of solution<\/li>\r\n \t<li>0.00080 g NaOH in 0.50 L of solution<\/li>\r\n \t<li>1 \u00d7 10<sup>\u22127<\/sup><em>M<\/em> Ca(OH)<sub>2<\/sub><\/li>\r\n \t<li>0.0245 <em>M<\/em> KNO<sub>3<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Even though both NH<sub>3<\/sub> and C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub> are weak bases, NH<sub>3<\/sub> is a much stronger acid than C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>. Which of the following is correct at equilibrium for a solution that is initially 0.10 <em>M<\/em> in NH<sub>3<\/sub> and 0.10 <em>M<\/em> in C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\left[{\\text{OH}}^{-}\\right]=\\left[{\\text{NH}}_{4}^{+}\\right][\/latex]<\/li>\r\n \t<li>[latex]\\left[{\\text{NH}}_{4}^{+}\\right]=\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex]<\/li>\r\n \t<li>[latex]\\left[{\\text{OH}}^{-}\\right]=\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex]<\/li>\r\n \t<li>[NH<sub>3<\/sub>] = [C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>]<\/li>\r\n \t<li>both a and b are correct<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"789235\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"789235\"]\r\n\r\n1.\u00a0The equilibrium is [latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>The addition of NaOH adds OH<sup>\u2212<\/sup> to the system and, according to Le Ch\u00e2telier\u2019s principle, the equilibrium will shift to the left. Thus, the percent of converted NH<sub>3<\/sub> will decrease.<\/li>\r\n \t<li>The addition of HCl will add [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex] ions, which will then react with the OH<sup>\u2212<\/sup> ions. Thus, the equilibrium will shift to the right, and the percent will increase.<\/li>\r\n \t<li>The addition of NH<sub>4<\/sub>Cl adds [latex]{\\text{NH}}_{4}^{+}[\/latex] ions, shifting the equilibrium to the left. Thus, the percent will decrease.<\/li>\r\n<\/ol>\r\n3.\u00a0The effects are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Adding HCl will add [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex] ions, which will then react with the OH<sup>\u2212<\/sup> ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO<sub>2<\/sub>, and decreasing the concentration of [latex]{\\text{NO}}_{2}^{-}[\/latex] ions.<\/li>\r\n \t<li>Adding HNO<sub>2<\/sub> increases the concentration of HNO<sub>2<\/sub> and shifts the equilibrium to the left, increasing the concentration of [latex]{\\text{NO}}_{2}^{-}[\/latex] ions and decreasing the concentration of OH<sup>\u2212<\/sup> ions.<\/li>\r\n \t<li>Adding NaOH adds OH<sup>\u2212<\/sup> ions, which shifts the equilibrium to the left, increasing the concentration of [latex]{\\text{NO}}_{2}^{-}[\/latex] ions and decreasing the concentrations of HNO<sub>2<\/sub>.<\/li>\r\n \t<li>Adding NaCl has no effect on the concentrations of the ions.<\/li>\r\n \t<li>Adding KNO<sub>2<\/sub> adds [latex]{\\text{NO}}_{2}^{-}[\/latex] ions and shifts the equilibrium to the right, increasing the HNO<sub>2<\/sub> and OH<sup>\u2212<\/sup> ion concentrations.<\/li>\r\n<\/ol>\r\n5.\u00a0The equations of the occurring chemical processes are:\r\n<p style=\"text-align: center;\">[latex]\\text{HCl}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}\\text{COOH}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{COO}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\nThis is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO<sub>2<\/sub>H exists primarily as HCO<sub>2<\/sub>H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO<sub>2<\/sub>H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] produced by the stronger acid.\r\n\r\n7.\u00a0<em>K<sub>a<\/sub><\/em> and\u00a0<em>K<sub>b<\/sub><\/em> are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>The reaction is [latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{OH}}^{-}\\left(aq\\right)+{\\text{NH}}_{4}^{+}\\left(aq\\right)[\/latex]\r\n[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(3.1\\times {10}^{-3}\\right)\\left(3.1\\times {10}^{-3}\\right)}{\\left(0.533\\right)}=1.8\\times {10}^{-5}[\/latex]<\/li>\r\n \t<li>The reaction is [latex]{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{NO}}_{2}^{-}\\left(aq\\right)[\/latex]\r\n[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{NO}}_{2}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{HNO}}_{2}\\right]}=\\frac{\\left(0.0438\\right)\\left(0.011\\right)}{\\left(1.07\\right)}=4.5\\times {10}^{-4}[\/latex]<\/li>\r\n \t<li>The reaction is [latex]{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{OH}}^{-}\\left(aq\\right)+{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\left(aq\\right)[\/latex]\r\n[latex]{K}_{\\text{b}}=\\frac{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\right]}=\\frac{\\left(4.3\\times {10}^{-3}\\right)\\left(4.3\\times {10}^{-3}\\right)}{\\left(0.25\\right)}=7.4\\times {10}^{-5}[\/latex]<\/li>\r\n \t<li>The reaction is [latex]{\\text{NH}}_{4}^{+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{NH}}_{3}\\left(aq\\right)[\/latex]\r\n[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{NH}}_{3}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{NH}}_{4}^{+}\\right]}=\\frac{\\left(7.5\\times {10}^{-6}\\right)\\left(7.5\\times {10}^{-6}\\right)}{\\left(0.100\\right)}=5.6\\times {10}^{-10}[\/latex]<\/li>\r\n<\/ol>\r\n9.\u00a0The reaction is [latex]{\\text{HSO}}_{4}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{SO}}_{4}^{2-}\\left(aq\\right)[\/latex].\r\n\r\nThe concentrations at equilibrium are:\r\n<ul>\r\n \t<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = (0.29)(0.10 <em>M<\/em>) = 0.029 <em>M<\/em><\/li>\r\n \t<li>[latex]\\left[{\\text{HSO}}_{4}^{-}\\right][\/latex] = 0.10 <em>M<\/em> \u2212 0.029 <em>M<\/em> = 0.071 <em>M<\/em><\/li>\r\n \t<li>[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{SO}}_{4}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{HSO}}_{4}^{-}\\right]}=\\frac{\\left(0.029\\right)\\left(0.029\\right)}{\\left(0.071\\right)}=1.2\\times {10}^{-2}[\/latex]<\/li>\r\n<\/ul>\r\n11.\u00a0The ionization constants are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{\\left(1.00\\times {10}^{-14}\\right)}{2.3\\times {10}^{-3}}=4.3\\times {10}^{-12}[\/latex]<\/li>\r\n \t<li>[latex]{K}_{\\text{a}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{b}}}=\\frac{\\left(1.00\\times {10}^{-14}\\right)}{7.4\\times {10}^{-5}}=1.4\\times {10}^{-10}[\/latex]<\/li>\r\n \t<li>[latex]{K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{\\left(1.00\\times {10}^{-14}\\right)}{1\\times {10}^{-7}}=1\\times {10}^{-7}[\/latex]<\/li>\r\n \t<li>[latex]{K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{\\left(1.00\\times {10}^{-14}\\right)}{2.4\\times {10}^{-12}}=4.2\\times {10}^{-3}[\/latex]<\/li>\r\n \t<li>[latex]{K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{\\left(1.00\\times {10}^{-14}\\right)}{2.4\\times {10}^{-12}}=4.2\\times {10}^{-3}[\/latex]<\/li>\r\n \t<li>[latex]{K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{\\left(1.00\\times {10}^{-14}\\right)}{1.2\\times {10}^{-2}}=8.3\\times {10}^{-13}[\/latex]<\/li>\r\n<\/ol>\r\n13.\u00a0The reactions are:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\\\ {\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\end{array}[\/latex]<\/p>\r\nBecause NH<sub>3<\/sub> is much stronger than C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>, it dissociates more. As the initial concentrations of both bases are the same, at equilibrium, [NH<sub>3<\/sub>] &lt; [C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>], [latex]\\left[{\\text{NH}}_{4}^{+}\\right][\/latex] &gt; [latex]\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex], and [OH<sup>\u2212<\/sup>] &gt; [latex]\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex]. Therefore, (a) is the correct statement. The contribution to the total [OH<sup>\u2212<\/sup>] at equilibrium from C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub> is negligible compared to HN<sub>3<\/sub>. Therefore [latex]\\left[{\\text{OH}}^{-}\\right]=\\left[{\\text{NH}}_{4}^{+}\\right][\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>acid ionization constant (<em>K<\/em><sub>a<\/sub>): <\/strong>equilibrium constant for the ionization of a weak acid\r\n\r\n<strong>base ionization constant (<em>K<\/em><sub>b<\/sub>): <\/strong>equilibrium constant for the ionization of a weak base\r\n\r\n<strong>leveling effect of water: <\/strong>any acid stronger than [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex], or any base stronger than OH<sup>\u2212<\/sup> will react with water to form [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex], or OH<sup>\u2212<\/sup>, respectively; water acts as a base to make all strong acids appear equally strong, and it acts as an acid to make all strong bases appear equally strong\r\n\r\n<strong>oxyacid: <\/strong>compound containing a nonmetal and one or more hydroxyl groups\r\n\r\n<strong>percent ionization: <\/strong>ratio of the concentration of the ionized acid to the initial acid concentration, times 100","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Assess the relative strengths of acids and bases according to their ionization constants<\/li>\n<li>Rationalize trends in acid\u2013base strength in relation to molecular structure<\/li>\n<li>Carry out equilibrium calculations for weak acid\u2013base systems<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idm193375984\">The relative strength of an acid or base is the extent to which it ionizes when dissolved in water. If the ionization reaction is essentially complete, the acid or base is termed\u00a0<em data-effect=\"italics\">strong<\/em>; if relatively little ionization occurs, the acid or base is weak. As will be evident throughout the remainder of this chapter, there are many more weak acids and bases than strong ones. The most common strong acids and bases are listed in Table 1.<\/p>\n<table>\n<thead>\n<tr>\n<th colspan=\"2\">Table\u00a01. Some Common Strong acids and \u00a0Strong Bases<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Strong Acids<\/th>\n<th>Strong Bases<\/th>\n<\/tr>\n<tr>\n<td>HClO<sub>4<\/sub> perchloric acid<\/td>\n<td>LiOH lithium hydroxide<\/td>\n<\/tr>\n<tr>\n<td>HCl hydrochloric acid<\/td>\n<td>NaOH sodium\u00a0hydroxide<\/td>\n<\/tr>\n<tr>\n<td>HBr hydrobromic acid<\/td>\n<td>KOH potassium\u00a0hydroxide<\/td>\n<\/tr>\n<tr>\n<td>HI hydroiodic acid<\/td>\n<td>Ca(OH)<sub>2<\/sub> calcium\u00a0hydroxide<\/td>\n<\/tr>\n<tr>\n<td>HNO<sub>3<\/sub> nitric acid<\/td>\n<td>Sr(OH)<sub>2<\/sub> strontium\u00a0hydroxide<\/td>\n<\/tr>\n<tr>\n<td>H<sub>2<\/sub>SO<sub>4<\/sub> sulfuric acid<\/td>\n<td>Ba(OH)<sub>2<\/sub> barium hydroxide<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"CNX_Chem_14_03_strong\" class=\"os-figure\">The relative strengths of acids may be quantified by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the\u00a0<strong><span id=\"term586\" data-type=\"term\">acid-ionization constant,\u00a0<em data-effect=\"italics\">K<\/em><sub>a<\/sub><\/span><\/strong>. For the reaction of an acid HA:<\/div>\n<div><\/div>\n<div class=\"os-figure\" style=\"text-align: center;\">[latex]\\text{HA}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{A}}^{-}\\left(aq\\right)[\/latex]<\/div>\n<p style=\"text-align: left;\">the acid ionization constant is written<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{A}}^{-}\\right]}{\\text{[HA]}}[\/latex]<\/p>\n<p>where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include [H<sub>2<\/sub>O] in the equation. The larger the\u00a0<em data-effect=\"italics\">K<\/em><sub>a<\/sub>\u00a0of an acid, the larger the concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex]and A<sup>\u2212<\/sup>\u00a0relative to the concentration of the nonionized acid, HA, in an equilibrium mixture, and the stronger the acid. An acid is classified as \u201cstrong\u201d when it undergoes complete ionization, in which case the concentration of HA is zero and the acid ionization constant is immeasurably large (<em data-effect=\"italics\">K<\/em><sub>a<\/sub>\u00a0\u2248 \u221e). Acids that are partially ionized are called \u201cweak,\u201d and their acid ionization constants may be experimentally measured. A table of ionization constants for weak acids is provided in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids.<\/a><\/p>\n<p>To illustrate this idea, three acid ionization equations and\u00a0<em data-effect=\"italics\">K<\/em><sub>a<\/sub>\u00a0values are shown below. The ionization constants increase from first to last of the listed equations, indicating the relative acid strength increases in the order [latex]\\text{CH}_{3}\\text{CO}_{2}\\text{H}[\/latex] &lt; [latex]\\text{HNO}_{2}[\/latex]\u00a0&lt; [latex]{\\text{HSO}}_{4}^{-}:[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 90px;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+\\text{H}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\left(aq\\right)\\qquad{K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 90px;\">[latex]{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{NO}}_{2}^{-}\\left(aq\\right)\\qquad{K}_{\\text{a}}=4.6\\times {10}^{-4}[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 90px;\">[latex]{\\text{HSO}}_{4}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(aq\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{SO}}_{4}^{2-}\\left(aq\\right)\\qquad{K}_{\\text{a}}=1.2\\times {10}^{-2}[\/latex]<\/p>\n<p>Another measure of the strength of an acid is its percent ionization. The <b>percent ionization<\/b> of a weak acid is defined in terms of the compostion of an equilibrium mixture:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{% ionization}=\\dfrac{{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}_{\\text{eq}}}{{\\left[\\text{HA}\\right]}_{0}}\\times 100[\/latex]<\/p>\n<p>where the numerator is equivalent to the concentration of the acid&#8217;s conjugate base (per stoichiometry, [A<sup>\u2212<\/sup>] = [H<sub>3<\/sub>O<sup>+<\/sup>]). Unlike the\u00a0<em>K<\/em><sub>a<\/sub>\u00a0value, the percent ionization of a weak acid varies with the initial concentration of acid, typically decreasing as concentration increases. Equilibrium calculations of the sort described later in this chapter can be used to confirm this behavior.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0Calculation of Percent Ionization from pH<\/h3>\n<p>Calculate the percent ionization of a 0.125-<em>M<\/em> solution of nitrous acid (a weak acid), with a pH of 2.09.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q996427\">Show Solution<\/span><\/p>\n<div id=\"q996427\" class=\"hidden-answer\" style=\"display: none\">\n<p>The percent ionization for an acid is:<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}_{\\text{eq}}}{{\\left[{\\text{HNO}}_{2}\\right]}_{0}}\\times 100[\/latex]<\/p>\n<p>The chemical equation for the dissociation of the nitrous acid is: [latex]{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NO}}_{2}^{-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)[\/latex]. Since 10<sup>\u2212pH<\/sup> = [latex]\\left[\\text{H}_{3}\\text{O}^{+}\\right][\/latex], we find that 10<sup>\u22122.09<\/sup> = 8.1 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>, so that percent ionization is:<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{8.1\\times {10}^{-3}}{0.125}\\times 100=6.5\\%[\/latex]<\/p>\n<p>Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idp51316352\">Check Your Learning<\/h4>\n<p>Calculate the percent ionization of a 0.10-<em>M<\/em> solution of acetic acid with a pH of 2.89.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q74260\">Show Solution<\/span><\/p>\n<div id=\"q74260\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.3% ionized<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">Watch this\u00a0<a href=\"https:\/\/phet.colorado.edu\/en\/simulation\/acid-base-solutions\" target=\"_blank\" rel=\"noopener\">simulation of strong and weak acids and bases at the molecular level<\/a>.<\/div>\n<p>As we did with acids, we can measure the relative strengths of bases by measuring their <b>base-ionization constant, (<em>K<\/em><sub>b<\/sub>)<\/b> in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, B:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{B}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{HB}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex],<\/p>\n<p>we write the equation for the ionization constant as:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{b}}=\\dfrac{\\left[{\\text{HB}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[\\text{B}\\right]}[\/latex]<\/p>\n<p>where the concentrations are those at equilibrium. Again, we do not include [H<sub>2<\/sub>O] in the equation because water is the solvent. The chemical reactions and ionization constants of the three bases shown are:<\/p>\n<p style=\"text-align: left; padding-left: 90px;\">[latex]{\\text{NO}}_{2}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{HNO}}_{2}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\qquad{K}_{\\text{b}}=2.22\\times {10}^{-11}[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 90px;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\qquad{K}_{\\text{b}}=5.6\\times {10}^{-10}[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 90px;\">[latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\qquad{K}_{\\text{b}}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p style=\"text-align: left;\">A table of ionization constants of weak bases appears in <a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-bases-2\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Bases<\/a>. As for acids, the relative strength of a base is also reflected in its percent ionization, computed as<\/p>\n<p style=\"text-align: center;\">[latex]\\%\\text{ ionization}=[\\text{OH}^{-}]_{eq}\/[\\text{B}]_{0}\\times{100}\\%[\/latex]<\/p>\n<p>but will vary depending on the base ionization constant and the initial concentration of the solution.<\/p>\n<h2>Relative Strengths of Conjugate Acid-Base Pairs<\/h2>\n<p>Br\u00f8nsted-Lowry acid-base chemistry is the transfer of protons; thus, logic suggests a relation between the relative strengths of conjugate acid-base pairs. The strength of an acid or base is quantified in its ionization constant,\u00a0<em>K<\/em><sub>a<\/sub>\u00a0or\u00a0<em>K<\/em><sub>b<\/sub>, which represents the extent of the acid or base ionization reaction. For the conjugate acid-base pair HA \/ A<sup>\u2212<\/sup>, ionization equilibrium equations and ionization constant expressions are<\/p>\n<p style=\"padding-left: 90px;\">[latex]\\text{HA}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{A}}^{-}\\left(aq\\right)\\qquad{K}_{\\text{a}}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{A}}^{-}\\right]}{\\left[\\text{HA}\\right]}[\/latex]<\/p>\n<p style=\"padding-left: 90px;\">[latex]{\\text{A}}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{OH}}^{-}\\left(aq\\right)+\\text{HA}\\left(aq\\right)\\qquad{K}_{\\text{b}}=\\dfrac{\\left[\\text{HA}\\right]\\left[\\text{OH}\\right]}{\\left[{\\text{A}}^{-}\\right]}[\/latex]<\/p>\n<p>Adding these two chemical equations yields the equation for the autoionization for water:<\/p>\n<p style=\"text-align: center;\">[latex]\\cancel{\\text{HA}\\left(aq\\right)}+\\text{H}_{2}\\text{O}\\left(l\\right)+\\cancel{\\text{A}^{-}\\left(aq\\right)}+\\text{H}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons\\text{H}_{3}\\text{O}^{+}\\left(aq\\right)+\\cancel{\\text{A}^{-}\\left(aq\\right)}+\\text{OH}^{-}\\left(aq\\right)+\\cancel{\\text{HA}\\left(aq\\right)}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{2H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>As discussed in another chapter on equilibrium, the equilibrium constant for a summed reaction is equal to the mathematical product of the equilibrium constants for the added reactions, and so<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}\\times {K}_{\\text{b}}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{A}}^{-}\\right]}{\\text{[HA]}}\\times \\dfrac{\\text{[HA]}\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{A}}^{-}\\right]}=\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]={K}_{\\text{w}}[\/latex]<\/p>\n<p id=\"fs-idm224463760\">This equation states the relation between ionization constants for any conjugate acid-base pair, namely, their mathematical product is equal to the ion product of water,\u00a0<em data-effect=\"italics\">K<\/em><sub>w<\/sub>. By rearranging this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident:<\/p>\n<p style=\"text-align: center;\">[latex]K_{\\text{a}}=K_{\\text{w}}\/K_{\\text{b}}\\qquad\\text{ or }\\qquad{K}_{\\text{b}}=K_{\\text{w}}\/K_{\\text{a}}[\/latex]<\/p>\n<p id=\"fs-idm225241056\">The inverse proportional relation between\u00a0<em>K<\/em><sub>a<\/sub>\u00a0and\u00a0<em>K<\/em><sub>b<\/sub>\u00a0means\u00a0<em>the stronger the acid or base, the weaker its conjugate partner<\/em>. Figure 1 illustrates this relation for several conjugate acid-base pairs.<\/p>\n<div style=\"width: 810px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213731\/CNX_Chem_14_03_strengths.jpg\" alt=\"The diagram shows two horizontal bars. The first, labeled, \u201cRelative acid strength,\u201d at the top is red on the left and gradually changes to purple on the right. The red end at the left is labeled, \u201cStronger acids.\u201d The purple end at the right is labeled, \u201cWeaker acids.\u201d Just outside the bar to the lower left is the label, \u201cK subscript a.\u201d The bar is marked off in increments with a specific acid listed above each increment. The first mark is at 1.0 with H subscript 3 O superscript positive sign. The second is ten raised to the negative two with H C l O subscript 2. The third is ten raised to the negative 4 with H F. The fourth is ten raised to the negative 6 with H subscript 2 C O subscript 3. The fifth is ten raised to a negative 8 with C H subscript 3 C O O H. The sixth is ten raised to the negative ten with N H subscript 4 superscript positive sign. The seventh is ten raised to a negative 12 with H P O subscript 4 superscript 2 negative sign. The eighth is ten raised to the negative 14 with H subscript 2 O. Similarly the second bar, which is labeled \u201cRelative conjugate base strength,\u201d is purple at the left end and gradually becomes blue at the right end. Outside the bar to the left is the label, \u201cWeaker bases.\u201d Outside the bar to the right is the label, \u201cStronger bases.\u201d Below and to the left of the bar is the label, \u201cK subscript b.\u201d The bar is similarly marked at increments with bases listed above each increment. The first is at ten raised to the negative 14 with H subscript 2 O above it. The second is ten raised to the negative 12 C l O subscript 2 superscript negative sign. The third is ten raised to the negative ten with F superscript negative sign. The fourth is ten raised to a negative eight with H C O subscript 3 superscript negative sign. The fifth is ten raised to the negative 6 with C H subscript 3 C O O superscript negative sign. The sixth is ten raised to the negative 4 with N H subscript 3. The seventh is ten raised to the negative 2 with P O subscript 4 superscript three negative sign. The eighth is 1.0 with O H superscript negative sign.\" width=\"800\" height=\"277\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a01. This diagram shows the relative strengths of conjugate acid-base pairs, as indicated by their ionization constants in aqueous solution.<\/p>\n<\/div>\n<p>The listing of conjugate acid\u2013base pairs shown in Figure 2 is arranged to show the relative strength of each species as compared with water, whose entries are highlighted in each of the table\u2019s columns. In the acid column, those species listed below water are weaker acids than water. These species do not undergo acid ionization in water; they are not Bronsted-Lowry acids. All the species listed above water are stronger acids, transferring protons to water to some extent when dissolved in an aqueous solution to generate hydronium ions. Species above water but below hydronium ion are\u00a0<em data-effect=\"italics\">weak acids<\/em>, undergoing partial acid ionization, wheres those above hydronium ion are\u00a0<em data-effect=\"italics\">strong acids<\/em>\u00a0that are completely ionized in aqueous solution.<\/p>\n<div style=\"width: 810px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213733\/CNX_Chem_14_03_Corresp.jpg\" alt=\"This figure includes a table separated into a left half which is labeled \u201cAcids\u201d and a right half labeled \u201cBases.\u201d A red arrow points up the left side, which is labeled \u201cIncreasing acid strength.\u201d Similarly, a blue arrow points downward along the right side, which is labeled \u201cIncreasing base strength.\u201d Names of acids and bases are listed next to each arrow toward the center of the table, followed by chemical formulas. Acids listed top to bottom are sulfuric acid, H subscript 2 S O subscript 4, hydrogen iodide, H I, hydrogen bromide, H B r, hydrogen chloride, H C l, nitric acid, H N O subscript 3, hydronium ion ( in pink text) H subscript 3 O superscript plus, hydrogen sulfate ion, H S O subscript 4 superscript negative, phosphoric acid, H subscript 3 P O subscript 4, hydrogen fluoride, H F, nitrous acid, H N O subscript 2, acetic acid, C H subscript 3 C O subscript 2 H, carbonic acid H subscript 2 C O subscript 3, hydrogen sulfide, H subscript 2 S, ammonium ion, N H subscript 4 superscript +, hydrogen cyanide, H C N, hydrogen carbonate ion, H C O subscript 3 superscript negative, water (shaded in beige) H subscript 2 O, hydrogen sulfide ion, H S superscript negative, ethanol, C subscript 2 H subscript 5 O H, ammonia, N H subscript 3, hydrogen, H subscript 2, methane, and C H subscript 4. The acids at the top of the listing from sulfuric acid through nitric acid are grouped with a bracket to the right labeled \u201cUndergo complete acid ionization in water.\u201d Similarly, the acids at the bottom from hydrogen sulfide ion through methane are grouped with a bracket and labeled, \u201cDo not undergo acid ionization in water.\u201d The right half of the figure lists bases and formulas. From top to bottom the bases listed are hydrogen sulfate ion, H S O subscript 4 superscript negative, iodide ion, I superscript negative, bromide ion, B r superscript negative, chloride ion, C l superscript negative, nitrate ion, N O subscript 3 superscript negative, water (shaded in beige), H subscript 2 O, sulfate ion, S O subscript 4 superscript 2 negative, dihydrogen phosphate ion, H subscript 2 P O subscript 4 superscript negative, fluoride ion, F superscript negative, nitrite ion, N O subscript 2 superscript negative, acetate ion, C H subscript 3 C O subscript 2 superscript negative, hydrogen carbonate ion, H C O subscript 3 superscript negative, hydrogen sulfide ion, H S superscript negative, ammonia, N H subscript 3, cyanide ion, C N superscript negative, carbonate ion, C O subscript 3 superscript 2 negative, hydroxide ion (in blue), O H superscript negative, sulfide ion, S superscript 2 negative, ethoxide ion, C subscript 2 H subscript 5 O superscript negative, amide ion N H subscript 2 superscript negative, hydride ion, H superscript negative, and methide ion C H subscript 3 superscript negative. The bases at the top, from perchlorate ion through nitrate ion are group with a bracket which is labeled \u201cDo not undergo base ionization in water.\u201d Similarly, the lower 5 in the listing, from sulfide ion through methide ion are grouped and labeled \u201cUndergo complete base ionization in water.\u201d\" width=\"800\" height=\"717\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a02. The chart shows the relative strengths of conjugate acid-base pairs.<\/p>\n<\/div>\n<p>If all these strong acids are completely ionized in water, why does the column indicate they vary in strength, with nitric acid being the weakest and perchloric acid the strongest? Notice that the sole acid species present in an aqueous solution of any strong acid is H<sub>3<\/sub>O<sup>+<\/sup>(<em>aq<\/em>), meaning that hydronium ion is the strongest acid that may exist in water; any stronger acid will react completely with water to generate hydronium ions. This limit on the acid strength of solutes in a solution is called a\u00a0<strong>leveling effect<\/strong>. To measure the differences in acid strength for \u201cstrong\u201d acids, the acids must be dissolved in a solvent that is\u00a0<em>less basic<\/em>\u00a0than water. In such solvents, the acids will be \u201cweak,\u201d and so any differences in the extent of their ionization can be determined. For example, the binary hydrogen halides HCl, HBr, and HI are strong acids in water but weak acids in ethanol (strength increasing HCl &lt; HBr &lt; HI).<\/p>\n<p id=\"fs-idm167383056\">The right column of Figure 2 lists a number of substances in order of increasing base strength from top to bottom. Following the same logic as for the left column, species listed above water are weaker bases and so they don\u2019t undergo base ionization when dissolved in water. Species listed between water and its conjugate base, hydroxide ion, are weak bases that partially ionize. Species listed below hydroxide ion are strong bases that completely ionize in water to yield hydroxide ions (i.e., they are\u00a0<em>leveled<\/em>\u00a0to hydroxide). A comparison of the acid and base columns in this table supports the reciprocal relation between the strengths of conjugate acid-base pairs. For example, the conjugate bases of the strong acids (top of table) are all of negligible strength. A strong acid exhibits an immeasurably large\u00a0<em>K<\/em><sub>a<\/sub>, and so its conjugate base will exhibit a\u00a0<em>K<\/em><sub>b<\/sub>\u00a0that is essentially zero:<\/p>\n<p style=\"padding-left: 90px;\">strong acid:\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K_a[\/latex] \u2248 [latex]\\infty[\/latex]<\/p>\n<p style=\"padding-left: 90px;\">conjugate base:\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K_b = K_w\/K_a = K_w\/\\infty[\/latex] \u2248 0<\/p>\n<p id=\"fs-idm225937888\">A similar approach can be used to support the observation that conjugate acids of strong bases (<em>K<\/em><sub>b<\/sub>\u00a0\u2248 \u221e) are of negligible strength (<em>K<\/em><sub>a<\/sub>\u00a0\u2248 0).<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2: Calculating Ionization Constants for Conjugate Acid-Base Pairs<\/h3>\n<p>Use the <em>K<\/em><sub>b<\/sub> for the nitrite ion, [latex]{\\text{NO}}_{2}^{-}[\/latex], to calculate the <em>K<\/em><sub>a<\/sub> for its conjugate acid.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q534789\">Show Solution<\/span><\/p>\n<div id=\"q534789\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>K<\/em><sub>b<\/sub> for [latex]{\\text{NO}}_{2}^{-}[\/latex] is given in this section as 2.22 \u00d7 10<sup>\u221211<\/sup>. The conjugate acid of [latex]{\\text{NO}}_{2}^{-}[\/latex] is HNO<sub>2<\/sub>; <em>K<\/em><sub>a<\/sub> for HNO<sub>2<\/sub> can be calculated using the relationship:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}\\times {K}_{\\text{b}}=1.0\\times {10}^{-14}={K}_{\\text{w}}[\/latex]<\/p>\n<p>Solving for <em>K<\/em><sub>a<\/sub>, we get:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\dfrac{{K}_{\\text{w}}}{{K}_{\\text{b}}}=\\dfrac{1.0\\times {10}^{-14}}{2.22\\times {10}^{-11}}=4.5\\times {10}^{-4}[\/latex]<\/p>\n<p>This answer can be verified by finding the <em>K<\/em><sub>a<\/sub> for HNO<sub>2<\/sub> in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids<\/a>.<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idp28811616\">Check Your Learning<\/h4>\n<p>We can determine the relative acid strengths of [latex]{\\text{NH}}_{4}^{+}[\/latex] and HCN by comparing their ionization constants. The ionization constant of HCN is given in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids<\/a>\u00a0as 4 \u00d7 10<sup>\u221210<\/sup>. The ionization constant of [latex]{\\text{NH}}_{4}^{+}[\/latex] is not listed, but the ionization constant of its conjugate base, NH<sub>3<\/sub>, is listed as 1.8 \u00d7 10<sup>\u22125<\/sup>. Determine the ionization constant of [latex]{\\text{NH}}_{4}^{+}[\/latex], and decide which is the stronger acid, HCN or [latex]{\\text{NH}}_{4}^{+}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q218282\">Show Solution<\/span><\/p>\n<div id=\"q218282\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\text{NH}}_{4}^{+}[\/latex] is the slightly stronger acid (<em>K<\/em><sub>a<\/sub> for [latex]{\\text{NH}}_{4}^{+}[\/latex] = 5.6 \u00d7 10<sup>\u221210<\/sup>).<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.<\/li>\n<li>Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.<\/li>\n<li>Use this list of important industrial compounds (and Figure\u00a02) to answer the following questions regarding: CaO, Ca(OH)<sub>2<\/sub>, CH<sub>3<\/sub>CO<sub>2<\/sub>H, CO<sub>2,<\/sub> HCl, H<sub>2<\/sub>CO<sub>3<\/sub>, HF, HNO<sub>2<\/sub>, HNO<sub>3<\/sub>, H<sub>3<\/sub>PO<sub>4<\/sub>, H<sub>2<\/sub>SO<sub>4<\/sub>, NH<sub>3<\/sub>, NaOH, Na<sub>2<\/sub>CO<sub>3<\/sub>.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Identify the strong Br\u00f8nsted-Lowry acids and strong Br\u00f8nsted-Lowry bases.<\/li>\n<li>List those compounds in (a) that can behave as Br\u00f8nsted-Lowry acids with strengths lying between those of [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex] and H<sub>2<\/sub>O.<\/li>\n<li>List those compounds in (a) that can behave as Br\u00f8nsted-Lowry bases with strengths lying between those of H<sub>2<\/sub>O and OH<sup>\u2212<\/sup>.<\/li>\n<\/ol>\n<\/li>\n<li>Explain why the ionization constant, <em>K<\/em><sub>a<\/sub>, for H<sub>2<\/sub>SO<sub>4<\/sub> is larger than the ionization constant for H<sub>2<\/sub>SO<sub>3<\/sub>.<\/li>\n<li>Explain why the ionization constant, <em>K<\/em><sub>a<\/sub>, for HI is larger than the ionization constant for HF.<\/li>\n<li>What is the ionization constant at 25 \u00b0C for the weak acid [latex]{\\text{CH}}_{3}{\\text{NH}}_{3}^{+}[\/latex], the conjugate acid of the weak base CH<sub>3<\/sub>NH<sub>2<\/sub>, <em>K<\/em><sub>b<\/sub> = 4.4 \u00d7 10<sup>\u22124<\/sup>.<\/li>\n<li>What is the ionization constant at 25 \u00b0C for the weak acid [latex]{\\left({\\text{CH}}_{3}\\right)}_{2}{\\text{NH}}_{2}^{+}[\/latex], the conjugate acid of the weak base (CH<sub>3<\/sub>)<sub>2<\/sub>NH, <em>K<\/em><sub>b<\/sub> = 7.4 \u00d7 10<sup>\u22124<\/sup>?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q314309\">Show Selected Solutions<\/span><\/p>\n<div id=\"q314309\" class=\"hidden-answer\" style=\"display: none\">\n<p>2.\u00a0The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH<sup>\u2212<\/sup>, which causes the solution to be basic. An example is NaCN. The CN<sup>\u2212<\/sup> reacts with water as follows:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CN}}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons \\text{HCN}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>4.\u00a0The oxidation state of the sulfur in H<sub>2<\/sub>SO<sub>4<\/sub> is greater than the oxidation state of the sulfur in H<sub>2<\/sub>SO<sub>3<\/sub>.<\/p>\n<p>6. <em>K<\/em><sub>w<\/sub> = <em>K<\/em><sub>a<\/sub> \u00d7 <em>K<\/em><sub>b<\/sub>; thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{K}_{\\text{a}}&=&\\frac{{K}_{\\text{w}}}{{K}_{\\text{b}}}\\\\{K}_{\\text{a}}&=&\\frac{1.0\\times {10}^{-14}}{4.4\\times {10}^{-4}}=2.3\\times {10}^{-11}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Acid- Base Equilibrium Calculations<\/h2>\n<p>The chapter on chemical equilibria introduced several types of equilibrium calculations and the various mathematical strategies that are helpful in performing them. These strategies are generally useful for equilibrium systems regardless of chemical reaction class, and so they may be effectively applied to acid-base equilibrium problems. This section presents several example exercises involving equilibrium calculations for acid-base systems.<\/p>\n<div style=\"width: 708px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/openstax.org\/resources\/31169c8dd8328325b22c0ea1eb5ca58a61b9d659\" alt=\"This image shows two bottles containing clear colorless solutions. Each bottle contains a single p H indicator strip. The strip in the bottle on the left is red, and a similar red strip is placed on a filter paper circle in front of the bottle on surface on which the bottles are resting. Similarly, the second bottle on the right contains and orange strip and an orange strip is placed in front of it on a filter paper circle. Between the two bottles is a pack of p Hydrion papers with a p H color scale on its cover.\" width=\"698\" height=\"531\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a03. pH paper indicates that a 0.l-<em>M<\/em> solution of HCl (beaker on left) has a pH of 1. The acid is fully ionized and [H<sub>3<\/sub>O<sup>+<\/sup>] = 0.1 <em>M<\/em>. A 0.1-<em>M<\/em> solution of CH<sub>3<\/sub>CO<sub>2<\/sub>H (beaker on right) is has a pH of 3 ([H<sub>3<\/sub>O<sup>+<\/sup>] = 0.001 <em>M<\/em>) because the weak acid CH<sub>3<\/sub>CO<sub>2<\/sub>H is only partially ionized. In this solution, [H<sub>3<\/sub>O<sup>+<\/sup>] &lt; [CH<sub>3<\/sub>CO<sub>2<\/sub>H]. (credit: modification of work by Sahar Atwa)<\/p>\n<\/div>\n<table id=\"fs-idm41095184\" summary=\"This table has two columns and ten rows. The first row is a header row and it labels each column, \u201cIonization Reaction,\u201d and \u201cK subscript a at 25 degrees C.\u201d Under the \u201cIonization Reaction\u201d column are the following equations: \u201cH S O subscript 4 superscript negative sign plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus S O subscript 4 superscript 2 negative sign,\u201d \u201cH F plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus F superscript negative sign,\u201d \u201cH N O subscript 2 plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus N O subscript 2 superscript negative sign,\u201d \u201cH N C O plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sing plus H C O subscript 2 superscript negative sign,\u201d \u201cC H subscript 3 C O subscript 2 H plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus C I O superscript negative sign,\u201d \u201c H B r O plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus B r O superscript negative sign,\u201d and \u201cH C N plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus C N superscript negative sign.\u201d Under the \u201cK subscript a at 25 degrees C\u201d column are the following values: 1.2 times ten to the negative two, 7.2 times ten to the negative 4, 4.5 times ten to the negative 4, 3.46 times ten to the negative 4, 1.8 times ten to the negative 4, 1.8 times ten to the negative 5, 3.5 times ten to the negative 8, 2 times ten to the negative 9, and 4 times ten to the negative ten.\">\n<thead>\n<tr>\n<th colspan=\"2\">Table\u00a02. Ionization Constants of Some Weak Acids<\/th>\n<\/tr>\n<tr valign=\"top\">\n<th>Ionization Reaction<\/th>\n<th><em>K<\/em><sub>a<\/sub> at 25 \u00b0C<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex]{\\text{HSO}}_{4}^{-}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{SO}}_{4}^{2-}[\/latex]<\/td>\n<td>1.2 \u00d7 10<sup>\u22122<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\text{HF}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{F}}^{-}[\/latex]<\/td>\n<td>7.2 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{HNO}}_{2}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{NO}}_{2}^{-}[\/latex]<\/td>\n<td>4.5 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\text{HNCO}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{NCO}}^{-}[\/latex]<\/td>\n<td>3.46 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{HCO}}_{2}\\text{H}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{HCO}}_{2}^{-}[\/latex]<\/td>\n<td>1.8 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}[\/latex]<\/td>\n<td>1.8 \u00d7 10<sup>\u22125<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\text{HCIO}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{CIO}}^{-}[\/latex]<\/td>\n<td>3.5 \u00d7 10<sup>\u22128<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\text{HBrO}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{BrO}}^{-}[\/latex]<\/td>\n<td>2 \u00d7 10<sup>\u22129<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\text{HCN}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{CN}}^{-}[\/latex]<\/td>\n<td>4 \u00d7 10<sup>\u221210<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Table\u00a02 gives the ionization constants for several weak acids; additional ionization constants can be found in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids<\/a>.<\/p>\n<div style=\"width: 651px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/openstax.org\/resources\/cfacb6d6b36c83b23129cf49fc39d1384ff6d850\" alt=\"This photo shows two glass containers filled with a transparent liquid. In between the containers is a p H strip indicator guide. There are p H strips placed in front of each glass container. The liquid in the container on the left appears to have a p H of 10 or 11. The liquid in the container on the right appears to have a p H of about 13 or 14.\" width=\"641\" height=\"368\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a04. pH paper indicates that a 0.1-<em>M<\/em> solution of NH<sub>3<\/sub> (left) is weakly basic. The solution has a pOH of 3 ([OH<sup>\u2212<\/sup>] = 0.001 <em>M<\/em>) because the weak base NH<sub>3<\/sub> only partially reacts with water. A 0.1-<em>M<\/em> solution of NaOH (right) has a pOH of 1 because NaOH is a strong base. (credit: modification of work by Sahar Atwa)<\/p>\n<\/div>\n<p>The ionization constants of several weak bases are given in Table\u00a03 and in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-bases-2\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Bases<\/a>.<\/p>\n<table id=\"fs-idm84795184\" summary=\"This table has two columns and six rows. The first row is a header row and it labels each column: \u201cIonization Reaction,\u201d and \u201cK subscript b at 25 degrees C.\u201d Under the \u201cIonization Reaction\u201d column are the following reactions: \u201c( C H subscript 3 ) subscript 2 N H plus H subscript 2 O equilibrium arrow ( C H subscript 3 ) subscript 2 N H subscript 2 superscript plus sign plus O H superscript negative sign,\u201d \u201cC H subscript 3 N H subscript 2 plus H subscript 2 O equilibrium arrow C H subscript 3 N H subscript 3 superscript plus sign plus O H superscript negative sign,\u201d \u201c( C H subscript 3 ) subscript 3 N plus H subscript 2 O equilibrium arrow ( C H subscript 3 ) subscript 3 N H superscript plus sign plus O H superscript negative sign,\u201d \u201cN H subscript 3 plus H subscript 2 O equilibrium arrow N H subscript 4 superscript plus sign plus O H superscript negative sign,\u201d and \u201cC subscript 6 H subscript 5 N H subscript 2 plus H subscript 2 O equilibrium arrow C subscript 6 N subscript 5 N H subscript 3 superscript plus sign plus O H superscript negative sign.\u201d Under the \u201cK subscript b at 25 degrees C\u201d column are the following values: \u201c7.4 times ten to the negative 4,\u201d \u201c4.4 times ten to the negative 4,\u201d \u201c7.4 times ten to the negative 5,\u201d \u201c1.8 times ten to the negative 5,\u201d and \u201c4.6 times ten to the negative 10.\u201d\">\n<thead>\n<tr>\n<th colspan=\"2\">Table\u00a03. Ionization Constants of Some Weak Bases<\/th>\n<\/tr>\n<tr valign=\"top\">\n<th>Ionization Reaction<\/th>\n<th><em>K<\/em><sub>b<\/sub> at 25 \u00b0C<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex]{\\left({\\text{CH}}_{3}\\right)}_{2}\\text{NH}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\left({\\text{CH}}_{3}\\right)}_{2}{\\text{NH}}_{2}^{+}+{\\text{OH}}^{-}[\/latex]<\/td>\n<td>7.4 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{CH}}_{3}{\\text{NH}}_{2}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{CH}}_{3}{\\text{NH}}_{3}^{+}+{\\text{OH}}^{-}[\/latex]<\/td>\n<td>4.4 \u00d7 10<sup>\u22124<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}+{\\text{OH}}^{-}[\/latex]<\/td>\n<td>6.3 \u00d7 10<sup>\u22125<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{NH}}_{3}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{NH}}_{4}^{+}+{\\text{OH}}^{-}[\/latex]<\/td>\n<td>1.8 \u00d7 10<sup>\u22125<\/sup><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{C}}_{6}{\\text{N}}_{5}{\\text{NH}}_{3}^{+}+{\\text{OH}}^{-}[\/latex]<\/td>\n<td>4.6 \u00d7 10<sup>\u221210<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox exercises\">\n<h3>Think about It<\/h3>\n<p>Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F<sup>\u2212<\/sup> or CN<sup>\u2212<\/sup>, is the stronger base? See Table\u00a03.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"4\"><\/textarea><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 <em>M<\/em> in HCO<sub>2<\/sub>H and 0.10 <em>M<\/em> in HClO.<\/li>\n<li>Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 <em>M<\/em> in HNO<sub>2<\/sub> and 0.120 <em>M<\/em> in HBrO.<\/li>\n<li>Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 <em>M<\/em> in CH<sub>3<\/sub>NH<sub>2<\/sub> and 0.10 <em>M<\/em> in C<sub>5<\/sub>H<sub>5<\/sub>N (<em>K<\/em><sub>b<\/sub> = 1.7 \u00d7 10<sup>\u22129<\/sup>).<\/li>\n<li>Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 <em>M<\/em> in NH<sub>3<\/sub> and 0.100 <em>M<\/em> in C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>.<\/li>\n<li>Using the <em>K<\/em><sub>a<\/sub> values in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids<\/a>, place [latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{3+}[\/latex] in the correct location in Figure\u00a03.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q588268\">Show Solution to Question 2<\/span><\/p>\n<div id=\"q588268\" class=\"hidden-answer\" style=\"display: none\">\n<h4>Question 2<\/h4>\n<p>The reactions and equilibrium constants are:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{NO}}_{2}^{-}\\left(aq\\right){K}_{\\text{a}}=4.5\\times {10}^{-4}\\\\ \\text{HBrO}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{BrO}}^{-}\\left(aq\\right){K}_{\\text{a}}=2\\times {10}^{-9}\\end{array}[\/latex]<\/p>\n<p>As <em>K<\/em><sub>a<\/sub> is much larger for HNO<sub>2<\/sub> than for HBrO, the first equilibrium will dominate. The equilibrium expression is [latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{NO}}_{2}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{HNO}}_{2}\\right]}=4.5\\times {10}^{-5}[\/latex]<\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[HNO<sub>2<\/sub>]<\/th>\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\n<th>[NO<sub>2<\/sub><sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.134<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.134 \u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.134 \u2212 <em>x<\/em>) \u2248 0.134 gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{NO}}_{2}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{HNO}}_{2}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.134-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.134}=4.5\\times {10}^{-4}[\/latex]<\/p>\n<p>Solving for <em>x<\/em> gives 7.77 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>. Because this value is 5.8% of 0.134, our assumption is incorrect. Therefore, we must use the quadratic formula. Using the data gives the following equation: <em>x<\/em><sup>2<\/sup> + 4.5 \u00d7 10<sup>\u22127<\/sup><em>x<\/em> \u2212 6.03 \u00d7 10<sup>\u22125<\/sup> = 0<\/p>\n<p>Using the quadratic formula gives (<em>a<\/em> = 1, <em>b<\/em> = 4.5 \u00d7 10<sup>\u22124<\/sup>, and <em>c<\/em> = \u22126.03 \u00d7 10<sup>\u22125<\/sup>)<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{ }x&=&\\frac{-b\\pm \\sqrt{{b}^{\\text{2+}}-4ac}}{2a}=\\frac{-\\left(4.5\\times {10}^{-4}\\right)\\pm \\sqrt{{\\left(4.5\\times {10}^{-4}\\right)}^{\\text{2+}}-4\\left(1\\right)\\left(-6.03\\times {10}^{-5}\\right)}}{2\\left(1\\right)}\\\\{}&=&\\frac{-\\left(4.5\\times {10}^{-4}\\right)\\pm \\left(1.55\\times {10}^{-2}\\right)}{2}=7.54\\times {10}^{-3}M\\left(\\text{positive root}\\right)\\end{array}[\/latex]<\/p>\n<p>The equilibrium concentrations are therefore:<\/p>\n<ul>\n<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\left[{\\text{NO}}_{2}^{-}\\right]=7.54\\times {10}^{-3}=7.5\\times {10}^{-3}M[\/latex]<\/li>\n<li>[HNO<sub>2<\/sub>] = 0.134 \u2212 7.54 \u00d7 10<sup>\u22123<\/sup> = 0.1264 = 0.126<\/li>\n<\/ul>\n<p>[OH<sup>\u2212<\/sup>] can be calculated using <em>K<\/em><sub>w<\/sub>:<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{OH}}^{-}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{1.0\\times {10}^{-14}}{7.54\\times {10}^{-3}}=1.33\\times {10}^{-12}=1.3\\times {10}^{-12}M[\/latex]<\/p>\n<p>Finally, use the other equilibrium to find the other concentrations. Assume for [HBrO] that (0.120 \u2212 <em>x<\/em>) \u2248 0.124 <em>M<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{BrO}}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[\\text{HBrO}\\right]}=2\\times {10}^{-9}\\approx \\frac{\\left[{\\text{BrO}}^{-}\\right]\\left(7.54\\times {10}^{-3}\\right)}{0.120}[\/latex]<\/p>\n<p>Solving for [BrO<sup>\u2212<\/sup>] gives:<\/p>\n<ul>\n<li>[BrO<sup>\u2212<\/sup>] = 3.2 \u00d7 10<sup>\u22128<\/sup> = 3.2 \u00d7 10<sup>\u22128<\/sup><em>M<\/em><\/li>\n<li>[HBrO] = 0.120 \u2212 3.2 \u00d7 10<sup>\u22128<\/sup> = 0.120 <em>M<\/em><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q399954\">Show Solution to Question 4<\/span><\/p>\n<div id=\"q399954\" class=\"hidden-answer\" style=\"display: none\">\n<h4>Question 4<\/h4>\n<p>The reactions and equilibrium constants are:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{}{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{OH}}^{-}\\left(aq\\right)+{\\text{NH}}_{4}^{+}\\left(aq\\right){K}_{\\text{b}}=1.8\\times {10}^{-5}\\\\ {\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{OH}}^{-}\\left(aq\\right)+{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\left(aq\\right){K}_{\\text{b}}=4.6\\times {10}^{-10}\\end{array}[\/latex]<\/p>\n<p>As <em>K<\/em><sub>b<\/sub> is much larger for NH<sub>3<\/sub> than for C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2,<\/sub> the first equilibrium will dominate. The equilibrium expression is [latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[NH<sub>3<\/sub>]<\/th>\n<th>[OH<sup>\u2212<\/sup>]<\/th>\n<th>[NH<sub>4<\/sub><sup>+<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.115<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.115 \u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.115 \u2212 <em>x<\/em>) \u2248 0.115 gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{NH}}_{4}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.115-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.115}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>Solving for <em>x<\/em> gives 1.44 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>. Because this value is less than 5% of 0.115 <em>M<\/em>, our assumption is correct. The equilibrium concentrations are therefore:<\/p>\n<ul>\n<li>[OH<sup>\u2212<\/sup>] = [latex]\\left[{\\text{NO}}_{4}^{+}\\right][\/latex] = 1.44 \u00d7 10<sup>\u22123<\/sup> = 0.0014 <em>M<\/em><\/li>\n<li>[NH<sub>3<\/sub>] = 0.115 \u2212 0.00144 = 0.1136 = 0.144 <em>M<\/em><\/li>\n<\/ul>\n<p>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] can be calculated using <em>K<\/em><sub>w<\/sub>:<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{OH}}^{-}\\right]}=\\frac{1.0\\times {10}^{-14}}{0.00144}=6.94\\times {10}^{-12}=6.9\\times {10}^{-12}M[\/latex]<\/p>\n<p>Finally, use the other equilibrium to find the other concentrations. Assume for [C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>] that (0.100 \u2212 <em>x<\/em>) \u2248 0.100 <em>M<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{OH}}^{-}\\right]}=\\frac{1.0\\times {10}^{-14}}{0.00144}=6.94\\times {10}^{-12}=6.9\\times {10}^{-12}M[\/latex]<\/p>\n<p>Solving for [latex]\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex] gives:<\/p>\n<ul>\n<li>[latex]\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex] = 3.9 \u00d7 10<sup>\u22128<\/sup><em>M<\/em><\/li>\n<li>[C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>] = 0.100 \u2212 3.19 \u00d7 10<sup>\u22128<\/sup> = 0.100 <em>M<\/em><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 3:\u00a0Determination of <em>K<\/em><sub>a<\/sub> from Equilibrium Concentrations<\/h3>\n<p>Acetic acid is the principal ingredient in vinegar (Figure\u00a05); that&#8217;s why it tastes sour. At equilibrium, a solution contains [CH<sub>3<\/sub>CO<sub>2<\/sub>H] = 0.0787 <em>M<\/em> and [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right]=0.00118M[\/latex]. What is the value of <em>K<\/em><sub>a<\/sub> for acetic acid?<\/p>\n<div style=\"width: 360px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213737\/CNX_Chem_14_03_Vinegar.jpg\" alt=\"An image shows the label of a bottle of distilled white vinegar. The label states that the contents have been reduced with water to 5 percent acidity.\" width=\"350\" height=\"333\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a05. Vinegar is a solution of acetic acid, a weak acid. (credit: modification of work by \u201cHomeSpot HQ\u201d\/Flickr)<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q846674\">Show Solution<\/span><\/p>\n<div id=\"q846674\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-idm119280448\">We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}=\\dfrac{\\left(0.00118\\right)\\left(0.00118\\right)}{0.0787}=1.77\\times {10}^{-5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idm68555680\">Check Your Learning<\/h4>\n<p>What is the equilibrium constant for the ionization of the [latex]{\\text{HSO}}_{4}^{-}[\/latex] ion, the weak acid used in some household cleansers:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{HSO}}_{4}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{SO}}_{4}^{2-}\\left(aq\\right)[\/latex]<\/p>\n<p>In one mixture of NaHSO<sub>4<\/sub> and Na<sub>2<\/sub>SO<sub>4<\/sub> at equilibrium, [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 0.027 <em>M<\/em>; [latex]\\left[{\\text{HSO}}_{4}^{-}\\right]=0.29M[\/latex]; and [latex]\\left[{\\text{SO}}_{4}^{2-}\\right]=0.13M[\/latex].<\/p>\n<p><em><br \/>\n<\/em><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q139813\">Show Solution<\/span><\/p>\n<div id=\"q139813\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>K<\/em><sub>a<\/sub> for [latex]{\\text{HSO}}_{4}^{-}[\/latex] = 1.2 \u00d7 10<sup>\u22122<\/sup><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 4:\u00a0Determination of <em>K<\/em><sub>b<\/sub> from Equilibrium Concentrations<\/h3>\n<p>Caffeine, C<sub>8<\/sub>H<sub>10<\/sub>N<sub>4<\/sub>O<sub>2<\/sub> is a weak base. What is the value of <em>K<\/em><sub>b<\/sub> for caffeine if a solution at equilibrium has [C<sub>8<\/sub>H<sub>10<\/sub>N<sub>4<\/sub>O<sub>2<\/sub>] = 0.050 <em>M<\/em>, [latex]\\left[{\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}{\\text{H}}^{+}\\right][\/latex] = 5.0 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>, and [OH<sup>\u2212<\/sup>] = 2.5 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q358312\">Show Solution<\/span><\/p>\n<div id=\"q358312\" class=\"hidden-answer\" style=\"display: none\">\n<p>At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}{\\text{H}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<br \/>\n[latex]{K}_{\\text{b}}=\\dfrac{\\left[{\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}{\\text{H}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}\\right]}=\\dfrac{\\left(5.0\\times {10}^{-3}\\right)\\left(2.5\\times {10}^{-3}\\right)}{0.050}=2.5\\times {10}^{-4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idm122074528\">Check Your Learning<\/h4>\n<p>What is the equilibrium constant for the ionization of the [latex]{\\text{HPO}}_{4}^{2-}[\/latex] ion, a weak base:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{HPO}}_{4}^{2-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{2}{\\text{PO}}_{4}^{-}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>In a solution containing a mixture of NaH<sub>2<\/sub>PO<sub>4<\/sub> and Na<sub>2<\/sub>HPO<sub>4<\/sub> at equilibrium, [OH<sup>\u2212<\/sup>] = 1.3 \u00d7 10<sup>\u22126<\/sup><em>M<\/em>; [latex]\\left[{\\text{H}}_{2}{\\text{PO}}_{4}^{-}\\right]=0.042M[\/latex]; and [latex]\\left[{\\text{HPO}}_{4}^{2-}\\right]=0.341M[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q153773\">Show Solution<\/span><\/p>\n<div id=\"q153773\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>K<sub>b<\/sub><\/em> for [latex]{\\text{HPO}}_{4}^{2-}=1.6\\times {10}^{-7}[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 5:\u00a0Determination of <em>K<\/em><sub>a<\/sub> or <em>K<\/em><sub>b<\/sub> from pH<\/h3>\n<p>The pH of a 0.0516-<em>M<\/em> solution of nitrous acid, HNO<sub>2<\/sub>, is 2.34. What is its <em>K<\/em><sub>a<\/sub>?<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{NO}}_{2}^{-}\\left(aq\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q978442\">Show Solution<\/span><\/p>\n<div id=\"q978442\" class=\"hidden-answer\" style=\"display: none\">\n<p>The nitrous acid concentration provided is a\u00a0<em>formal<\/em>\u00a0concentration, one that does not account for any chemical equilibria that may be established in solution. Such concentrations are treated as \u201cinitial\u201d values for equilibrium calculations using the ICE table approach. Notice the initial value of hydronium ion is listed as\u00a0<em>approximately<\/em>\u00a0zero because a small concentration of H<sub>3<\/sub>O<sup>+<\/sup>\u00a0is present (1 \u00d7 10<sup>\u22127<\/sup>\u00a0<em>M<\/em>) due to the autoprotolysis of water. In many cases, such as all the ones presented in this chapter, this concentration is much less than that generated by ionization of the acid (or base) in question and may be neglected.<\/p>\n<p id=\"fs-idm204817712\">The pH provided is a logarithmic measure of the hydronium ion concentration resulting from the acid ionization of the nitrous acid, and so it represents an \u201cequilibrium\u201d value for the ICE table:<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]={10}^{-2.34}=0.0046M[\/latex]<\/p>\n<p>The ICE table for this system is then:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/openstax.org\/resources\/fa4bb024e3500e0f2df46b5d3f916db6e44b1d9f\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of \u201cH N O subscript 2 plus sign H subscript 2 O equilibrium sign H subscript 3 O superscript positive sign plus sign N O subscript 2 superscript negative sign.\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.0516, negative 0.0046, 0.0470. The second column is blank in all three rows. The third column has the following: approximately 0, positive 0.0046, 0.0046. The fourth column has the following: 0, positive 0.0046, 0.0046.\" width=\"1000\" height=\"262\" \/><\/p>\n<p>Finally, we calculate the value of the equilibrium constant using the data in the table:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{NO}}_{2}^{-}\\right]}{\\left[{\\text{HNO}}_{2}\\right]}=\\dfrac{\\left(0.0046\\right)\\left(0.0046\\right)}{\\left(0.0470\\right)}=4.5\\times {10}^{-4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idm166688304\">Check Your Learning<\/h4>\n<p>The pH of a solution of household ammonia, a 0.950-<em>M<\/em> solution of NH<sub>3,<\/sub> is 11.612. What is <em>K<\/em><sub>b<\/sub> for NH<sub>3<\/sub>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q487528\">Show Solution<\/span><\/p>\n<div id=\"q487528\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>K<sub>b<\/sub><\/em> = 1.8 \u00d7 10<sup>\u22125<\/sup><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 6:\u00a0Equilibrium Concentrations in a Solution of a Weak Acid<\/h3>\n<p>Formic acid, HCO<sub>2<\/sub>H, is the irritant that causes the body\u2019s reaction to ant stings (Figure\u00a06).<\/p>\n<div style=\"width: 360px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213744\/CNX_Chem_14_03_AntSting.jpg\" alt=\"A photograph is shown of a large black ant on the end of a human finger.\" width=\"350\" height=\"197\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a06. The pain of an ant\u2019s sting is caused by formic acid. (credit: John Tann)<\/p>\n<\/div>\n<p>What is the concentration of hydronium ion and the pH in a 0.534-<em>M<\/em> solution of formic acid?<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{HCO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{HCO}}_{2}^{-}\\left(aq\\right){K}_{\\text{a}}=1.8\\times {10}^{-4}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q135863\">Show Solution<\/span><\/p>\n<div id=\"q135863\" class=\"hidden-answer\" style=\"display: none\">\n<h4>Step 1: Determine <em>x<\/em> and equilibrium concentrations<\/h4>\n<p>The ICE table for this system is<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/openstax.org\/resources\/cbf133e6eed622c35dfb6c5c537670c8dc581a7b\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of \u201cH C O subscript 2 H plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign.\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.534, blank, 0.534 minus x. The second column is blank in all three rows. The third column has the following: approximately 0, positive x, x. The fourth column has the following: 0, positive x, x.\" width=\"1170\" height=\"267\" \/><\/p>\n<p>Substituting the equilibrium concentration terms into the K<sub>a<\/sub> expression gives<\/p>\n<p style=\"text-align: center;\"><span style=\"text-align: center;\">[latex]\\begin{array}{rcl}{K}_{\\text{a}}&=&1.8\\times {10}^{-4}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{HCO}}_{2}^{-}\\right]}{\\left[{\\text{HCO}}_{2}\\text{H}\\right]}\\\\{}&=&\\dfrac{\\left(x\\right)\\left(x\\right)}{0.534-x}=1.8\\times {10}^{-4}\\end{array}[\/latex]<\/span><\/p>\n<p>Because the initial concentration of acid is reasonably large and <em>K<\/em><sub>a<\/sub> is very small, we assume that <em>x<\/em> &lt;&lt; 0.534, which <em>permits<\/em> us to simplify the denominator term as (0.534 \u2212 <em>x<\/em>) = 0.534. This gives:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=1.8\\times {10}^{-4}=\\dfrac{{x}^{\\text{2}}}{0.534}[\/latex]<\/p>\n<p>Solve for <em>x<\/em> as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{x}^{\\text{2}}&=&0.534\\times \\left(1.8\\times {10}^{-4}\\right)=9.6\\times {10}^{-5}\\\\x&=&\\sqrt{9.6\\times {10}^{-5}}\\\\&=&9.8\\times {10}^{-3} M\\end{array}[\/latex]<\/p>\n<p>To check the assumption that <em>x<\/em> is small compared to 0.534, we calculate:<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{x}{0.534}=\\dfrac{9.8\\times{10}^{-3}}{0.534}=1.8\\times{10}^{-2}\\left(1.8\\%\\text{ of 0.534}\\right)[\/latex]<\/p>\n<p><em>x<\/em> is less than 5% of the initial concentration; the assumption is valid.<\/p>\n<p id=\"fs-idm490302912\">As defined in the ICE table,\u00a0<em>x<\/em>\u00a0is equal to the equilibrium concentration of hydronium ion:<\/p>\n<p style=\"text-align: center;\">[latex]x=[\\text{H}_{3}\\text{O}^{+}]=0.0098 M[\/latex]<\/p>\n<p style=\"text-align: left;\"><span style=\"font-size: 1rem; text-align: initial;\">Finally, the pH is calculated to be<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\text{pH}=\\text{log}[\\text{H}_{3}\\text{O}^{+}]=-\\text{log}(0.0098)=2.01[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idp80326224\">Check Your Learning<\/h4>\n<p>Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-<em>M<\/em> solution of acetic acid, CH<sub>3<\/sub>CO<sub>2<\/sub>H?<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\left(aq\\right)\\qquad{K}_{\\text{a}}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>(Hint: Determine [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right][\/latex] at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized \u00d7 100, or [latex]\\dfrac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right]}{{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}_{\\text{initial}}}\\times 100[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q26536\">Show Solution<\/span><\/p>\n<div id=\"q26536\" class=\"hidden-answer\" style=\"display: none\">\n<p>percent ionization = 1.3%<\/p><\/div>\n<\/div>\n<\/div>\n<p>Example 7 shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 7:\u00a0Equilibrium Concentrations in a Solution of a Weak Base<\/h3>\n<p>Find the concentration of hydroxide ion in a 0.25-<em>M<\/em> solution of trimethylamine, a weak base:<\/p>\n<p style=\"text-align: center;\">[latex]{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\qquad{K}_{\\text{b}}=6.3\\times {10}^{-5}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q636372\">Show Solution<\/span><\/p>\n<div id=\"q636372\" class=\"hidden-answer\" style=\"display: none\">\n<p>The ICE table for this system is<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213748\/CNX_Chem_14_03_ICETable4_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of \u201c( C H subscript 3 ) subscript 3 N plus sign H subscript 2 O equilibrium arrow ( C H subscript 3 ) subscript 3 N H superscript positive sign plus sign O H superscript positive sign.\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.25, negative x, 0.25 plus sign negative x. The second column is blank in all three rows. The third column has the following: 0, x, 0 plus x. The fourth column has the following: approximately 0, x, and approximately 0 plus x.\" width=\"697\" height=\"183\" \/><\/p>\n<h4><\/h4>\n<p>At equilibrium:<\/p>\n<p style=\"text-align: center;\"><span style=\"text-align: center;\">[latex]{K}_{\\text{b}}=\\dfrac{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\right]}=\\dfrac{\\left(x\\right)\\left(x\\right)}{0.25-x}=6.3\\times {10}^{-5}[\/latex]<\/span><\/p>\n<p>If we assume that <em>x<\/em> is small relative to 0.25, then we can replace (0.25 \u2212 <em>x<\/em>) in the preceding equation with 0.25. Solving the simplified equation gives\u00a0[latex]x=4.0\\times {10}^{-3}[\/latex].<\/p>\n<p>This change is less than 5% of the initial concentration (0.25), so the assumption is justified. Recall that, for this computation, <em>x<\/em> is equal to the equilibrium concentration of <em>hydroxide ion<\/em> in the solution (see earlier tabulation):<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{OH}}^{-}\\right]=\\text{~}0+x=x=4.0\\times {10}^{-3}M[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]=4.0\\times {10}^{-3}M[\/latex]<\/p>\n<p>Then calculate pOH as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{pOH}=-\\text{log}\\left(4.0\\times {10}^{-3}\\right)=2.40[\/latex]<\/p>\n<p>Using the relation introduced in the previous section of this chapter, we get<\/p>\n<p style=\"text-align: center;\">[latex]\\text{pH}+\\text{pOH}=\\text{p}{K}_{\\text{w}}=14.00[\/latex]<\/p>\n<p>which\u00a0permits the computation of pH:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{pH}=14.00-\\text{pOH}=14.00 - 2.40=11.60[\/latex]<\/p>\n<h4>Step 3: Check the work<\/h4>\n<p>A check of our arithmetic shows that <em>K<\/em><sub>b<\/sub> = 6.3 \u00d7 10<sup>\u22125<\/sup>.<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idm83886080\">Check Your Learning<\/h4>\n<p>Calculate the hydroxide ion concentration and the percent ionization of a 0.0325-<em data-effect=\"italics\">M<\/em>\u00a0solution of ammonia, a weak base with a\u00a0<em data-effect=\"italics\">K<\/em><sub>b<\/sub>\u00a0of 1.76 \u00d7 10<sup>\u22125<\/sup>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q145683\">Show Solution<\/span><\/p>\n<div id=\"q145683\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>7.56 \u00d7 10<sup>\u22124<\/sup><em>M<\/em><\/li>\n<li>2.33%<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>In some cases, the strength of the weak acid or base and its formal (initial) concentration result in an appreciable ionization. Though the ICE strategy remains effective for these systems, the algebra is a bit more involved because the simplifying assumption that\u00a0<em>x<\/em>\u00a0is negligible can not be made. Calculations of this sort are demonstrated in Example 8 below.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 8: Calculating Equilibrium Concentrations without simplifying assumptions<\/h3>\n<p>Sodium bisulfate, NaHSO<sub>4<\/sub>, is used in some household cleansers because it contains the [latex]{\\text{HSO}}_{4}^{-}[\/latex] ion, a weak acid. What is the pH of a 0.50-<em>M<\/em> solution of [latex]{\\text{HSO}}_{4}^{-}?[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{HSO}}_{4}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{SO}}_{4}^{2-}\\left(aq\\right)\\qquad{K}_{\\text{a}}=1.2\\times {10}^{-2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q23029\">Show Solution<\/span><\/p>\n<div id=\"q23029\" class=\"hidden-answer\" style=\"display: none\">\n<p>The ICE table for this system is:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213751\/CNX_Chem_14_03_ICETable5_img.jpg\" alt=\"This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium ( M ). The second column has the header of \u201cH S O subscript 4 superscript negative sign plus sign H subscript 2 O equilibrium sign H subscript 3 O superscript positive sign plus sign S O subscript 4 superscript 2 superscript negative sign.\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.50, negative x, 0.50 plus sign negative x equals 0.50 minus x. The second column is blank for all three rows. The third column has the following: approximately 0, x, 0 plus sign x equals x. The fourth column has the following: 0, x, 0 plus sign x equals x.\" width=\"700\" height=\"224\" \/><\/p>\n<p>As we begin solving for <em>x<\/em>, we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of <em>x<\/em>. At equilibrium:<\/p>\n<p style=\"text-align: center;\"><span style=\"text-align: center;\">[latex]{K}_{\\text{a}}=1.2\\times {10}^{-2}=\\dfrac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{SO}}_{4}^{2-}\\right]}{\\left[{\\text{HSO}}_{4}^{-}\\right]}=\\dfrac{\\left(x\\right)\\left(x\\right)}{0.50-x}[\/latex]<\/span><\/p>\n<p>If we assume that <em>x<\/em> is small and approximate (0.50 \u2212 <em>x<\/em>) as 0.50, we find [latex]x=7.7\\times {10}^{-2}M[\/latex].\u00a0When we check the assumption, we calculate:<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{x}{{\\left[{\\text{HSO}}_{4}^{-}\\right]}_{\\text{i}}}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{x}{0.50}=\\dfrac{7.7\\times {10}^{-2}}{0.50}=0.15\\left(15\\%\\right)[\/latex]<\/p>\n<p>The value of <em>x<\/em> is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find <em>x<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=1.2\\times {10}^{-2}=\\dfrac{\\left(x\\right)\\left(x\\right)}{0.50-x}[\/latex]<\/p>\n<p>Rearranging this equation yields<\/p>\n<p style=\"text-align: center;\">[latex]6.0\\times {10}^{-3}-1.2\\times {10}^{-2}x={x}^{2}[\/latex]<\/p>\n<p>Writing the equation in quadratic form gives<\/p>\n<p style=\"text-align: center;\">[latex]{x}^{2}+1.2\\times {10}^{-2}x - 6.0\\times {10}^{-3}=0[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Solving for the two roots of this quadratic equation results in a negative value that may be discarded as physically irrelevant and a positive value equal to\u00a0<em data-effect=\"italics\">x<\/em>. As defined in the ICE table,\u00a0<em data-effect=\"italics\">x<\/em>\u00a0is equal to the hydronium concentration.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=~0+x=0+7.2\\times {10}^{-2}M=7.2\\times {10}^{-2}M[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\text{pH}=-\\text{log}\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=-\\text{log}(7.2\\times {10}^{-2})=1.14[\/latex]<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idm108532480\">Check Your Learning<\/h4>\n<ol>\n<li>Calculate the pH in a 0.010-<em>M<\/em> solution of caffeine, a weak base: [latex]{\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{8}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{O}}_{2}{\\text{H}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right){K}_{\\text{b}}=2.5\\times {10}^{-4}[\/latex]<\/li>\n<\/ol>\n<p>(Hint: It will be necessary to convert [OH<sup>\u2212<\/sup>] to [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] or pOH to pH toward the end of the calculation.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q774358\">Show Solution<\/span><\/p>\n<div id=\"q774358\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\text{pH }11.16[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<p>Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Acids<\/a>\u00a0and\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-bases-2\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Bases<\/a>.<\/p>\n<h4>Solution 1<\/h4>\n<p>0.0092 <em>M<\/em> HClO, a weak acid<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q936215\">Show Solution<\/span><\/p>\n<div id=\"q936215\" class=\"hidden-answer\" style=\"display: none\">\n<p>The reaction is: [latex]\\text{HClO}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{ClO}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>The equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{ClO}}^{-}\\right]}{\\left[\\text{HClO}\\right]}=3.5\\times {10}^{-8}[\/latex]<\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[HClO]<\/th>\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\n<th>[ClO<sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.0092<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.0092\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0092 \u2212 <em>x<\/em>) \u2248 0.0092 gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{ClO}}^{-}\\right]}{\\left[\\text{HClO}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.0092-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.0092}=3.5\\times {10}^{-8}[\/latex]<\/p>\n<p>Solving for <em>x<\/em> gives 1.79 \u00d7 10<sup>\u22125<\/sup><em>M<\/em>. This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:<\/p>\n<ul>\n<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = [ClO] = 1.8 \u00d7 10<sup>\u22125<\/sup><em>M<\/em><\/li>\n<li>[HClO] = 0.0092 \u2212 1.79 \u00d7 10<sup>\u22125<\/sup> = 0.00918 = 0.00092 <em>M<\/em><\/li>\n<li>[latex]\\left[{\\text{OH}}^{-}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{1.0\\times {10}^{-14}}{1.79\\times {10}^{-5}}=5.6\\times {10}^{-10}M[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<h4>Solution 2<\/h4>\n<p>0.0784 <em>M<\/em> C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>, a weak base<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q996941\">Show Solution<\/span><\/p>\n<div id=\"q996941\" class=\"hidden-answer\" style=\"display: none\">\n<p>The reaction is\u00a0[latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>The equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\right]}=4.6\\times {10}^{-10}[\/latex]<\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>]<\/th>\n<th>[C<sub>5<\/sub>H<sub>5<\/sub>NH<sub>3<\/sub><sup>+<\/sup>]<\/th>\n<th>[OH<sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.0784<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.0784\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0784 \u2212 <em>x<\/em>) \u2248 0.0784 gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.0784-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.0784}=4.6\\times {10}^{-10}[\/latex]<\/p>\n<p>Solving for <em>x<\/em> gives 6.01 \u00d7 10<sup>\u22126<\/sup><em>M<\/em>. This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:<\/p>\n<ul>\n<li>[latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right][\/latex] = [OH<sup>\u2212<\/sup>] = 6.0\u00a0\u00d7 10<sup>\u22126<\/sup><em>M<\/em><\/li>\n<li>[C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>] = 0.0784 \u2212 6.01 \u00d7 10<sup>\u22126<\/sup> = 0.007839 = 0.00784<\/li>\n<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{OH}}^{-}\\right]}=\\frac{1.0\\times {10}^{-14}}{6.01\\times {10}^{-6}}=1.7\\times {10}^{-9}M[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<h4>Solution 3<\/h4>\n<p>0.0810 <em>M<\/em> HCN, a weak acid<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q342499\">Show Solution<\/span><\/p>\n<div id=\"q342499\" class=\"hidden-answer\" style=\"display: none\">\n<p>The reaction is [latex]\\text{HCN}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CN}}^{-}\\left(aq\\right)[\/latex].<\/p>\n<p>The equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CN}}^{-}\\right]}{\\left[\\text{HCN}\\right]}=4\\times {10}^{-10}[\/latex]<\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[HClO]<\/th>\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\n<th>[CN<sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.0810<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.0810\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.0810 \u2212 <em>x<\/em>) \u2248 0.0810 gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CN}}^{-}\\right]}{\\left[\\text{HCN}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.0810-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.0810}=4\\times {10}^{-10}[\/latex]<\/p>\n<p>Solving for <em>x<\/em> gives 5.69 \u00d7 10<sup>\u22126<\/sup><em>M<\/em>. This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:<\/p>\n<ul>\n<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = [CN<sup>\u2212<\/sup>] = 5.7 \u00d7 10<sup>\u22126<\/sup><em>M<\/em><\/li>\n<li>[HCN] = 0.0810 \u2212 5.69 \u00d7 10<sup>\u22126<\/sup> = 0.08099 = 0.0810 <em>M<\/em><\/li>\n<li>[latex]\\left[{\\text{OH}}^{-}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{1.0\\times {10}^{-14}}{5.69\\times {10}^{-6}}=1.8\\times {10}^{-9}M[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<h4>Solution 4<\/h4>\n<p>0.11 <em>M<\/em> (CH<sub>3<\/sub>)<sub>3<\/sub>N, a weak base<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q210337\">Show Solution<\/span><\/p>\n<div id=\"q210337\" class=\"hidden-answer\" style=\"display: none\">\n<p>The reaction is\u00a0[latex]{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>The equilibrium expression is\u00a0[latex]{K}_{\\text{b}}=\\frac{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\right]}=7.4\\times {10}^{-5}[\/latex]<\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[(CH<sub>3<\/sub>)<sub>3<\/sub>N]<\/th>\n<th>[(CH<sub>3<\/sub>)<sub>3<\/sub>NH<sup>+<\/sup>]<\/th>\n<th>[OH<sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.11<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.11\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.11 \u2212 <em>x<\/em>) \u2248 0.11 gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.11-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.11}=7.4\\times {10}^{-5}[\/latex]<\/p>\n<p>Solving for <em>x<\/em> gives 2.85 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>. This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:<\/p>\n<ul>\n<li>[latex]\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\right][\/latex] = [OH<sup>\u2212<\/sup>] = 2.9 \u00d7 10<sup>\u22123<\/sup><em>M<\/em><\/li>\n<li>[(CH<sub>3<\/sub>)<sub>3<\/sub>N] = 0.11 \u2212 2.85 \u00d7 10<sup>\u22123<\/sup> = 0.107 = 0.11 <em>M<\/em><\/li>\n<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{OH}}^{-}\\right]}=\\frac{1.0\\times {10}^{-14}}{2.85\\times {10}^{-3}}=3.5\\times {10}^{-12}M[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<h4>Solution 5<\/h4>\n<p>0.120 <em>M<\/em> [latex]\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{2+}[\/latex] a weak acid, <em>K<\/em><sub>a<\/sub> = 1.6 \u00d7 10<sup>\u22127<\/sup><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q874526\">Show Solution<\/span><\/p>\n<div id=\"q874526\" class=\"hidden-answer\" style=\"display: none\">\n<p>The reaction is\u00a0[latex]\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{2+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons \\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{+}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)[\/latex]<\/p>\n<p>The equilibrium expression is\u00a0[latex]{K}_{\\text{a}}=\\frac{\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{+}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{2+}\\right]}=1.6\\times {10}^{-7}[\/latex]<\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[Fe(H<sub>2<\/sub>O)<sub>6<\/sub><sup>2+<\/sup>]<\/th>\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\n<th>[Fe(H<sub>2<\/sub>O)<sub>5<\/sub>(OH)<sup>+<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.120<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.120\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium expression and making the assumption that (0.120 \u2212 <em>x<\/em>) \u2248 0.120 gives:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{+}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{2+}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.120-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.120}=1.6\\times {10}^{-7}[\/latex]<\/p>\n<p>Solving for <em>x<\/em> gives 1.39 \u00d7 10<sup>\u22124<\/sup><em>M<\/em>. This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:<\/p>\n<ul>\n<li style=\"text-align: left;\">[latex]\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{+}\\right][\/latex] = [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 1.4 \u00d7 10<sup>\u22124<\/sup><em>M<\/em><\/li>\n<li style=\"text-align: left;\">[latex]\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{2+}\\right][\/latex] = 0.120 \u2212 1.39 \u00d7 10<sup>\u22124<\/sup> = 0.1199 = 0.120 <em>M<\/em><\/li>\n<li style=\"text-align: left;\">[latex]\\left[{\\text{OH}}^{-}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{1.0\\times {10}^{-14}}{1.39\\times {10}^{-4}}=7.2\\times {10}^{-11}M[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<h2>Effect of Molecular Structure on Acid-Base Strength<\/h2>\n<p>In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 7A, the order of increasing acidity is HF &lt; HCl &lt; HBr &lt; HI. Likewise, for group 6A, the order of increasing acid strength is H<sub>2<\/sub>O &lt; H<sub>2<\/sub>S &lt; H<sub>2<\/sub>Se &lt; H<sub>2<\/sub>Te.<\/p>\n<p>Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is CH<sub>4<\/sub> &lt; NH<sub>3<\/sub> &lt; H<sub>2<\/sub>O &lt; HF; across the third row, it is SiH<sub>4<\/sub> &lt; PH<sub>3<\/sub> &lt; H<sub>2<\/sub>S &lt; HCl (see Figure\u00a07).<\/p>\n<div style=\"width: 710px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213752\/CNX_Chem_14_03_AcidpH.jpg\" alt=\"This diagram has two rows and four columns. Red arrows point left across the bottom of the figure and down at the right side and are labeled \u201cIncreasing acid strength.\u201d Blue arrows point left across the bottom and up at the right side of the figure and are labeled \u201cIncreasing base strength.\u201d The first column is labeled 14 at the top and two white squares are beneath it. The first has the number 6 in the upper left corner and the formula C H subscript 4 in the center along with designation Neither acid nor base. The second square contains the number 14 in the upper left corner, the formula C H subscript 4 at the center and the designation Neither acid nor base. The second column is labeled 15 at the top and two blue squares are beneath it. The first has the number 7 in the upper left corner and the formula N H subscript 3 in the center along with the designation Weak base and K subscript b equals 1.8 times 10 superscript negative 5. The second square contains the number 15 in the upper left corner, the formula P H subscript 3 at the center and the designation Very weak base and K subscript b equals 4 times 10 superscript negative 28. The third column is labeled 16 at the top and two squares are beneath it. The first is shaded tan and has the number 8 in the upper left corner and the formula H subscript 2 O in the center along with the designation neutral. The second square is shaded pink, contains the number 16 in the upper left corner, the formula H subscript 2 S at the center and the designation Weak acid and K subscript a equals 9.5 times 10 superscript negative 8. The fourth column is labeled 17 at the top and two squares are beneath it. The first is shaded pink, has the number 9 in the upper left corner and the formula H F in the center along with the designation Weak acid and K subscript a equals 6.8 times 10 superscript negative 4. The second square is shaded a deeper pink, contains the number 17 in the upper left corner, the formula H C l at the center, and the designation Strong acid.\" width=\"700\" height=\"444\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a07. As you move from left to right and down the periodic table, the acid strength increases. As you move from right to left and up, the base strength increases.<\/p>\n<\/div>\n<h2>Ternary Acids and Bases<\/h2>\n<p>Compounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric, depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group. Such compounds have the general formula O<sub>n<\/sub>E(OH)<sub>m<\/sub>, and include sulfuric acid, O<sub>2<\/sub>S(OH)<sub>2<\/sub>, sulfurous acid, OS(OH)<sub>2<\/sub>, nitric acid, O<sub>2<\/sub>NOH, perchloric acid, O<sub>3<\/sub>ClOH, aluminum hydroxide, Al(OH)<sub>3<\/sub>, calcium hydroxide, Ca(OH)<sub>2<\/sub>, and potassium hydroxide, KOH:<br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213754\/CNX_Chem_14_03_OHbonds_img.jpg\" alt=\"A diagram is shown that includes a central atom designated with the letter E. Single bonds extend above, below, left, and right of the E. An O atom is bonded to the right of the E, and an arrow points to the bond labeling it, \u201cBond a.\u201d An H atom is single bonded to the right of the O atom. An arrow pointing to this bond connects it to the label, \u201cBond b.\u201d\" \/><\/p>\n<p>If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond <em>a<\/em> between the element and oxygen is more readily broken than bond <em>b<\/em> between oxygen and hydrogen. Hence bond <em>a<\/em> is ionic, hydroxide ions are released to the solution, and the material behaves as a base\u2014this is the case with Ca(OH)<sub>2<\/sub> and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds.<\/p>\n<p>If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond <em>a<\/em> relatively strongly covalent. The oxygen-hydrogen bond, bond <em>b<\/em>, is thereby weakened because electrons are displaced toward E. Bond <em>b<\/em> is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic \u2212OH groups that are called <strong>oxyacids<\/strong>.<\/p>\n<p>Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H<sub>2<\/sub>SO<sub>4<\/sub>, or O<sub>2<\/sub>S(OH)<sub>2<\/sub> (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H<sub>2<\/sub>SO<sub>3<\/sub>, or OS(OH)<sub>2<\/sub> (with a sulfur oxidation number of +4). Likewise nitric acid, HNO<sub>3<\/sub>, or O<sub>2<\/sub>NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO<sub>2<\/sub>, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure\u00a08).<\/p>\n<div style=\"width: 890px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213755\/CNX_Chem_14_03_Oxyacid.jpg\" alt=\"A diagram is shown that includes four structural formulas for acids. A red, right pointing arrow is placed beneath the structures which is labeled \u201cIncreasing acid strength.\u201d At the top left, the structure of Nitrous acid is provided. It includes an H atom to which an O atom with two unshared electron pairs is connected with a single bond to the right. A single bond extends to the right and slightly below to a N atom with one unshared electron pair. A double bond extends up and to the right from this N atom to an O atom which has two unshared electron pairs. To the upper right is a structure for Nitric acid. This structure differs from the previous structure in that the N atom is directly to the right of the first O atom and a second O atom with three unshared electron pairs is connected with a single bond below and to the right of the N atom which has no unshared electron pairs. At the lower left, an O atom with two unshared electron pairs is double bonded to its right to an S atom with a single unshared electron pair. An O atom with two unshared electron pairs is bonded above and an H atom is single bonded to this O atom. To the right of the S atom is a single bond to another O atom with two unshared electron pairs to which an H atom is single bonded. This structure is labeled \u201cSulfurous acid.\u201d A similar structure which is labeled \u201cSulfuric acid\u201d is placed in the lower right region of the figure. This structure differs in that an H atom is single bonded to the left of the first O atom, leaving it with two unshared electron pairs and a fourth O atom with two unshared electron pairs is double bonded beneath the S atom, leaving it with no unshared electron pairs.\" width=\"880\" height=\"500\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a08. As the oxidation number of the central atom E increases, the acidity also increases.<\/p>\n<\/div>\n<p>Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate Al(H<sub>2<\/sub>O)<sub>3<\/sub>(OH)<sub>3<\/sub>, is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, Al(H<sub>2<\/sub>O)<sub>3<\/sub>(OH)<sub>3<\/sub>, is converted into the soluble ion, [latex]{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{2}{\\left(\\text{OH}\\right)}_{4}\\right]}^{-}[\/latex], by reaction with hydroxide ion:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{3}{\\left(\\text{OH}\\right)}_{3}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\rightleftharpoons {\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{2}{\\left(\\text{OH}\\right)}_{4}\\right]}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>In this reaction, a proton is transferred from one of the aluminum-bound H<sub>2<\/sub>O molecules to a hydroxide ion in solution. The Al(H<sub>2<\/sub>O)<sub>3<\/sub>(OH)<sub>3<\/sub> compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion [latex]{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}\\right]}^{3+}[\/latex] by reaction with hydronium ion:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{3H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{3}{\\left(\\text{OH}\\right)}_{3}\\left(aq\\right)\\rightleftharpoons \\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}^{3+}\\left(aq\\right)+{\\text{3H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<p>In this case, protons are transferred from hydronium ions in solution to Al(H<sub>2<\/sub>O)<sub>3<\/sub>(OH)<sub>3<\/sub>, and the compound functions as a base.<\/p>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>Propionic acid, C<sub>2<\/sub>H<sub>5<\/sub>CO<sub>2<\/sub>H (<em>K<\/em><sub>a<\/sub> = 1.34 \u00d7 10<sup>\u22125<\/sup>), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a 0.698-<em>M<\/em> solution of C<sub>2<\/sub>H<sub>5<\/sub>CO<sub>2<\/sub>H?<\/li>\n<li>White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g\/cm<sup>3<\/sup>, what is the pH?<\/li>\n<li>The ionization constant of lactic acid, CH<sub>3<\/sub>CH(OH)CO<sub>2<\/sub>H, an acid found in the blood after strenuous exercise, is 1.36 \u00d7\u00a010<sup>\u22124<\/sup>. If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?<\/li>\n<li>Nicotine, C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub>, is a base that will accept two protons (<em>K<\/em><sub>1<\/sub> = 7 \u00d7\u00a010<sup>\u22127<\/sup>, <em>K<\/em><sub>2<\/sub> = 1.4 \u00d7 10<sup>\u221211<\/sup>). What is the concentration of each species present in a 0.050-<em>M<\/em> solution of nicotine?<\/li>\n<li>The pH of a 0.20-<em>M<\/em> solution of HF is 1.92. Determine <em>K<\/em><sub>a<\/sub> for HF from these data.<\/li>\n<li>The pH of a 0.15-<em>M<\/em> solution of [latex]{\\text{HSO}}_{4}^{-}[\/latex] is 1.43. Determine <em>K<\/em><sub>a<\/sub> for [latex]{\\text{HSO}}_{4}^{-}[\/latex] from these data.<\/li>\n<li>The pH of a 0.10-<em>M<\/em> solution of caffeine is 11.16. Determine <em>K<\/em><sub>b<\/sub> for caffeine from these data:<\/li>\n<li>The pH of a solution of household ammonia, a 0.950 M solution of NH<sub>3,<\/sub> is 11.612. Determine <em>K<\/em><sub>b<\/sub> for NH<sub>3<\/sub> from these data.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q864039\">Show Solution to Question 2<\/span><\/p>\n<div id=\"q864039\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, find the mass of acetic acid. <em>d<\/em> = 1.007 g\/cm<sup>3<\/sup>. Take 1.0 L of solution to have the quantities on a mole basis. Then, since 1000 cm<sup>3<\/sup> = 1.0 L, 1000 cm<sup>3<\/sup>\u00a0\u00d7 1.007 g\/cm<sup>3<\/sup> = 1007 g in 1.0 L. Then, 5.00% of this is the mass of acetic acid:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Mass (acetic acid)}=\\text{1007 g}\\times \\frac{5.0%}{100%}=\\text{50.35 g}[\/latex]<\/p>\n<p>Now calculate the number of moles of acetic acid present. The molar mass of acetic acid is 60.053 g\/mol:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{mol acetic acid}=\\frac{50.35\\cancel{\\text{g}}}{60.053\\cancel{\\text{g}}{\\text{mol}}^{-1}}=\\text{0.838 mol}[\/latex]<\/p>\n<p>From the moles of acetic acid and <em>K<\/em><sub>a<\/sub>, calculate [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]:[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=1.8\\times {10}^{-5}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}[\/latex]<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub>H]<\/th>\n<th>[H<sub>3<\/sub>O<sup>+<\/sup>]<\/th>\n<th>[CH<sub>3<\/sub>CO<sub>2<\/sub><sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.838<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.838 \u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substitution gives:\u00a0[latex]{K}_{\\text{a}}=1.8\\times {10}^{-5}=\\frac{{x}^{\\text{2}}}{0.838-x}[\/latex]<\/p>\n<p>Drop <em>x<\/em> because it is small in comparison with 0.838 <em>M<\/em>.<\/p>\n<ul>\n<li><em>x<\/em><sup>2<\/sup> = 0.838(1.8 \u00d7 10<sup>\u22125<\/sup>) = 1.508 \u00d7 10<sup>\u22125<\/sup> = 3.88 \u00d7 10<sup>\u22123<\/sup> = 2.41<\/li>\n<li>pH = \u2212log(3.88 \u00d7 10<sup>\u22123<\/sup>) = 2.41<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q287147\">Show Solution to Question 4<\/span><\/p>\n<div id=\"q287147\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{ }{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\left({K}_{\\text{b1}}=7\\times {10}^{-7}\\right)\\\\ {\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\left({K}_{\\text{b2}}=1.4\\times {10}^{-11}\\right)\\end{array}[\/latex]<\/p>\n<p>First set up a concentration table:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub>]<\/th>\n<th>[C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub>H<sup>+<\/sup>]<\/th>\n<th>[OH<sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.050<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Change (<em>M<\/em>)<\/th>\n<td>\u2212<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.050 \u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting the equilibrium concentrations into the equilibrium equation and making the assumption that (0.050 \u2212 <em>x<\/em>) = 0.050, we get:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{K}_{\\text{b1}}&=&\\frac{\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}\\right]}=7\\times {10}^{-7}\\\\{}&=&\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.050-x\\right)}=\\frac{{x}^{\\text{2+}}}{0.050}=7\\times {10}^{-7}\\end{array}[\/latex]<\/p>\n<p>Solving for <em>x<\/em> gives 1.87 \u00d7 10<sup>\u22124<\/sup> = 2 \u00d7 10<sup>\u22124<\/sup><em>M<\/em> = [OH<sup>\u2212<\/sup>]<\/p>\n<p>Because <em>x<\/em> is less than 5% of 0.050 and [OH<sup>\u2212<\/sup>] is greater than 4.5 \u00d7 10<sup>\u22127\u00a0<\/sup><em>M<\/em>, our customary assumptions are justified. We can calculate<\/p>\n<ul>\n<li>[C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub>] = 0.050 \u2212 <em>x<\/em> = 0.050 \u2212 2 \u00d7 10<sup>\u22124<\/sup> = 0.048 <em>M<\/em><\/li>\n<li>[OH<sup>\u2212<\/sup>] = [latex]\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\right][\/latex] = <em>x<\/em> = 2 \u00d7 10<sup>\u22124<\/sup><em>M<\/em><\/li>\n<\/ul>\n<p>Now calculate the concentration of [latex]{\\text{C}}_{10}{\\text{H}}_{2}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}[\/latex] in a solution with [OH<sup>\u2212<\/sup>] and [latex]\\left[{\\text{C}}_{10}{\\text{H}}_{2}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}\\right][\/latex] equal to 2 \u00d7 10<sup>\u22124<\/sup><em>M<\/em>. The equilibrium between these species is [latex]{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}_{2}^{\\text{2+}}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]. We know [C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub>H<sup>+<\/sup>] and [OH<sup>\u2212<\/sup>], so we can calculate the concentration of [latex]{\\text{C}}_{10}{\\text{H}}_{2}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}[\/latex] from the equilibrium expression:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}{}{K}_{\\text{b2}}&=&\\frac{\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\right]}=1.4\\times {10}^{-11}\\\\{}&=&\\frac{\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}\\right]\\left[2\\times {10}^{-4}\\right]}{\\left[2\\times {10}^{-4}\\right]}\\\\ \\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}\\right]&=&1.4\\times {10}^{-11}M\\end{array}[\/latex]<\/p>\n<p>The concentration of OH<sup>\u2212<\/sup> produced in this ionization is equal to the concentration of [latex]{\\text{C}}_{10}{\\text{H}}_{2}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}[\/latex], 1.4 \u00d7 10<sup>\u221211<\/sup><em>M<\/em>, which is much smaller than the 2 \u00d710<sup>\u22124<\/sup><em>M<\/em> produced in the first ionization; therefore, we are justified in neglecting the OH<sup>\u2212<\/sup> formed from [latex]{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}[\/latex].<\/p>\n<p>We can now calculate the concentration of H<sub>2<\/sub>O<sup>+<\/sup> present from the ionization of water:<\/p>\n<p style=\"text-align: center;\"><em>K<\/em><sub>a<\/sub> = 1 \u00d7 10<sup>\u221214<\/sup> = [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] [OH<sup>\u2212<\/sup>]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{1\\times {10}^{-14}}{\\left[{\\text{OH}}^{-}\\right]}=\\frac{1\\times {10}^{-14}}{1.9\\times {10}^{-4}}=5.3\\times {10}^{-11}M[\/latex]<\/p>\n<p>We can now summarize the concentrations of all species in solution as follows:<\/p>\n<ul>\n<li>[C<sub>10<\/sub>H<sub>14<\/sub>N<sub>2<\/sub>] = 0.049 <em>M<\/em><\/li>\n<li>[latex]\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}^{+}\\right][\/latex] = 1.9 \u00d7 10<sup>\u22124\u00a0<\/sup><em>M<\/em><\/li>\n<li>[latex]\\left[{\\text{C}}_{10}{\\text{H}}_{14}{\\text{N}}_{2}{\\text{H}}_{2}^{2+}\\right]=1.4\\times {10}^{-11}M[\/latex]<\/li>\n<li>[OH<sup>\u2212<\/sup>] = 1.9 \u00d7 10<sup>\u22124<\/sup><em>M<\/em><\/li>\n<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 5.3 \u00d7 10<sup>\u221211<\/sup><em>M<\/em><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q969367\">Show Solution to Question 6<\/span><\/p>\n<div id=\"q969367\" class=\"hidden-answer\" style=\"display: none\">\n<p>The reaction is [latex]{\\text{HSO}}_{4}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{SO}}_{2}^{2-}\\left(aq\\right)[\/latex].<\/p>\n<p>The concentrations at equilibrium are [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 10<sup>\u2212pH<\/sup> = 10<sup>\u22121.43<\/sup> = 0.0372 <em>M<\/em><\/p>\n<ul>\n<li>[HF] = 0.15 \u2212 0.0372 <em>M<\/em> = 0.113 <em>M<\/em><\/li>\n<li>[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{SO}}_{4}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{HSO}}_{4}^{-}\\right]}=\\frac{\\left(0.0372\\right)\\left(0.0372\\right)}{\\left(0.113\\right)}=1.2\\times {10}^{-2}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q864327\">Show Solution to Question 8<\/span><\/p>\n<div id=\"q864327\" class=\"hidden-answer\" style=\"display: none\">\n<p>The reaction is [latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex].<\/p>\n<p>The pOH can be determined from pOH = 14.000 \u2212 pH = 14.000 \u2212 11.612 = 2.388. Therefore, the concentrations at equilibrium are [latex]\\left[{\\text{NH}}_{4}^{+}\\right][\/latex] = [OH<sup>\u2212<\/sup>] = 10<sup>\u2212pOH<\/sup> = 10<sup>\u22122.388<\/sup> = 0.004093 <em>M<\/em><\/p>\n<ul>\n<li>[NH<sub>3<\/sub>] = 0.950 \u2212 0.004093 = 0.9459 <em>M<\/em><\/li>\n<li>[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(0.004093\\right)\\left(0.004093\\right)}{\\left(0.9459\\right)}=1.77\\times {10}^{-5}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=5557241&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=ZNo6gfCAgWE&amp;video_target=tpm-plugin-965fa6qw-ZNo6gfCAgWE\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/AcidsAndBasesChemistry_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Acids and Bases Chemistry &#8211; Basic Introduction&#8221; here (opens in new window)<\/a>.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>The strengths of Br\u00f8nsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion. The strengths of the binary acids increase from left to right across a period of the periodic table (CH<sub>4<\/sub> &lt; NH<sub>3<\/sub> &lt; H<sub>2<\/sub>O &lt; HF), and they increase down a group (HF &lt; HCl &lt; HBr &lt; HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H<sub>2<\/sub>SO<sub>3<\/sub> &lt; H<sub>2<\/sub>SO<sub>4<\/sub>). The strengths of oxyacids also increase as the electronegativity of the central element increases [H<sub>2<\/sub>SeO<sub>4<\/sub> &lt; H<sub>2<\/sub>SO<sub>4<\/sub>].<\/p>\n<h4>Key Equations<\/h4>\n<ul>\n<li>[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{A}}^{-}\\right]}{\\left[\\text{HA}\\right]}[\/latex]<\/li>\n<li>[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{HB}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[\\text{B}\\right]}[\/latex]<\/li>\n<li><em>K<\/em><sub>a<\/sub> \u00d7 <em>K<\/em><sub>b<\/sub> = 1.0 \u00d7 10<sup>\u221214<\/sup> = <em>K<\/em><sub>w<\/sub><\/li>\n<li>[latex]\\text{Percent ionization}=\\frac{{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}_{\\text{eq}}}{{\\left[\\text{HA}\\right]}_{0}}\\times 100[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>The odor of vinegar is due to the presence of acetic acid, CH<sub>3<\/sub>CO<sub>2<\/sub>H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-<em>M<\/em> aqueous solution of this acid.<\/li>\n<li>Household ammonia is a solution of the weak base NH<sub>3<\/sub> in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1-<em>M<\/em> aqueous solution of this base.<\/li>\n<li>Which base, CH<sub>3<\/sub>NH<sub>2<\/sub> or (CH<sub>3<\/sub>)<sub>2<\/sub>NH, is the strongest base? Which conjugate acid, [latex]{\\left({\\text{CH}}_{3}\\right)}_{2}{\\text{NH}}_{2}^{+}[\/latex] or (CH<sub>3<\/sub>)<sub>2<\/sub>NH, is the strongest acid?<\/li>\n<li>Which is the stronger acid, [latex]{\\text{NH}}_{4}^{+}[\/latex] or HBrO?<\/li>\n<li>Which is the stronger base, (CH<sub>3<\/sub>)<sub>3<\/sub>N or [latex]{\\text{H}}_{2}{\\text{BO}}_{3}^{-}?[\/latex]<\/li>\n<li>Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>H<sub>2<\/sub>O or HF<\/li>\n<li>B(OH)<sub>3<\/sub> or Al(OH)<sub>3<\/sub><\/li>\n<li>[latex]{\\text{HSO}}_{3}^{-}[\/latex] or [latex]{\\text{HSO}}_{4}^{-}[\/latex]<\/li>\n<li>NH<sub>3<\/sub> or H<sub>2<\/sub>S<\/li>\n<li>H<sub>2<\/sub>O or H<sub>2<\/sub>Te<\/li>\n<\/ol>\n<\/li>\n<li>Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\text{HSO}}_{4}^{-}[\/latex] or [latex]{\\text{HSeO}}_{4}^{-}[\/latex]<\/li>\n<li>NH<sub>3<\/sub> or H<sub>2<\/sub>O<\/li>\n<li>PH<sub>3<\/sub> or HI<\/li>\n<li>NH<sub>3<\/sub> or PH<sub>3<\/sub><\/li>\n<li>H<sub>2<\/sub>S or HBr<\/li>\n<\/ol>\n<\/li>\n<li>Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>acidity: HCl, HBr, HI<\/li>\n<li>basicity: H<sub>2<\/sub>O, OH<sup>\u2212<\/sup>, H<sup>\u2212<\/sup>, Cl<sup>\u2212<\/sup><\/li>\n<li>basicity: Mg(OH)<sub>2<\/sub>, Si(OH)<sub>4<\/sub>, ClO<sub>3<\/sub>(OH) (Hint: Formula could also be written as HClO<sub>4<\/sub>).<\/li>\n<li>acidity: HF, H<sub>2<\/sub>O, NH<sub>3<\/sub>, CH<sub>4<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>acidity: NaHSO<sub>3<\/sub>, NaHSeO<sub>3<\/sub>, NaHSO<sub>4<\/sub><\/li>\n<li>basicity: [latex]{\\text{BrO}}_{2}^{-}[\/latex], [latex]{\\text{ClO}}_{2}^{-}[\/latex], [latex]{\\text{IO}}_{2}^{-}[\/latex]<\/li>\n<li>acidity: HOCl, HOBr, HOI<\/li>\n<li>acidity: HOCl, HOClO, HOClO<sub>2<\/sub>, HOClO<sub>3<\/sub><\/li>\n<li>basicity: [latex]{\\text{NH}}_{2}^{-}[\/latex], HS<sup>\u2212<\/sup>, HTe<sup>\u2212<\/sup>, [latex]{\\text{PH}}_{2}^{-}[\/latex]<\/li>\n<li>basicity: BrO<sup>\u2212<\/sup>, [latex]{\\text{BrO}}_{2}^{-}[\/latex], [latex]{\\text{BrO}}_{3}^{-}[\/latex], [latex]{\\text{BrO}}_{4}^{-}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>The active ingredient formed by aspirin in the body is salicylic acid, C<sub>6<\/sub>H<sub>4<\/sub>OH(CO<sub>2<\/sub>H). The carboxyl group (\u2212CO<sub>2<\/sub>H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001-<em>M<\/em> aqueous solution of C<sub>6<\/sub>H<sub>4<\/sub>OH(CO<sub>2<\/sub>H).<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q929665\">Show Solution<\/span><\/p>\n<div id=\"q929665\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.\u00a0[H<sub>2<\/sub>O] &gt; [CH<sub>3<\/sub>CO<sub>2<\/sub>H] &gt; [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] \u2248 [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right][\/latex] &gt; [OH<sup>\u2212<\/sup>]<\/p>\n<p>3.\u00a0The strongest base or strongest acid is the one with the larger <em>K<\/em><sub>b<\/sub> or <em>K<\/em><sub>a<\/sub>, respectively. In these two examples, they are (CH<sub>3<\/sub>)<sub>2<\/sub>NH and [latex]{\\text{CH}}_{3}{\\text{NH}}_{3}^{+}[\/latex].<\/p>\n<p>5.\u00a0Look up <a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-bases-2\/\" target=\"_blank\" rel=\"noopener\">Ionization Constants of Weak Bases<\/a> the value of <em>K<\/em><sub>b<\/sub> for (CH<sub>3<\/sub>)<sub>3<\/sub>N and the value of <em>K<\/em><sub>a<\/sub> for H<sub>3<\/sub>BO<sub>3<\/sub>. From the latter, calculate the value of <em>K<\/em><sub>b<\/sub> for [latex]{\\text{H}}_{2}{\\text{BO}}_{3}^{-}[\/latex]. Then compare values:<\/p>\n<ul>\n<li><em>K<\/em><sub>b<\/sub>(CH<sub>3<\/sub>)<sub>3<\/sub>N = 7.4 \u00d7 10<sup>\u22125<\/sup><\/li>\n<li>[latex]{K}_{\\text{a}}\\left({\\text{H}}_{3}{\\text{BO}}_{3}\\right)=5.8\\times {10}^{-10}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{b}}}[\/latex], [latex]{K}_{\\text{b}}=\\frac{1.0\\times {10}^{-14}}{5.8\\times {10}^{-10}}=1.7\\times {10}^{-5}[\/latex]<\/li>\n<\/ul>\n<p>A comparison shows that the larger <em>K<\/em><sub>b<\/sub> is that of triethylamine.<\/p>\n<p>7.\u00a0The more acidic compounds are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\text{HSO}}_{4}^{-}[\/latex]; higher electronegativity of the central ion.<\/li>\n<li>H<sub>2<\/sub>O; NH<sub>3<\/sub> is a base and water is neutral, or decide on the basis of <em>K<\/em><sub>a<\/sub> values.<\/li>\n<li>HI; PH<sub>3<\/sub> is weaker than HCl; HCl is weaker than HI. Thus, PH<sub>3<\/sub> is weaker than HI.<\/li>\n<li>PH<sub>3<\/sub>; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group.<\/li>\n<li>HBr; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S, so HBr is the stronger acid.<\/li>\n<\/ol>\n<p>9. The correct ordered lists are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>NaHSeO<sub>3<\/sub> &lt; NaHSO<sub>3<\/sub> &lt; NaHSO<sub>4<\/sub>; in polyoxy acids, the more electronegative central element\u2014S, in this case\u2014forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner.<\/li>\n<li>[latex]{\\text{ClO}}_{2}^{-}<{\\text{BrO}}_{2}^{-}<{\\text{IO}}_{2}^{-}[\/latex]; the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.<\/li>\n<li>HOI &lt; HOBr &lt; HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.<\/li>\n<li>HOCl &lt; HOClO &lt; HOClO<sub>2<\/sub> &lt; HOClO<sub>3<\/sub>; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases).<\/li>\n<li>[latex]{\\text{HTe}}^{-}<{\\text{HS}}^{-}<<{\\text{PH}}_{2}^{-}<{\\text{NH}}_{2}^{-}[\/latex]; [latex]{\\text{PH}}_{2}^{-}[\/latex] and [latex]{\\text{NH}}_{2}^{-}[\/latex] are anions of weak bases, so they act as strong bases toward H<sup>+<\/sup>. [latex]{\\text{HTe}}^{-}[\/latex] and HS<sup>\u2212<\/sup> are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion.<\/li>\n<li>[latex]{\\text{BrO}}_{4}^{-}<{\\text{BrO}}_{3}^{-}<{\\text{BrO}}_{2}^{-}<{\\text{BrO}}^{-}[\/latex]; with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic.<\/li>\n<\/ol>\n<p>10.\u00a0[latex]\\left[\\text{H}_{2}\\text{O}\\right]\\gt\\left[\\text{C}_{6}\\text{H}_{4}\\text{OH}\\left(\\text{CO}_{2}\\text{H}\\right)\\right]\\gt\\left[\\text{H}^{+}\\right]\\text{O}\\gt\\left[\\text{C}_{6}\\text{H}_{4}\\text{OH}\\left(\\text{CO}_{2}\\right)^{-}\\right]\\gt\\left[\\text{C}_{6}\\text{H}_{4}\\text{O}\\left(\\text{CO}_{2}\\text{H}\\right)^{-}\\right]\\gt\\left[\\text{OH}^{-}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li id=\"fs-idm94403824\">Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO<sub>3<\/sub>)<sub>2<\/sub>, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO<sub>3<\/sub> with CuO.<\/li>\n<li>Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)<sub>2<\/sub> in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs<\/li>\n<li>What do we represent when we write: [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\left(aq\\right)?[\/latex]<\/li>\n<li>Explain why equilibrium calculations are not necessary to determine ionic concentrations in solutions of certain strong electrolytes such as NaOH and HCl. Under what conditions are equilibrium calculations necessary as part of the determination of the concentrations of all ions of some other strong electrolytes in solution?<\/li>\n<li>Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.<\/li>\n<li>What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid?<\/li>\n<li>What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak base?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q405128\">Show Selected Solutions<\/span><\/p>\n<div id=\"q405128\" class=\"hidden-answer\" style=\"display: none\">\n<p>2.\u00a0[latex]\\begin{array}{cccccc}\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)+& \\text{HCl}\\left(aq\\right)& \\longrightarrow & {\\text{Mg}}^{2+}\\left(aq\\right)+& 2{\\text{Cl}}^{-}\\left(aq\\right)+& {\\text{2H}}_{2}\\text{O}\\left(l\\right)\\\\ \\text{BB}& \\text{BA}& & \\text{CB}& \\text{CA}& \\end{array}[\/latex]<\/p>\n<p>4.\u00a0Strong electrolytes are 100% ionized, and, as long as the component ions are neither weak acids nor weak bases, the ionic species present result from the dissociation of the strong electrolyte. Equilibrium calculations are necessary when one (or more) of the ions is a weak acid or a weak base.<\/p>\n<p>6. The two assumptions are<\/p>\n<ol>\n<li>Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration.<\/li>\n<li>Assume we can neglect the contribution of water to the equilibrium concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li>Which of the following will increase the percent of NH<sub>3<\/sub> that is converted to the ammonium ion in water (Hint: Use LeCh\u00e2telier\u2019s principle.)?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>addition of NaOH<\/li>\n<li>addition of HCl<\/li>\n<li>addition of NH<sub>4<\/sub>Cl<\/li>\n<\/ol>\n<\/li>\n<li>Which of the following will increase the percent of HF that is converted to the fluoride ion in water?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>addition of NaOH<\/li>\n<li>addition of HCl<\/li>\n<li>addition of NaF<\/li>\n<\/ol>\n<\/li>\n<li>What is the effect on the concentrations of [latex]{\\text{NO}}_{2}^{-}[\/latex], HNO<sub>2<\/sub>, and OH<sup>\u2212<\/sup> when the following are added to a solution of KNO<sub>2<\/sub> in water:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>HCl<\/li>\n<li>HNO<sub>2<\/sub><\/li>\n<li>NaOH<\/li>\n<li>NaCl<\/li>\n<li>KNO<\/li>\n<\/ol>\n<p>The equation for the equilibrium is: [latex]{\\text{NO}}_{2}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{HNO}}_{2}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/li>\n<li>What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>HCl<\/li>\n<li>KF<\/li>\n<li>NaCl<\/li>\n<li>KOH<\/li>\n<li>HF<\/li>\n<\/ol>\n<p>The equation for the equilibrium is: [latex]\\text{HF}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{F}}^{-}\\left(aq\\right)[\/latex]<\/li>\n<li>Why is the hydronium ion concentration in a solution that is 0.10 <em>M<\/em> in HCl and 0.10 <em>M<\/em> in HCOOH determined by the concentration of HCl?<\/li>\n<li>From the equilibrium concentrations given, calculate <em>K<\/em><sub>a<\/sub> for each of the weak acids and <em>K<\/em><sub>b<\/sub> for each of the weak bases.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>CH<sub>3<\/sub>CO<sub>2<\/sub>H: [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 1.34 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>; [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}^{-}\\right][\/latex] = 1.34 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>; [CH<sub>3<\/sub>CO<sub>2<\/sub>H] = 9.866 \u00d7 10<sup>\u22122<\/sup><em>M<\/em><\/li>\n<li>ClO<sup>\u2212<\/sup>: [OH<sup>\u2212<\/sup>] = 4.0 \u00d7 10<sup>\u22124<\/sup><em>M<\/em>;\u00a0[HClO] = 2.38 \u00d7 10<sup>\u22125<\/sup><em>M<\/em>;\u00a0[ClO<sup>\u2212<\/sup>] = 0.273 <em>M<\/em><\/li>\n<li>HCO<sub>2<\/sub>H: [HCO<sub>2<\/sub>H] = 0.524 <em>M<\/em>; [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 9.8 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>; [latex]\\left[{\\text{HCO}}_{2}^{-}\\right][\/latex] = 9.8 \u00d7 10<sup>\u22123<\/sup><em>M<\/em><\/li>\n<li>[latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}:[\/latex] [latex]\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex] = 0.233 <em>M<\/em>;\u00a0[C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>] = 2.3 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>;\u00a0[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 2.3 \u00d7 10<sup>\u22123<\/sup><em>M<\/em><\/li>\n<\/ol>\n<\/li>\n<li>From the equilibrium concentrations given, calculate <em>K<\/em><sub>a<\/sub> for each of the weak acids and <em>K<\/em><sub>b<\/sub> for each of the weak bases.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>NH<sub>3<\/sub>: [OH<sup>\u2212<\/sup>] = 3.1 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>;\u00a0[latex]\\left[{\\text{NH}}_{4}^{+}\\right][\/latex] = 3.1 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>;\u00a0[NH<sub>3<\/sub>] = 0.533 <em>M<\/em><\/li>\n<li>HNO<sub>2<\/sub>: [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 0.011 <em>M<\/em>; [latex]\\left[{\\text{NO}}_{2}^{-}\\right][\/latex] = 0.0438 <em>M<\/em>;\u00a0[HNO<sub>2<\/sub>] = 1.07 <em>M<\/em><\/li>\n<li>(CH<sub>3<\/sub>)<sub>3<\/sub>N: [(CH<sub>3<\/sub>)<sub>3<\/sub>N] = 0.25 <em>M<\/em>;\u00a0[latex]\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\right][\/latex] = 4.3 \u00d7 10<sup>\u22123<\/sup><em>M<\/em>;\u00a0[OH<sup>\u2212<\/sup>] = 4.3 \u00d7 10<sup>\u22123<\/sup><em>M<\/em><\/li>\n<li>[latex]{\\text{NH}}_{4}^{+}:[\/latex] [latex]\\left[{\\text{NH}}_{4}^{+}\\right][\/latex] = 0.100 <em>M<\/em>; [NH<sub>3<\/sub>] = 7.5 \u00d7 10<sup>\u22126<\/sup><em>M<\/em>;\u00a0[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = 7.5 \u00d7 10<sup>\u22126<\/sup><em>M<\/em><\/li>\n<\/ol>\n<\/li>\n<li>Determine <em>K<\/em><sub>b<\/sub> for the nitrite ion, [latex]{\\text{NO}}_{2}^{-}[\/latex]. In a 0.10-<em>M<\/em> solution this base is 0.0015% ionized.<\/li>\n<li>Determine <em>K<\/em><sub>a<\/sub> for hydrogen sulfate ion, [latex]{\\text{HSO}}_{4}^{-}[\/latex]. In a 0.10-<em>M<\/em> solution the acid is 29% ionized.<\/li>\n<li>Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>F<sup>\u2212<\/sup><\/li>\n<li>[latex]{\\text{NH}}_{4}^{+}[\/latex]<\/li>\n<li>[latex]{\\text{AsO}}_{4}^{3-}[\/latex]<\/li>\n<li>[latex]{\\left({\\text{CH}}_{3}\\right)}_{2}{\\text{NH}}_{2}^{+}[\/latex]<\/li>\n<li>[latex]{\\text{NO}}_{2}^{-}[\/latex]<\/li>\n<li>[latex]{\\text{HC}}_{2}{\\text{O}}_{4}^{-}[\/latex] (as a base)<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>HTe<sup>\u2212<\/sup> (as a base)<\/li>\n<li>[latex]{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}[\/latex]<\/li>\n<li>[latex]{\\text{HAsO}}_{4}^{3-}[\/latex] (as a base)<\/li>\n<li>[latex]{\\text{HO}}_{2}^{-}[\/latex] (as a base)<\/li>\n<li>[latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}[\/latex]<\/li>\n<li>[latex]{\\text{HSO}}_{3}^{-}[\/latex] (as a base)<\/li>\n<\/ol>\n<\/li>\n<li>For which of the following solutions must we consider the ionization of water when calculating the pH or pOH?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>3 \u00d7 10<sup>\u22128<\/sup><em>M<\/em> HNO<sub>3<\/sub><\/li>\n<li>0.10 g HCl in 1.0 L of solution<\/li>\n<li>0.00080 g NaOH in 0.50 L of solution<\/li>\n<li>1 \u00d7 10<sup>\u22127<\/sup><em>M<\/em> Ca(OH)<sub>2<\/sub><\/li>\n<li>0.0245 <em>M<\/em> KNO<sub>3<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Even though both NH<sub>3<\/sub> and C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub> are weak bases, NH<sub>3<\/sub> is a much stronger acid than C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>. Which of the following is correct at equilibrium for a solution that is initially 0.10 <em>M<\/em> in NH<sub>3<\/sub> and 0.10 <em>M<\/em> in C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\left[{\\text{OH}}^{-}\\right]=\\left[{\\text{NH}}_{4}^{+}\\right][\/latex]<\/li>\n<li>[latex]\\left[{\\text{NH}}_{4}^{+}\\right]=\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex]<\/li>\n<li>[latex]\\left[{\\text{OH}}^{-}\\right]=\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex]<\/li>\n<li>[NH<sub>3<\/sub>] = [C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>]<\/li>\n<li>both a and b are correct<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q789235\">Show Selected Solutions<\/span><\/p>\n<div id=\"q789235\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.\u00a0The equilibrium is [latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>The addition of NaOH adds OH<sup>\u2212<\/sup> to the system and, according to Le Ch\u00e2telier\u2019s principle, the equilibrium will shift to the left. Thus, the percent of converted NH<sub>3<\/sub> will decrease.<\/li>\n<li>The addition of HCl will add [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex] ions, which will then react with the OH<sup>\u2212<\/sup> ions. Thus, the equilibrium will shift to the right, and the percent will increase.<\/li>\n<li>The addition of NH<sub>4<\/sub>Cl adds [latex]{\\text{NH}}_{4}^{+}[\/latex] ions, shifting the equilibrium to the left. Thus, the percent will decrease.<\/li>\n<\/ol>\n<p>3.\u00a0The effects are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Adding HCl will add [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex] ions, which will then react with the OH<sup>\u2212<\/sup> ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO<sub>2<\/sub>, and decreasing the concentration of [latex]{\\text{NO}}_{2}^{-}[\/latex] ions.<\/li>\n<li>Adding HNO<sub>2<\/sub> increases the concentration of HNO<sub>2<\/sub> and shifts the equilibrium to the left, increasing the concentration of [latex]{\\text{NO}}_{2}^{-}[\/latex] ions and decreasing the concentration of OH<sup>\u2212<\/sup> ions.<\/li>\n<li>Adding NaOH adds OH<sup>\u2212<\/sup> ions, which shifts the equilibrium to the left, increasing the concentration of [latex]{\\text{NO}}_{2}^{-}[\/latex] ions and decreasing the concentrations of HNO<sub>2<\/sub>.<\/li>\n<li>Adding NaCl has no effect on the concentrations of the ions.<\/li>\n<li>Adding KNO<sub>2<\/sub> adds [latex]{\\text{NO}}_{2}^{-}[\/latex] ions and shifts the equilibrium to the right, increasing the HNO<sub>2<\/sub> and OH<sup>\u2212<\/sup> ion concentrations.<\/li>\n<\/ol>\n<p>5.\u00a0The equations of the occurring chemical processes are:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{HCl}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}\\text{COOH}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{COO}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p>This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO<sub>2<\/sub>H exists primarily as HCO<sub>2<\/sub>H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO<sub>2<\/sub>H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] produced by the stronger acid.<\/p>\n<p>7.\u00a0<em>K<sub>a<\/sub><\/em> and\u00a0<em>K<sub>b<\/sub><\/em> are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>The reaction is [latex]{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{OH}}^{-}\\left(aq\\right)+{\\text{NH}}_{4}^{+}\\left(aq\\right)[\/latex]<br \/>\n[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(3.1\\times {10}^{-3}\\right)\\left(3.1\\times {10}^{-3}\\right)}{\\left(0.533\\right)}=1.8\\times {10}^{-5}[\/latex]<\/li>\n<li>The reaction is [latex]{\\text{HNO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{NO}}_{2}^{-}\\left(aq\\right)[\/latex]<br \/>\n[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{NO}}_{2}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{HNO}}_{2}\\right]}=\\frac{\\left(0.0438\\right)\\left(0.011\\right)}{\\left(1.07\\right)}=4.5\\times {10}^{-4}[\/latex]<\/li>\n<li>The reaction is [latex]{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{OH}}^{-}\\left(aq\\right)+{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\left(aq\\right)[\/latex]<br \/>\n[latex]{K}_{\\text{b}}=\\frac{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}{\\text{NH}}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\left({\\text{CH}}_{3}\\right)}_{3}\\text{N}\\right]}=\\frac{\\left(4.3\\times {10}^{-3}\\right)\\left(4.3\\times {10}^{-3}\\right)}{\\left(0.25\\right)}=7.4\\times {10}^{-5}[\/latex]<\/li>\n<li>The reaction is [latex]{\\text{NH}}_{4}^{+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{NH}}_{3}\\left(aq\\right)[\/latex]<br \/>\n[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{NH}}_{3}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{NH}}_{4}^{+}\\right]}=\\frac{\\left(7.5\\times {10}^{-6}\\right)\\left(7.5\\times {10}^{-6}\\right)}{\\left(0.100\\right)}=5.6\\times {10}^{-10}[\/latex]<\/li>\n<\/ol>\n<p>9.\u00a0The reaction is [latex]{\\text{HSO}}_{4}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{SO}}_{4}^{2-}\\left(aq\\right)[\/latex].<\/p>\n<p>The concentrations at equilibrium are:<\/p>\n<ul>\n<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] = [latex]\\left[{\\text{SO}}_{4}^{2-}\\right][\/latex] = (0.29)(0.10 <em>M<\/em>) = 0.029 <em>M<\/em><\/li>\n<li>[latex]\\left[{\\text{HSO}}_{4}^{-}\\right][\/latex] = 0.10 <em>M<\/em> \u2212 0.029 <em>M<\/em> = 0.071 <em>M<\/em><\/li>\n<li>[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{SO}}_{4}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{HSO}}_{4}^{-}\\right]}=\\frac{\\left(0.029\\right)\\left(0.029\\right)}{\\left(0.071\\right)}=1.2\\times {10}^{-2}[\/latex]<\/li>\n<\/ul>\n<p>11.\u00a0The ionization constants are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{\\left(1.00\\times {10}^{-14}\\right)}{2.3\\times {10}^{-3}}=4.3\\times {10}^{-12}[\/latex]<\/li>\n<li>[latex]{K}_{\\text{a}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{b}}}=\\frac{\\left(1.00\\times {10}^{-14}\\right)}{7.4\\times {10}^{-5}}=1.4\\times {10}^{-10}[\/latex]<\/li>\n<li>[latex]{K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{\\left(1.00\\times {10}^{-14}\\right)}{1\\times {10}^{-7}}=1\\times {10}^{-7}[\/latex]<\/li>\n<li>[latex]{K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{\\left(1.00\\times {10}^{-14}\\right)}{2.4\\times {10}^{-12}}=4.2\\times {10}^{-3}[\/latex]<\/li>\n<li>[latex]{K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{\\left(1.00\\times {10}^{-14}\\right)}{2.4\\times {10}^{-12}}=4.2\\times {10}^{-3}[\/latex]<\/li>\n<li>[latex]{K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{\\left(1.00\\times {10}^{-14}\\right)}{1.2\\times {10}^{-2}}=8.3\\times {10}^{-13}[\/latex]<\/li>\n<\/ol>\n<p>13.\u00a0The reactions are:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{NH}}_{4}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\\\ {\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\end{array}[\/latex]<\/p>\n<p>Because NH<sub>3<\/sub> is much stronger than C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>, it dissociates more. As the initial concentrations of both bases are the same, at equilibrium, [NH<sub>3<\/sub>] &lt; [C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>], [latex]\\left[{\\text{NH}}_{4}^{+}\\right][\/latex] &gt; [latex]\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex], and [OH<sup>\u2212<\/sup>] &gt; [latex]\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}^{+}\\right][\/latex]. Therefore, (a) is the correct statement. The contribution to the total [OH<sup>\u2212<\/sup>] at equilibrium from C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub> is negligible compared to HN<sub>3<\/sub>. Therefore [latex]\\left[{\\text{OH}}^{-}\\right]=\\left[{\\text{NH}}_{4}^{+}\\right][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>acid ionization constant (<em>K<\/em><sub>a<\/sub>): <\/strong>equilibrium constant for the ionization of a weak acid<\/p>\n<p><strong>base ionization constant (<em>K<\/em><sub>b<\/sub>): <\/strong>equilibrium constant for the ionization of a weak base<\/p>\n<p><strong>leveling effect of water: <\/strong>any acid stronger than [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex], or any base stronger than OH<sup>\u2212<\/sup> will react with water to form [latex]{\\text{H}}_{3}{\\text{O}}^{+}[\/latex], or OH<sup>\u2212<\/sup>, respectively; water acts as a base to make all strong acids appear equally strong, and it acts as an acid to make all strong bases appear equally strong<\/p>\n<p><strong>oxyacid: <\/strong>compound containing a nonmetal and one or more hydroxyl groups<\/p>\n<p><strong>percent ionization: <\/strong>ratio of the concentration of the ionized acid to the initial acid concentration, times 100<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3479\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry 2e. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/\">https:\/\/openstax.org\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/chemistry-2e\/pages\/1-introduction<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Acids and Bases Chemistry - Basic Introduction. <strong>Authored by<\/strong>: The Organic Chemistry Tutor. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ZNo6gfCAgWE\">https:\/\/youtu.be\/ZNo6gfCAgWE<\/a>. <strong>License<\/strong>: <em>Other<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry 2e\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/chemistry-2e\/pages\/1-introduction\"},{\"type\":\"copyrighted_video\",\"description\":\"Acids and Bases Chemistry - Basic Introduction\",\"author\":\"The Organic Chemistry Tutor\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/ZNo6gfCAgWE\",\"project\":\"\",\"license\":\"other\",\"license_terms\":\"Standard YouTube License\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3479","chapter","type-chapter","status-publish","hentry"],"part":2988,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3479","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":29,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3479\/revisions"}],"predecessor-version":[{"id":7853,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3479\/revisions\/7853"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/parts\/2988"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3479\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/media?parent=3479"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapter-type?post=3479"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/contributor?post=3479"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/license?post=3479"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}