{"id":3486,"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=3486"},"modified":"2016-10-27T15:38:27","modified_gmt":"2016-10-27T15:38:27","slug":"hydrolysis-of-salt-solutions-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/chapter\/hydrolysis-of-salt-solutions-2\/","title":{"raw":"Hydrolysis of Salt Solutions","rendered":"Hydrolysis of Salt Solutions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this module, you will be able to:\r\n<ul>\r\n \t<li>Predict whether a salt solution will be acidic, basic, or neutral<\/li>\r\n \t<li>Calculate the concentrations of the various species in a salt solution<\/li>\r\n \t<li>Describe the process that causes solutions of certain metal ions to be acidic<\/li>\r\n<\/ul>\r\n<\/div>\r\nAs we have seen in the section on chemical reactions, when an acid and base are mixed, they undergo a neutralization reaction. The word \u201cneutralization\u201d seems to imply that a stoichiometrically equivalent solution of an acid and a base would be neutral. This is sometimes true, but the salts that are formed in these reactions may have acidic or basic properties of their own, as we shall now see.\r\n<h2 data-type=\"title\">Acid-Base Neutralization<\/h2>\r\nA solution is neutral when it contains equal concentrations of hydronium and hydroxide ions. When we mix solutions of an acid and a base, an acid-base neutralization reaction occurs. However, even if we mix stoichiometrically equivalent quantities, we may find that the resulting solution is not neutral. It could contain either an excess of hydronium ions or an excess of hydroxide ions because the nature of the salt formed determines whether the solution is acidic, neutral, or basic. The following four situations illustrate how solutions with various pH values can arise following a neutralization reaction using stoichiometrically equivalent quantities:\r\n<ol>\r\n \t<li>A strong acid and a strong base, such as HCl(<em>aq<\/em>) and NaOH(<em>aq<\/em>) will react to form a neutral solution since the conjugate partners produced are of negligible strength: [latex]\\text{HCl}\\left(aq\\right)+\\text{NaOH}\\left(aq\\right)\\rightleftharpoons \\text{NaCl}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/li>\r\n \t<li>A strong acid and a weak base yield a weakly acidic solution, not because of the strong acid involved, but because of the conjugate acid of the weak base.<\/li>\r\n \t<li>A weak acid and a strong base yield a weakly basic solution. A solution of a weak acid reacts with a solution of a strong base to form the conjugate base of the weak acid and the conjugate acid of the strong base. The conjugate acid of the strong base is a weaker acid than water and has no effect on the acidity of the resulting solution. However, the conjugate base of the weak acid is a weak base and ionizes slightly in water. This increases the amount of hydroxide ion in the solution produced in the reaction and renders it slightly basic.<\/li>\r\n \t<li>A weak acid plus a weak base can yield either an acidic, basic, or neutral solution. This is the most complex of the four types of reactions. When the conjugate acid and the conjugate base are of unequal strengths, the solution can be either acidic or basic, depending on the relative strengths of the two conjugates. Occasionally the weak acid and the weak base will have the <em>same<\/em> strength, so their respective conjugate base and acid will have the same strength, and the solution will be neutral. To predict whether a particular combination will be acidic, basic or neutral, tabulated <em>K<\/em> values of the conjugates must be compared.<\/li>\r\n<\/ol>\r\n<div class=\"textbox shaded\">\r\n<h3>Stomach Antacids<\/h3>\r\nOur stomachs contain a solution of roughly 0.03 <em>M<\/em> HCl, which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, CaCO<sub>3<\/sub>. The reaction,\r\n<p style=\"text-align: center;\">[latex]{\\text{CaCO}}_{3}\\left(s\\right)+2\\text{HCl}\\left(aq\\right)\\rightleftharpoons {\\text{aCl}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\text{CO}}_{2}\\left(g\\right)[\/latex]<\/p>\r\nnot only neutralizes stomach acid, it also produces CO<sub>2<\/sub>(<em>g<\/em>), which may result in a satisfying belch.\r\n\r\nMilk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, Mg(OH)<sub>2<\/sub>. It works according to the reaction:\r\n<p style=\"text-align: center;\">[latex]\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{2+}\\left(aq\\right)+2{\\text{OH}}^{\\text{-}}\\left(aq\\right)[\/latex]<\/p>\r\nThe hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that\r\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}+{\\text{OH}}^{\\text{-}}\\rightleftharpoons 2{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\r\nThis reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect.\r\n\r\nSeveral antacids have aluminum hydroxide, Al(OH)<sub>3<\/sub>, as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Culinary Aspects of Chemistry<\/h3>\r\nCooking is essentially synthetic chemistry that happens to be safe to eat. There are a number of examples of acid-base chemistry in the culinary world. One example is the use of baking soda, or sodium bicarbonate in baking. NaHCO<sub>3<\/sub> is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter \u201crises.\u201d Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.\r\n\r\nMany people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure\u00a01). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a \u201csour\u201d taste that we seem to enjoy.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"550\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213817\/CNX_Chem_14_03_FishLemon.jpg\" alt=\"An image is shown of two fish with heads removed and skin on with lemon slices placed in the body cavity.\" width=\"550\" height=\"358\" data-media-type=\"image\/jpeg\" \/> Figure\u00a01. A neutralization reaction takes place between citric acid in lemons or acetic acid in vinegar, and the bases in the flesh of fish.[\/caption]\r\n\r\n&nbsp;\r\n\r\nPickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour.\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Salts of Weak Bases and Strong Acids<\/h2>\r\nWhen we neutralize a weak base with a strong acid, the product is a salt containing the conjugate acid of the weak base. This conjugate acid is a weak acid. For example, ammonium chloride, NH<sub>4<\/sub>Cl, is a salt formed by the reaction of the weak base ammonia with the strong acid HCl:\r\n<p style=\"text-align: center;\">[latex]{\\text{NH}}_{3}\\left(aq\\right)+\\text{HCl}\\left(aq\\right)\\longrightarrow {\\text{NH}}_{4}\\text{Cl}\\left(aq\\right)[\/latex]<\/p>\r\nA solution of this salt contains ammonium ions and chloride ions. The chloride ion has no effect on the acidity of the solution since HCl is a strong acid. Chloride is a very weak base and will not accept a proton to a measurable extent. However, the ammonium ion, the conjugate acid of ammonia, reacts with water and increases the hydronium ion concentration:\r\n<p style=\"text-align: center;\">[latex]{\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{NH}}_{3}\\left(aq\\right)[\/latex]<\/p>\r\nThe equilibrium equation for this reaction is simply the ionization constant. <em>K<\/em><sub>a<\/sub>, for the acid [latex]{\\text{NH}}_{4}{}^{\\text{+}}:[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{NH}}_{3}\\right]}{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]}={K}_{\\text{a}}[\/latex]<\/p>\r\nWe will not find a value of <em>K<\/em><sub>a<\/sub> for the ammonium ion in <a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\">Ionization Constants of Weak Acids<\/a>. However, it is not difficult to determine <em>K<\/em><sub>a<\/sub> for [latex]{\\text{NH}}_{4}{}^{\\text{+}}[\/latex] from the value of the ionization constant of water, <em>K<\/em><sub>w<\/sub>, and <em>K<\/em><sub>b<\/sub>, the ionization constant of its conjugate base, NH<sub>3<\/sub>, using the following relationship:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{w}}={K}_{\\text{a}}\\times {K}_{\\text{b}}[\/latex]<\/p>\r\nThis relation holds for any base and its conjugate acid or for any acid and its conjugate base.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0The pH of a Solution of a Salt of a Weak Base and a Strong Acid<\/h3>\r\nAniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, [latex]\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}\\right]\\text{Cl}[\/latex], a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 <em>M<\/em> solution of aniline hydrochloride?\r\n<p style=\"text-align: center;\">[latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\left(aq\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"766004\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"766004\"]\r\n\r\nThe new step in this example is to determine <em>K<\/em><sub>a<\/sub> for the [latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}[\/latex] ion. The [latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}[\/latex] ion is the conjugate acid of a weak base. The value of <em>K<\/em><sub>a<\/sub> for this acid is not listed in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\">Ionization Constants of Weak Acids<\/a>, but we can determine it from the value of <em>K<\/em><sub>b<\/sub> for aniline, C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>, which is given as 4.6 [latex]\\times [\/latex] 10<sup>\u221210<\/sup> (<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Relative Strengths of Acids and Bases<\/a>\u00a0and <a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-bases-2\/\" target=\"_blank\">Ionization Constants of Weak Bases<\/a>):\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}\\left(\\text{ for }{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}\\right)\\times {K}_{\\text{b}}\\left(\\text{ for }{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\right)={K}_{\\text{w}}=1.0\\times {10}^{-14}[\/latex]\r\n[latex]{K}_{\\text{a}}\\left(\\text{for}{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}\\right)=\\frac{{K}_{\\text{w}}}{{K}_{\\text{b}}\\left(\\text{ for }{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\right)}=\\frac{1.0\\times {10}^{-14}}{4.6\\times {10}^{-10}}=2.2\\times {10}^{-5}[\/latex]<\/p>\r\nNow we have the ionization constant and the initial concentration of the weak acid, the information necessary to determine the equilibrium concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex], and the pH:\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213819\/CNX_Chem_14_04_steps1_img.jpg\" alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cDetermine the direction of change.\u201d The second is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth is labeled \u201cCheck the math.\u201d\" data-media-type=\"image\/jpeg\" \/>\r\nWith these steps we find [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 2.3 [latex]\\times [\/latex] 10<sup>\u22123<\/sup><em>M<\/em> and pH = 2.64\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\n<ol>\r\n \t<li>Do the calculations and show that the hydronium ion concentration for a 0.233-<em>M<\/em> solution of [latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}[\/latex] is 2.3 [latex]\\times [\/latex] 10<sup>\u22123<\/sup> and the pH is 2.64.<\/li>\r\n \t<li>What is the hydronium ion concentration in a 0.100-<em>M<\/em> solution of ammonium nitrate, NH<sub>4<\/sub>NO<sub>3<\/sub>, a salt composed of the ions [latex]{\\text{NH}}_{4}{}^{\\text{+}}[\/latex] and [latex]{\\text{NO}}_{3}{}^{\\text{-}}[\/latex]. Use the data in\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Table\u00a02 on Relative Strengths of Acids and Bases<\/a>\u00a0to\u00a0determine <em>K<\/em><sub>b<\/sub> for the ammonium ion. Which is the stronger acid [latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}[\/latex] or [latex]{\\text{NH}}_{4}{}^{\\text{+}}?[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"407406\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"407406\"]\r\n<ol>\r\n \t<li>[latex]{K}_{a}\\left(\\text{for}{\\text{NH}}_{4}{}^{\\text{+}}\\right)=5.6\\times {10}^{-10}[\/latex], [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 7.5 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><em>M<\/em><\/li>\r\n \t<li>[latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}[\/latex] is the stronger acid.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Salts of Weak Acids and Strong Bases<\/h2>\r\nWhen we neutralize a weak acid with a strong base, we get a salt that contains the conjugate base of the weak acid. This conjugate base is usually a weak base. For example, sodium acetate, NaCH<sub>3<\/sub>CO<sub>2<\/sub>, is a salt formed by the reaction of the weak acid acetic acid with the strong base sodium hydroxide:\r\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+\\text{NaOH}\\left(aq\\right)\\longrightarrow {\\text{NaCH}}_{3}{\\text{CO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(aq\\right)[\/latex]<\/p>\r\nA solution of this salt contains sodium ions and acetate ions. The sodium ion, as the conjugate acid of a strong base, has no effect on the acidity of the solution. However, the acetate ion, the conjugate base of acetic acid, reacts with water and increases the concentration of hydroxide ion:\r\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\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}}^{\\text{-}}\\left(aq\\right)[\/latex]<\/p>\r\nThe equilibrium equation for this reaction is the ionization constant, <em>K<\/em><sub>b<\/sub>, for the base [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}[\/latex]. The value of <em>K<\/em><sub>b<\/sub> can be calculated from the value of the ionization constant of water, <em>K<\/em><sub>w<\/sub>, and <em>K<\/em><sub>a<\/sub>, the ionization constant of the conjugate acid of the anion using the equation:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{w}}={K}_{\\text{a}}\\times {K}_{\\text{b}}[\/latex]<\/p>\r\nFor the acetate ion and its conjugate acid we have:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{b}}\\left(\\text{for}{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right)=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}\\left(\\text{for}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right)}=\\frac{1.0\\times {10}^{-14}}{1.8\\times {10}^{-5}}=5.6\\times {10}^{-10}[\/latex]<\/p>\r\nSome handbooks do not report values of <em>K<\/em><sub>b<\/sub>. They only report ionization constants for acids. If we want to determine a <em>K<\/em><sub>b<\/sub> value using one of these handbooks, we must look up the value of <em>K<\/em><sub>a<\/sub> for the conjugate acid and convert it to a <em>K<\/em><sub>b<\/sub> value.\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base<\/h3>\r\nDetermine the acetic acid concentration in a solution with [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]=0.050M[\/latex] and [OH<sup>\u2212<\/sup>] = 2.5 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><em>M<\/em> at equilibrium. The reaction is:\r\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\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}}^{\\text{-}}\\left(aq\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"597974\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"597974\"]\r\n\r\nWe are given two of three equilibrium concentrations and asked to find the missing concentration. If we can find the equilibrium constant for the reaction, the process is straightforward.\r\n\r\nThe acetate ion behaves as a base in this reaction; hydroxide ions are a product. We determine <em>K<\/em><sub>b<\/sub> as follows:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{b}}\\left(\\text{ for }{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right)=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}\\left(\\text{ for }{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right)}=\\frac{1.0\\times {10}^{-14}}{1.8\\times {10}^{-5}}=5.6\\times {10}^{-10}[\/latex]<\/p>\r\nNow find the missing concentration:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]}=5.6\\times {10}^{-10}[\/latex]\r\n[latex]=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]\\left(2.5\\times {10}^{-6}\\right)}{\\left(0.050\\right)}=5.6\\times {10}^{-10}[\/latex]<\/p>\r\nSolving this equation we get [CH<sub>3<\/sub>CO<sub>2<\/sub>H] = 1.1 [latex]\\times [\/latex] 10<sup>\u22125<\/sup><em>M<\/em>.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nWhat is the pH of a 0.083-<em>M<\/em> solution of CN<sup>\u2212<\/sup>? Use 4.0 [latex]\\times [\/latex] 10<sup>\u221210<\/sup> as <em>K<\/em><sub>a<\/sub> for HCN. Hint: We will probably need to convert pOH to pH or find [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] using [OH<sup>\u2212<\/sup>] in the final stages of this problem.\r\n[reveal-answer q=\"680253\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"680253\"]11.16[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Equilibrium in a Solution of a Salt of a Weak Acid and a Weak Base<\/h2>\r\nIn a solution of a salt formed by the reaction of a weak acid and a weak base, to predict the pH, we must know both the <em>K<\/em><sub>a<\/sub> of the weak acid and the <em>K<\/em><sub>b<\/sub> of the weak base. If <em>K<\/em><sub>a<\/sub> &gt; <em>K<\/em><sub>b<\/sub>, the solution is acidic, and if <em>K<\/em><sub>b<\/sub> &gt; <em>K<\/em><sub>a<\/sub>, the solution is basic.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3:\u00a0Determining the Acidic or Basic Nature of Salts<\/h3>\r\nDetermine whether aqueous solutions of the following salts are acidic, basic, or neutral:\r\n<ol>\r\n \t<li>KBr<\/li>\r\n \t<li>NaHCO<sub>3<\/sub><\/li>\r\n \t<li>NH<sub>4<\/sub>Cl<\/li>\r\n \t<li>Na<sub>2<\/sub>HPO<sub>4<\/sub><\/li>\r\n \t<li>NH<sub>4<\/sub>F<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"54649\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"54649\"]\r\n\r\nConsider each of the ions separately in terms of its effect on the pH of the solution, as shown here:\r\n<ol>\r\n \t<li>The K<sup>+<\/sup> cation and the Br<sup>\u2212<\/sup> anion are both spectators, since they are the cation of a strong base (KOH) and the anion of a strong acid (HBr), respectively. The solution is neutral.<\/li>\r\n \t<li>The Na<sup>+<\/sup> cation is a spectator, and will not affect the pH of the solution; while the [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] anion is amphiprotic, it could either behave as an acid or a base. The <em>K<\/em><sub>a<\/sub> of [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] is 4.7 [latex]\\times [\/latex] 10<sup>\u221211<\/sup>, so the <em>K<\/em><sub>b<\/sub> of its conjugate base is [latex]\\frac{1.0\\times {10}^{-14}}{4.7\\times {10}^{-11}}=2.1\\times {10}^{-4}[\/latex].\r\nSince <em>K<\/em><sub>b<\/sub> &gt;&gt; <em>K<\/em><sub>a<\/sub>, the solution is basic.<\/li>\r\n \t<li>The [latex]{\\text{NH}}_{4}{}^{\\text{+}}[\/latex] ion is acidic and the Cl<sup>\u2212<\/sup> ion is a spectator. The solution will be acidic.<\/li>\r\n \t<li>The Na<sup>+<\/sup> ion is a spectator, while the [latex]{\\text{HPO}}_{4}{}^{-}[\/latex] ion is amphiprotic, with a <em>K<\/em><sub>a<\/sub> of 3.6 [latex]\\times [\/latex] 10<sup>\u221213<\/sup>\r\nso that the <em>K<\/em><sub>b<\/sub> of its conjugate base is [latex]\\frac{1.0\\times {10}^{-14}}{3.6\\times {10}^{-13}}=2.8\\times {10}^{-2}[\/latex]. Because <em>K<\/em><sub>b<\/sub> &gt;&gt; <em>K<\/em><sub>a<\/sub>, the solution is basic.<\/li>\r\n \t<li>The [latex]{\\text{NH}}_{4}{}^{\\text{+}}[\/latex] ion is listed as being acidic, and the F<sup>\u2212<\/sup> ion is listed as a base, so we must directly compare the <em>K<\/em><sub>a<\/sub> and the <em>K<\/em><sub>b<\/sub> of the two ions. <em>K<\/em><sub>a<\/sub> of [latex]{\\text{NH}}_{4}{}^{\\text{+}}[\/latex] is 5.6 [latex]\\times [\/latex] 10<sup>\u221210<\/sup>, which seems very small, yet the <em>K<\/em><sub>b<\/sub> of F<sup>\u2212<\/sup> is 1.4 [latex]\\times [\/latex] 10<sup>\u221211<\/sup>, so the solution is acidic, since <em>K<\/em><sub>a<\/sub> &gt; <em>K<\/em><sub>b<\/sub>.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nDetermine whether aqueous solutions of the following salts are acidic, basic, or neutral:\r\n<ol>\r\n \t<li>K<sub>2<\/sub>CO<sub>3<\/sub><\/li>\r\n \t<li>CaCl<sub>2<\/sub><\/li>\r\n \t<li>KH<sub>2<\/sub>PO<sub>4<\/sub><\/li>\r\n \t<li>(NH<sub>4<\/sub>)<sub>2<\/sub>CO<sub>3<\/sub><\/li>\r\n \t<li>AlBr<sub>3<\/sub><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"774628\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"774628\"](1) basic; (2) neutral; (3) basic; (4) basic; (5) acidic[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">The Ionization of Hydrated Metal Ions<\/h2>\r\nIf we measure the pH of the solutions of a variety of metal ions we will find that these ions act as weak acids when in solution. The aluminum ion is an example. When aluminum nitrate dissolves in water, the aluminum ion reacts with water to give a hydrated aluminum ion, [latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}[\/latex], dissolved in bulk water. What this means is that the aluminum ion has the strongest interactions with the six closest water molecules (the so-called first solvation shell), even though it does interact with the other water molecules surrounding this [latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}[\/latex] cluster as well:\r\n<p style=\"text-align: center;\">[latex]\\text{Al}{\\left({\\text{NO}}_{3}\\right)}_{3}\\left(s\\right)+6{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow \\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}\\left(aq\\right)+3{\\text{NO}}_{3}{}^{\\text{-}}\\left(aq\\right)[\/latex]<\/p>\r\nWe frequently see the formula of this ion simply as \u201cAl<sup>3+<\/sup>(<em>aq<\/em>)\u201d, without explicitly noting the six water molecules that are the closest ones to the aluminum ion and just describing the ion as being solvated in water (hydrated). This is similar to the simplification of the formula of the hydronium ion, [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] to H<sup>+<\/sup>. However, in this case, the hydrated aluminum ion is a weak acid (Figure\u00a02) and donates a proton to a water molecule. Thus, the hydration becomes important and we may use formulas that show the extent of hydration:\r\n<p style=\"text-align: center;\">[latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\left(aq\\right){K}_{\\text{a}}=1.4\\times {10}^{-5}[\/latex]<\/p>\r\nAs with other polyprotic acids, the hydrated aluminum ion ionizes in stages, as shown by:\r\n<p style=\"text-align: center;\">[latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\left(aq\\right)[\/latex]\r\n[latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{4}{\\left(\\text{OH}\\right)}_{2}{}^{\\text{+}}\\left(aq\\right)[\/latex]\r\n[latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{4}{\\left(\\text{OH}\\right)}_{2}{}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{3}{\\left(\\text{OH}\\right)}_{3}\\left(aq\\right)[\/latex]<\/p>\r\nNote that some of these aluminum species are exhibiting amphiprotic behavior, since they are acting as acids when they appear on the right side of the equilibrium expressions and as bases when they appear on the left side.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"878\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213820\/CNX_Chem_14_04_hydronium.jpg\" alt=\"A reaction is shown using ball and stick models. On the left, inside brackets with a superscript of 3 plus outside to the right is structure labeled \u201c[ A l ( H subscript 2 O ) subscript 6 ] superscript 3 plus.\u201d Inside the brackets is s central grey atom to which 6 red atoms are bonded in an arrangement that distributes them evenly about the central grey atom. Each red atom has two smaller white atoms attached in a forked or bent arrangement. Outside the brackets to the right is a space-filling model that includes a red central sphere with two smaller white spheres attached in a bent arrangement. Beneath this structure is the label \u201cH subscript 2 O.\u201d A double sided arrow follows. Another set of brackets follows to the right of the arrows which have a superscript of two plus outside to the right. The structure inside the brackets is similar to that on the left, except a white atom is removed from the structure. The label below is also changed to \u201c[ A l ( H subscript 2 O ) subscript 5 O H ] superscript 2 plus.\u201d To the right of this structure and outside the brackets is a space filling model with a central red sphere to which 3 smaller white spheres are attached. This structure is labeled \u201cH subscript 3 O superscript plus.\u201d\" width=\"878\" height=\"299\" data-media-type=\"image\/jpeg\" \/> Figure\u00a03. When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.[\/caption]However, the ionization of a cation carrying more than one charge is usually not extensive beyond the first stage. Additional examples of the first stage in the ionization of hydrated metal ions are:\r\n<p style=\"text-align: center;\">[latex]\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\left(aq\\right){K}_{\\text{a}}=2.74[\/latex]\r\n[latex]\\text{Cu}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{2+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Cu}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{\\text{+}}\\left(aq\\right){K}_{\\text{a}}=~6.3[\/latex]\r\n[latex]\\text{Zn}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{4}{}^{2+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Zn}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{3}{\\left(\\text{OH}\\right)}^{\\text{+}}\\left(aq\\right){K}_{\\text{a}}=9.6[\/latex]<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 4:\u00a0Hydrolysis of [Al(H<sub>2<\/sub>O)<sub>6<\/sub>]<sup>3+<\/sup><\/h3>\r\nCalculate the pH of a 0.10-<em>M<\/em> solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion [latex]{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}\\right]}^{3+}[\/latex] in solution.\r\n[reveal-answer q=\"103892\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"103892\"]\r\n\r\nIn spite of the unusual appearance of the acid, this is a typical acid ionization problem.\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\/23213822\/CNX_Chem_14_04_steps2_img.jpg\" alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cDetermine the direction of change.\u201d The second is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth is labeled \u201cCheck the math.\u201d\" width=\"876\" height=\"155\" data-media-type=\"image\/jpeg\" \/>\r\n<ol>\r\n \t<li><em>Determine the direction of change<\/em>. The equation for the reaction and <em>K<\/em><sub>a<\/sub> are:[latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\left(aq\\right){K}_{\\text{a}}=1.4\\times {10}^{-5}[\/latex]The reaction shifts to the right to reach equilibrium.<\/li>\r\n \t<li><em>Determine x and equilibrium concentrations.<\/em> Use the table:<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213823\/CNX_Chem_14_04_ICETable13_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 \u201cA l ( H subscript 2 O ) subscript 6 superscript 3 positive sign plus H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus A l ( H subscript 2 O ) subscript 5 ( O H ) superscript 2 positive sign.\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.10 (which appears in red), negative x, 0.10 minus x. The second column is blank. The third column has the following: approximately 0, x, x. The fourth column has the following: 0, x, x.\" width=\"839\" height=\"227\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n \t<li><em>Solve for x and the equilibrium concentrations<\/em>. Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\right]}{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}\\right]}[\/latex][latex]=\\frac{\\left(x\\right)\\left(x\\right)}{0.10-x}=1.4\\times {10}^{-5}[\/latex]Solving this equation gives:[latex]x=1.2\\times {10}^{-3}M[\/latex]From this we find:[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=0+x=1.2\\times {10}^{-3}M[\/latex][latex]\\text{pH}=\\text{-log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=2.92\\left(\\text{an acidic solution}\\right)[\/latex]<\/li>\r\n \t<li><em>Check the work<\/em>. The arithmetic checks; when 1.2 [latex]\\times [\/latex] 10<sup>\u22123<\/sup><em>M<\/em> is substituted for <em>x<\/em>, the result = <em>K<\/em><sub>a<\/sub>.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nWhat is [latex]\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\right][\/latex] in a 0.15-<em>M<\/em> solution of Al(NO<sub>3<\/sub>)<sub>3<\/sub> that contains enough of the strong acid HNO<sub>3<\/sub> to bring [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] to 0.10 <em>M<\/em>?\r\n[reveal-answer q=\"548370\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"548370\"]2.1 [latex]\\times [\/latex] 10<sup>\u22125<\/sup><em>M<\/em>[\/hidden-answer]\r\n\r\n<\/div>\r\nThe constants for the different stages of ionization are not known for many metal ions, so we cannot calculate the extent of their ionization. However, practically all hydrated metal ions other than those of the alkali metals ionize to give acidic solutions. Ionization increases as the charge of the metal ion increases or as the size of the metal ion decreases.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nThe characteristic properties of aqueous solutions of Br\u00f8nsted-Lowry acids are due to the presence of hydronium ions; those of aqueous solutions of Br\u00f8nsted-Lowry bases are due to the presence of hydroxide ions. The neutralization that occurs when aqueous solutions of acids and bases are combined results from the reaction of the hydronium and hydroxide ions to form water. Some salts formed in neutralization reactions may make the product solutions slightly acidic or slightly basic.\r\n\r\nSolutions that contain salts or hydrated metal ions have a pH that is determined by the extent of the hydrolysis of the ions in the solution. The pH of the solutions may be calculated using familiar equilibrium techniques, or it may be qualitatively determined to be acidic, basic, or neutral depending on the relative <em>K<\/em><sub>a<\/sub> and <em>K<\/em><sub>b<\/sub> of the ions involved.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li>Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Al(NO<sub>3<\/sub>)<sub>3<\/sub><\/li>\r\n \t<li>RbI<\/li>\r\n \t<li>KHCO<sub>2<\/sub><\/li>\r\n \t<li>CH<sub>3<\/sub>NH<sub>3<\/sub>Br<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>FeCl<sub>3<\/sub><\/li>\r\n \t<li>K<sub>2<\/sub>CO<sub>3<\/sub><\/li>\r\n \t<li>NH<sub>4<\/sub>Br<\/li>\r\n \t<li>KClO<sub>4<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Novocaine, C<sub>13<\/sub>H<sub>21<\/sub>O<sub>2<\/sub>N<sub>2<\/sub>Cl, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is 7 [latex]\\times [\/latex] 10<sup>\u22126<\/sup>. Is a solution of novocaine acidic or basic? What are [H<sub>3<\/sub>O<sup>+<\/sup>], [OH<sup>\u2212<\/sup>], and pH of a 2.0% solution by mass of novocaine, assuming that the density of the solution is 1.0 g\/mL.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"434384\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"434384\"]\r\n\r\n2.\u00a0The solutions can be categorized as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>FeCl<sub>3<\/sub> dissociates into Fe<sup>3+<\/sup> ions (acidic metal cation) and Cl<sup>\u2212<\/sup> ions (the conjugate base of a strong acid and therefore essentially neutral). The aqueous solution is therefore acidic.<\/li>\r\n \t<li>K<sub>2<\/sub>CO<sub>3<\/sub> dissociates into K<sup>+<\/sup> ions (neutral metal cation) and [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] ions (the conjugate base of a weak acid and therefore basic). The aqueous solution is therefore basic.<\/li>\r\n \t<li>NH<sub>4<\/sub>Br dissociates into NH<sub>4<\/sub><sup>+<\/sup> ions (a weak acid) and Br<sup>\u2212<\/sup> ions (the conjugate base of a strong acid and therefore essentially neutral). The aqueous solution is therefore acidic.<\/li>\r\n \t<li>KClO<sub>4<\/sub> dissociates into K<sup>+<\/sup> ions (neutral metal cation) and [latex]{\\text{ClO}}_{4}{}^{-}[\/latex] ions (the conjugate base of a strong acid and therefore neutral). The aqueous solution is therefore neutral.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this module, you will be able to:<\/p>\n<ul>\n<li>Predict whether a salt solution will be acidic, basic, or neutral<\/li>\n<li>Calculate the concentrations of the various species in a salt solution<\/li>\n<li>Describe the process that causes solutions of certain metal ions to be acidic<\/li>\n<\/ul>\n<\/div>\n<p>As we have seen in the section on chemical reactions, when an acid and base are mixed, they undergo a neutralization reaction. The word \u201cneutralization\u201d seems to imply that a stoichiometrically equivalent solution of an acid and a base would be neutral. This is sometimes true, but the salts that are formed in these reactions may have acidic or basic properties of their own, as we shall now see.<\/p>\n<h2 data-type=\"title\">Acid-Base Neutralization<\/h2>\n<p>A solution is neutral when it contains equal concentrations of hydronium and hydroxide ions. When we mix solutions of an acid and a base, an acid-base neutralization reaction occurs. However, even if we mix stoichiometrically equivalent quantities, we may find that the resulting solution is not neutral. It could contain either an excess of hydronium ions or an excess of hydroxide ions because the nature of the salt formed determines whether the solution is acidic, neutral, or basic. The following four situations illustrate how solutions with various pH values can arise following a neutralization reaction using stoichiometrically equivalent quantities:<\/p>\n<ol>\n<li>A strong acid and a strong base, such as HCl(<em>aq<\/em>) and NaOH(<em>aq<\/em>) will react to form a neutral solution since the conjugate partners produced are of negligible strength: [latex]\\text{HCl}\\left(aq\\right)+\\text{NaOH}\\left(aq\\right)\\rightleftharpoons \\text{NaCl}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/li>\n<li>A strong acid and a weak base yield a weakly acidic solution, not because of the strong acid involved, but because of the conjugate acid of the weak base.<\/li>\n<li>A weak acid and a strong base yield a weakly basic solution. A solution of a weak acid reacts with a solution of a strong base to form the conjugate base of the weak acid and the conjugate acid of the strong base. The conjugate acid of the strong base is a weaker acid than water and has no effect on the acidity of the resulting solution. However, the conjugate base of the weak acid is a weak base and ionizes slightly in water. This increases the amount of hydroxide ion in the solution produced in the reaction and renders it slightly basic.<\/li>\n<li>A weak acid plus a weak base can yield either an acidic, basic, or neutral solution. This is the most complex of the four types of reactions. When the conjugate acid and the conjugate base are of unequal strengths, the solution can be either acidic or basic, depending on the relative strengths of the two conjugates. Occasionally the weak acid and the weak base will have the <em>same<\/em> strength, so their respective conjugate base and acid will have the same strength, and the solution will be neutral. To predict whether a particular combination will be acidic, basic or neutral, tabulated <em>K<\/em> values of the conjugates must be compared.<\/li>\n<\/ol>\n<div class=\"textbox shaded\">\n<h3>Stomach Antacids<\/h3>\n<p>Our stomachs contain a solution of roughly 0.03 <em>M<\/em> HCl, which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, CaCO<sub>3<\/sub>. The reaction,<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CaCO}}_{3}\\left(s\\right)+2\\text{HCl}\\left(aq\\right)\\rightleftharpoons {\\text{aCl}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\text{CO}}_{2}\\left(g\\right)[\/latex]<\/p>\n<p>not only neutralizes stomach acid, it also produces CO<sub>2<\/sub>(<em>g<\/em>), which may result in a satisfying belch.<\/p>\n<p>Milk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, Mg(OH)<sub>2<\/sub>. It works according to the reaction:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{2+}\\left(aq\\right)+2{\\text{OH}}^{\\text{-}}\\left(aq\\right)[\/latex]<\/p>\n<p>The hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}+{\\text{OH}}^{\\text{-}}\\rightleftharpoons 2{\\text{H}}_{2}\\text{O}\\left(l\\right)[\/latex]<\/p>\n<p>This reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect.<\/p>\n<p>Several antacids have aluminum hydroxide, Al(OH)<sub>3<\/sub>, as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Culinary Aspects of Chemistry<\/h3>\n<p>Cooking is essentially synthetic chemistry that happens to be safe to eat. There are a number of examples of acid-base chemistry in the culinary world. One example is the use of baking soda, or sodium bicarbonate in baking. NaHCO<sub>3<\/sub> is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter \u201crises.\u201d Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.<\/p>\n<p>Many people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure\u00a01). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a \u201csour\u201d taste that we seem to enjoy.<\/p>\n<div style=\"width: 560px\" 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\/23213817\/CNX_Chem_14_03_FishLemon.jpg\" alt=\"An image is shown of two fish with heads removed and skin on with lemon slices placed in the body cavity.\" width=\"550\" height=\"358\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a01. A neutralization reaction takes place between citric acid in lemons or acetic acid in vinegar, and the bases in the flesh of fish.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Pickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour.<\/p>\n<\/div>\n<h2 data-type=\"title\">Salts of Weak Bases and Strong Acids<\/h2>\n<p>When we neutralize a weak base with a strong acid, the product is a salt containing the conjugate acid of the weak base. This conjugate acid is a weak acid. For example, ammonium chloride, NH<sub>4<\/sub>Cl, is a salt formed by the reaction of the weak base ammonia with the strong acid HCl:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{NH}}_{3}\\left(aq\\right)+\\text{HCl}\\left(aq\\right)\\longrightarrow {\\text{NH}}_{4}\\text{Cl}\\left(aq\\right)[\/latex]<\/p>\n<p>A solution of this salt contains ammonium ions and chloride ions. The chloride ion has no effect on the acidity of the solution since HCl is a strong acid. Chloride is a very weak base and will not accept a proton to a measurable extent. However, the ammonium ion, the conjugate acid of ammonia, reacts with water and increases the hydronium ion concentration:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{NH}}_{3}\\left(aq\\right)[\/latex]<\/p>\n<p>The equilibrium equation for this reaction is simply the ionization constant. <em>K<\/em><sub>a<\/sub>, for the acid [latex]{\\text{NH}}_{4}{}^{\\text{+}}:[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{NH}}_{3}\\right]}{\\left[{\\text{NH}}_{4}{}^{\\text{+}}\\right]}={K}_{\\text{a}}[\/latex]<\/p>\n<p>We will not find a value of <em>K<\/em><sub>a<\/sub> for the ammonium ion in <a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\">Ionization Constants of Weak Acids<\/a>. However, it is not difficult to determine <em>K<\/em><sub>a<\/sub> for [latex]{\\text{NH}}_{4}{}^{\\text{+}}[\/latex] from the value of the ionization constant of water, <em>K<\/em><sub>w<\/sub>, and <em>K<\/em><sub>b<\/sub>, the ionization constant of its conjugate base, NH<sub>3<\/sub>, using the following relationship:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{w}}={K}_{\\text{a}}\\times {K}_{\\text{b}}[\/latex]<\/p>\n<p>This relation holds for any base and its conjugate acid or for any acid and its conjugate base.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0The pH of a Solution of a Salt of a Weak Base and a Strong Acid<\/h3>\n<p>Aniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride, [latex]\\left[{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}\\right]\\text{Cl}[\/latex], a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 <em>M<\/em> solution of aniline hydrochloride?<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\left(aq\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q766004\">Show Answer<\/span><\/p>\n<div id=\"q766004\" class=\"hidden-answer\" style=\"display: none\">\n<p>The new step in this example is to determine <em>K<\/em><sub>a<\/sub> for the [latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}[\/latex] ion. The [latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}[\/latex] ion is the conjugate acid of a weak base. The value of <em>K<\/em><sub>a<\/sub> for this acid is not listed in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-acids\/\" target=\"_blank\">Ionization Constants of Weak Acids<\/a>, but we can determine it from the value of <em>K<\/em><sub>b<\/sub> for aniline, C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub>, which is given as 4.6 [latex]\\times[\/latex] 10<sup>\u221210<\/sup> (<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Relative Strengths of Acids and Bases<\/a>\u00a0and <a class=\"target-chapter\" href=\".\/chapter\/ionization-constants-of-weak-bases-2\/\" target=\"_blank\">Ionization Constants of Weak Bases<\/a>):<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}\\left(\\text{ for }{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}\\right)\\times {K}_{\\text{b}}\\left(\\text{ for }{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\right)={K}_{\\text{w}}=1.0\\times {10}^{-14}[\/latex]<br \/>\n[latex]{K}_{\\text{a}}\\left(\\text{for}{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}\\right)=\\frac{{K}_{\\text{w}}}{{K}_{\\text{b}}\\left(\\text{ for }{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{2}\\right)}=\\frac{1.0\\times {10}^{-14}}{4.6\\times {10}^{-10}}=2.2\\times {10}^{-5}[\/latex]<\/p>\n<p>Now we have the ionization constant and the initial concentration of the weak acid, the information necessary to determine the equilibrium concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex], and the pH:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213819\/CNX_Chem_14_04_steps1_img.jpg\" alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cDetermine the direction of change.\u201d The second is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth is labeled \u201cCheck the math.\u201d\" data-media-type=\"image\/jpeg\" \/><br \/>\nWith these steps we find [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 2.3 [latex]\\times[\/latex] 10<sup>\u22123<\/sup><em>M<\/em> and pH = 2.64<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<ol>\n<li>Do the calculations and show that the hydronium ion concentration for a 0.233-<em>M<\/em> solution of [latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}[\/latex] is 2.3 [latex]\\times[\/latex] 10<sup>\u22123<\/sup> and the pH is 2.64.<\/li>\n<li>What is the hydronium ion concentration in a 0.100-<em>M<\/em> solution of ammonium nitrate, NH<sub>4<\/sub>NO<sub>3<\/sub>, a salt composed of the ions [latex]{\\text{NH}}_{4}{}^{\\text{+}}[\/latex] and [latex]{\\text{NO}}_{3}{}^{\\text{-}}[\/latex]. Use the data in\u00a0<a href=\".\/chapter\/relative-strengths-of-acids-and-bases-2\/\" target=\"_blank\">Table\u00a02 on Relative Strengths of Acids and Bases<\/a>\u00a0to\u00a0determine <em>K<\/em><sub>b<\/sub> for the ammonium ion. Which is the stronger acid [latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}[\/latex] or [latex]{\\text{NH}}_{4}{}^{\\text{+}}?[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q407406\">Show Answer<\/span><\/p>\n<div id=\"q407406\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]{K}_{a}\\left(\\text{for}{\\text{NH}}_{4}{}^{\\text{+}}\\right)=5.6\\times {10}^{-10}[\/latex], [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 7.5 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><em>M<\/em><\/li>\n<li>[latex]{\\text{C}}_{6}{\\text{H}}_{5}{\\text{NH}}_{3}{}^{\\text{+}}[\/latex] is the stronger acid.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">Salts of Weak Acids and Strong Bases<\/h2>\n<p>When we neutralize a weak acid with a strong base, we get a salt that contains the conjugate base of the weak acid. This conjugate base is usually a weak base. For example, sodium acetate, NaCH<sub>3<\/sub>CO<sub>2<\/sub>, is a salt formed by the reaction of the weak acid acetic acid with the strong base sodium hydroxide:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+\\text{NaOH}\\left(aq\\right)\\longrightarrow {\\text{NaCH}}_{3}{\\text{CO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(aq\\right)[\/latex]<\/p>\n<p>A solution of this salt contains sodium ions and acetate ions. The sodium ion, as the conjugate acid of a strong base, has no effect on the acidity of the solution. However, the acetate ion, the conjugate base of acetic acid, reacts with water and increases the concentration of hydroxide ion:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\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}}^{\\text{-}}\\left(aq\\right)[\/latex]<\/p>\n<p>The equilibrium equation for this reaction is the ionization constant, <em>K<\/em><sub>b<\/sub>, for the base [latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}[\/latex]. The value of <em>K<\/em><sub>b<\/sub> can be calculated from the value of the ionization constant of water, <em>K<\/em><sub>w<\/sub>, and <em>K<\/em><sub>a<\/sub>, the ionization constant of the conjugate acid of the anion using the equation:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{w}}={K}_{\\text{a}}\\times {K}_{\\text{b}}[\/latex]<\/p>\n<p>For the acetate ion and its conjugate acid we have:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{b}}\\left(\\text{for}{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right)=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}\\left(\\text{for}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right)}=\\frac{1.0\\times {10}^{-14}}{1.8\\times {10}^{-5}}=5.6\\times {10}^{-10}[\/latex]<\/p>\n<p>Some handbooks do not report values of <em>K<\/em><sub>b<\/sub>. They only report ionization constants for acids. If we want to determine a <em>K<\/em><sub>b<\/sub> value using one of these handbooks, we must look up the value of <em>K<\/em><sub>a<\/sub> for the conjugate acid and convert it to a <em>K<\/em><sub>b<\/sub> value.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base<\/h3>\n<p>Determine the acetic acid concentration in a solution with [latex]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]=0.050M[\/latex] and [OH<sup>\u2212<\/sup>] = 2.5 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><em>M<\/em> at equilibrium. The reaction is:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\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}}^{\\text{-}}\\left(aq\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q597974\">Show Answer<\/span><\/p>\n<div id=\"q597974\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are given two of three equilibrium concentrations and asked to find the missing concentration. If we can find the equilibrium constant for the reaction, the process is straightforward.<\/p>\n<p>The acetate ion behaves as a base in this reaction; hydroxide ions are a product. We determine <em>K<\/em><sub>b<\/sub> as follows:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{b}}\\left(\\text{ for }{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right)=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}\\left(\\text{ for }{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right)}=\\frac{1.0\\times {10}^{-14}}{1.8\\times {10}^{-5}}=5.6\\times {10}^{-10}[\/latex]<\/p>\n<p>Now find the missing concentration:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{\\text{-}}\\right]}=5.6\\times {10}^{-10}[\/latex]<br \/>\n[latex]=\\frac{\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]\\left(2.5\\times {10}^{-6}\\right)}{\\left(0.050\\right)}=5.6\\times {10}^{-10}[\/latex]<\/p>\n<p>Solving this equation we get [CH<sub>3<\/sub>CO<sub>2<\/sub>H] = 1.1 [latex]\\times[\/latex] 10<sup>\u22125<\/sup><em>M<\/em>.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>What is the pH of a 0.083-<em>M<\/em> solution of CN<sup>\u2212<\/sup>? Use 4.0 [latex]\\times[\/latex] 10<sup>\u221210<\/sup> as <em>K<\/em><sub>a<\/sub> for HCN. Hint: We will probably need to convert pOH to pH or find [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] using [OH<sup>\u2212<\/sup>] in the final stages of this problem.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q680253\">Show Answer<\/span><\/p>\n<div id=\"q680253\" class=\"hidden-answer\" style=\"display: none\">11.16<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">Equilibrium in a Solution of a Salt of a Weak Acid and a Weak Base<\/h2>\n<p>In a solution of a salt formed by the reaction of a weak acid and a weak base, to predict the pH, we must know both the <em>K<\/em><sub>a<\/sub> of the weak acid and the <em>K<\/em><sub>b<\/sub> of the weak base. If <em>K<\/em><sub>a<\/sub> &gt; <em>K<\/em><sub>b<\/sub>, the solution is acidic, and if <em>K<\/em><sub>b<\/sub> &gt; <em>K<\/em><sub>a<\/sub>, the solution is basic.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3:\u00a0Determining the Acidic or Basic Nature of Salts<\/h3>\n<p>Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:<\/p>\n<ol>\n<li>KBr<\/li>\n<li>NaHCO<sub>3<\/sub><\/li>\n<li>NH<sub>4<\/sub>Cl<\/li>\n<li>Na<sub>2<\/sub>HPO<sub>4<\/sub><\/li>\n<li>NH<sub>4<\/sub>F<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q54649\">Show Answer<\/span><\/p>\n<div id=\"q54649\" class=\"hidden-answer\" style=\"display: none\">\n<p>Consider each of the ions separately in terms of its effect on the pH of the solution, as shown here:<\/p>\n<ol>\n<li>The K<sup>+<\/sup> cation and the Br<sup>\u2212<\/sup> anion are both spectators, since they are the cation of a strong base (KOH) and the anion of a strong acid (HBr), respectively. The solution is neutral.<\/li>\n<li>The Na<sup>+<\/sup> cation is a spectator, and will not affect the pH of the solution; while the [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] anion is amphiprotic, it could either behave as an acid or a base. The <em>K<\/em><sub>a<\/sub> of [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] is 4.7 [latex]\\times[\/latex] 10<sup>\u221211<\/sup>, so the <em>K<\/em><sub>b<\/sub> of its conjugate base is [latex]\\frac{1.0\\times {10}^{-14}}{4.7\\times {10}^{-11}}=2.1\\times {10}^{-4}[\/latex].<br \/>\nSince <em>K<\/em><sub>b<\/sub> &gt;&gt; <em>K<\/em><sub>a<\/sub>, the solution is basic.<\/li>\n<li>The [latex]{\\text{NH}}_{4}{}^{\\text{+}}[\/latex] ion is acidic and the Cl<sup>\u2212<\/sup> ion is a spectator. The solution will be acidic.<\/li>\n<li>The Na<sup>+<\/sup> ion is a spectator, while the [latex]{\\text{HPO}}_{4}{}^{-}[\/latex] ion is amphiprotic, with a <em>K<\/em><sub>a<\/sub> of 3.6 [latex]\\times[\/latex] 10<sup>\u221213<\/sup><br \/>\nso that the <em>K<\/em><sub>b<\/sub> of its conjugate base is [latex]\\frac{1.0\\times {10}^{-14}}{3.6\\times {10}^{-13}}=2.8\\times {10}^{-2}[\/latex]. Because <em>K<\/em><sub>b<\/sub> &gt;&gt; <em>K<\/em><sub>a<\/sub>, the solution is basic.<\/li>\n<li>The [latex]{\\text{NH}}_{4}{}^{\\text{+}}[\/latex] ion is listed as being acidic, and the F<sup>\u2212<\/sup> ion is listed as a base, so we must directly compare the <em>K<\/em><sub>a<\/sub> and the <em>K<\/em><sub>b<\/sub> of the two ions. <em>K<\/em><sub>a<\/sub> of [latex]{\\text{NH}}_{4}{}^{\\text{+}}[\/latex] is 5.6 [latex]\\times[\/latex] 10<sup>\u221210<\/sup>, which seems very small, yet the <em>K<\/em><sub>b<\/sub> of F<sup>\u2212<\/sup> is 1.4 [latex]\\times[\/latex] 10<sup>\u221211<\/sup>, so the solution is acidic, since <em>K<\/em><sub>a<\/sub> &gt; <em>K<\/em><sub>b<\/sub>.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:<\/p>\n<ol>\n<li>K<sub>2<\/sub>CO<sub>3<\/sub><\/li>\n<li>CaCl<sub>2<\/sub><\/li>\n<li>KH<sub>2<\/sub>PO<sub>4<\/sub><\/li>\n<li>(NH<sub>4<\/sub>)<sub>2<\/sub>CO<sub>3<\/sub><\/li>\n<li>AlBr<sub>3<\/sub><\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q774628\">Show Answer<\/span><\/p>\n<div id=\"q774628\" class=\"hidden-answer\" style=\"display: none\">(1) basic; (2) neutral; (3) basic; (4) basic; (5) acidic<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">The Ionization of Hydrated Metal Ions<\/h2>\n<p>If we measure the pH of the solutions of a variety of metal ions we will find that these ions act as weak acids when in solution. The aluminum ion is an example. When aluminum nitrate dissolves in water, the aluminum ion reacts with water to give a hydrated aluminum ion, [latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}[\/latex], dissolved in bulk water. What this means is that the aluminum ion has the strongest interactions with the six closest water molecules (the so-called first solvation shell), even though it does interact with the other water molecules surrounding this [latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}[\/latex] cluster as well:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Al}{\\left({\\text{NO}}_{3}\\right)}_{3}\\left(s\\right)+6{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow \\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}\\left(aq\\right)+3{\\text{NO}}_{3}{}^{\\text{-}}\\left(aq\\right)[\/latex]<\/p>\n<p>We frequently see the formula of this ion simply as \u201cAl<sup>3+<\/sup>(<em>aq<\/em>)\u201d, without explicitly noting the six water molecules that are the closest ones to the aluminum ion and just describing the ion as being solvated in water (hydrated). This is similar to the simplification of the formula of the hydronium ion, [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] to H<sup>+<\/sup>. However, in this case, the hydrated aluminum ion is a weak acid (Figure\u00a02) and donates a proton to a water molecule. Thus, the hydration becomes important and we may use formulas that show the extent of hydration:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\left(aq\\right){K}_{\\text{a}}=1.4\\times {10}^{-5}[\/latex]<\/p>\n<p>As with other polyprotic acids, the hydrated aluminum ion ionizes in stages, as shown by:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\left(aq\\right)[\/latex]<br \/>\n[latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{4}{\\left(\\text{OH}\\right)}_{2}{}^{\\text{+}}\\left(aq\\right)[\/latex]<br \/>\n[latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{4}{\\left(\\text{OH}\\right)}_{2}{}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{3}{\\left(\\text{OH}\\right)}_{3}\\left(aq\\right)[\/latex]<\/p>\n<p>Note that some of these aluminum species are exhibiting amphiprotic behavior, since they are acting as acids when they appear on the right side of the equilibrium expressions and as bases when they appear on the left side.<\/p>\n<div style=\"width: 888px\" 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\/23213820\/CNX_Chem_14_04_hydronium.jpg\" alt=\"A reaction is shown using ball and stick models. On the left, inside brackets with a superscript of 3 plus outside to the right is structure labeled \u201c[ A l ( H subscript 2 O ) subscript 6 ] superscript 3 plus.\u201d Inside the brackets is s central grey atom to which 6 red atoms are bonded in an arrangement that distributes them evenly about the central grey atom. Each red atom has two smaller white atoms attached in a forked or bent arrangement. Outside the brackets to the right is a space-filling model that includes a red central sphere with two smaller white spheres attached in a bent arrangement. Beneath this structure is the label \u201cH subscript 2 O.\u201d A double sided arrow follows. Another set of brackets follows to the right of the arrows which have a superscript of two plus outside to the right. The structure inside the brackets is similar to that on the left, except a white atom is removed from the structure. The label below is also changed to \u201c[ A l ( H subscript 2 O ) subscript 5 O H ] superscript 2 plus.\u201d To the right of this structure and outside the brackets is a space filling model with a central red sphere to which 3 smaller white spheres are attached. This structure is labeled \u201cH subscript 3 O superscript plus.\u201d\" width=\"878\" height=\"299\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a03. When an aluminum ion reacts with water, the hydrated aluminum ion becomes a weak acid.<\/p>\n<\/div>\n<p>However, the ionization of a cation carrying more than one charge is usually not extensive beyond the first stage. Additional examples of the first stage in the ionization of hydrated metal ions are:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\left(aq\\right){K}_{\\text{a}}=2.74[\/latex]<br \/>\n[latex]\\text{Cu}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{2+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Cu}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{\\text{+}}\\left(aq\\right){K}_{\\text{a}}=~6.3[\/latex]<br \/>\n[latex]\\text{Zn}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{4}{}^{2+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Zn}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{3}{\\left(\\text{OH}\\right)}^{\\text{+}}\\left(aq\\right){K}_{\\text{a}}=9.6[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4:\u00a0Hydrolysis of [Al(H<sub>2<\/sub>O)<sub>6<\/sub>]<sup>3+<\/sup><\/h3>\n<p>Calculate the pH of a 0.10-<em>M<\/em> solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion [latex]{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}\\right]}^{3+}[\/latex] in solution.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q103892\">Show Answer<\/span><\/p>\n<div id=\"q103892\" class=\"hidden-answer\" style=\"display: none\">\n<p>In spite of the unusual appearance of the acid, this is a typical acid ionization problem.<\/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\/23213822\/CNX_Chem_14_04_steps2_img.jpg\" alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cDetermine the direction of change.\u201d The second is labeled \u201cDetermine x and the equilibrium concentrations.\u201d The third is labeled \u201cSolve for x and the equilibrium concentrations.\u201d The fourth is labeled \u201cCheck the math.\u201d\" width=\"876\" height=\"155\" data-media-type=\"image\/jpeg\" \/><\/p>\n<ol>\n<li><em>Determine the direction of change<\/em>. The equation for the reaction and <em>K<\/em><sub>a<\/sub> are:[latex]\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\left(aq\\right){K}_{\\text{a}}=1.4\\times {10}^{-5}[\/latex]The reaction shifts to the right to reach equilibrium.<\/li>\n<li><em>Determine x and equilibrium concentrations.<\/em> Use the table:<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\/23213823\/CNX_Chem_14_04_ICETable13_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 \u201cA l ( H subscript 2 O ) subscript 6 superscript 3 positive sign plus H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus A l ( H subscript 2 O ) subscript 5 ( O H ) superscript 2 positive sign.\u201d Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.10 (which appears in red), negative x, 0.10 minus x. The second column is blank. The third column has the following: approximately 0, x, x. The fourth column has the following: 0, x, x.\" width=\"839\" height=\"227\" data-media-type=\"image\/jpeg\" \/><\/li>\n<li><em>Solve for x and the equilibrium concentrations<\/em>. Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\right]}{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}{}^{3+}\\right]}[\/latex][latex]=\\frac{\\left(x\\right)\\left(x\\right)}{0.10-x}=1.4\\times {10}^{-5}[\/latex]Solving this equation gives:[latex]x=1.2\\times {10}^{-3}M[\/latex]From this we find:[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=0+x=1.2\\times {10}^{-3}M[\/latex][latex]\\text{pH}=\\text{-log}\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=2.92\\left(\\text{an acidic solution}\\right)[\/latex]<\/li>\n<li><em>Check the work<\/em>. The arithmetic checks; when 1.2 [latex]\\times[\/latex] 10<sup>\u22123<\/sup><em>M<\/em> is substituted for <em>x<\/em>, the result = <em>K<\/em><sub>a<\/sub>.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>What is [latex]\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}{\\left(\\text{OH}\\right)}^{2+}\\right][\/latex] in a 0.15-<em>M<\/em> solution of Al(NO<sub>3<\/sub>)<sub>3<\/sub> that contains enough of the strong acid HNO<sub>3<\/sub> to bring [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] to 0.10 <em>M<\/em>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q548370\">Show Answer<\/span><\/p>\n<div id=\"q548370\" class=\"hidden-answer\" style=\"display: none\">2.1 [latex]\\times[\/latex] 10<sup>\u22125<\/sup><em>M<\/em><\/div>\n<\/div>\n<\/div>\n<p>The constants for the different stages of ionization are not known for many metal ions, so we cannot calculate the extent of their ionization. However, practically all hydrated metal ions other than those of the alkali metals ionize to give acidic solutions. Ionization increases as the charge of the metal ion increases or as the size of the metal ion decreases.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>The characteristic properties of aqueous solutions of Br\u00f8nsted-Lowry acids are due to the presence of hydronium ions; those of aqueous solutions of Br\u00f8nsted-Lowry bases are due to the presence of hydroxide ions. The neutralization that occurs when aqueous solutions of acids and bases are combined results from the reaction of the hydronium and hydroxide ions to form water. Some salts formed in neutralization reactions may make the product solutions slightly acidic or slightly basic.<\/p>\n<p>Solutions that contain salts or hydrated metal ions have a pH that is determined by the extent of the hydrolysis of the ions in the solution. The pH of the solutions may be calculated using familiar equilibrium techniques, or it may be qualitatively determined to be acidic, basic, or neutral depending on the relative <em>K<\/em><sub>a<\/sub> and <em>K<\/em><sub>b<\/sub> of the ions involved.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li>Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Al(NO<sub>3<\/sub>)<sub>3<\/sub><\/li>\n<li>RbI<\/li>\n<li>KHCO<sub>2<\/sub><\/li>\n<li>CH<sub>3<\/sub>NH<sub>3<\/sub>Br<\/li>\n<\/ol>\n<\/li>\n<li>Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>FeCl<sub>3<\/sub><\/li>\n<li>K<sub>2<\/sub>CO<sub>3<\/sub><\/li>\n<li>NH<sub>4<\/sub>Br<\/li>\n<li>KClO<sub>4<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Novocaine, C<sub>13<\/sub>H<sub>21<\/sub>O<sub>2<\/sub>N<sub>2<\/sub>Cl, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is 7 [latex]\\times[\/latex] 10<sup>\u22126<\/sup>. Is a solution of novocaine acidic or basic? What are [H<sub>3<\/sub>O<sup>+<\/sup>], [OH<sup>\u2212<\/sup>], and pH of a 2.0% solution by mass of novocaine, assuming that the density of the solution is 1.0 g\/mL.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q434384\">Show Selected Answers<\/span><\/p>\n<div id=\"q434384\" class=\"hidden-answer\" style=\"display: none\">\n<p>2.\u00a0The solutions can be categorized as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>FeCl<sub>3<\/sub> dissociates into Fe<sup>3+<\/sup> ions (acidic metal cation) and Cl<sup>\u2212<\/sup> ions (the conjugate base of a strong acid and therefore essentially neutral). The aqueous solution is therefore acidic.<\/li>\n<li>K<sub>2<\/sub>CO<sub>3<\/sub> dissociates into K<sup>+<\/sup> ions (neutral metal cation) and [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] ions (the conjugate base of a weak acid and therefore basic). The aqueous solution is therefore basic.<\/li>\n<li>NH<sub>4<\/sub>Br dissociates into NH<sub>4<\/sub><sup>+<\/sup> ions (a weak acid) and Br<sup>\u2212<\/sup> ions (the conjugate base of a strong acid and therefore essentially neutral). The aqueous solution is therefore acidic.<\/li>\n<li>KClO<sub>4<\/sub> dissociates into K<sup>+<\/sup> ions (neutral metal cation) and [latex]{\\text{ClO}}_{4}{}^{-}[\/latex] ions (the conjugate base of a strong acid and therefore neutral). The aqueous solution is therefore neutral.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\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-3486\">\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. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.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>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/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":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3486","chapter","type-chapter","status-publish","hentry"],"part":2988,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3486","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3486\/revisions"}],"predecessor-version":[{"id":6062,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3486\/revisions\/6062"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/parts\/2988"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/3486\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/media?parent=3486"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapter-type?post=3486"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/contributor?post=3486"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/license?post=3486"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}