{"id":3496,"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=3496"},"modified":"2015-08-26T18:50:26","modified_gmt":"2015-08-26T18:50:26","slug":"polyprotic-acids-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/chapter\/polyprotic-acids-2\/","title":{"raw":"Polyprotic Acids","rendered":"Polyprotic Acids"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\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>Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idp98649872\">We can classify acids by the number of protons per molecule that they can give up in a reaction. Acids such as HCl, HNO<sub>3<\/sub>, and HCN that contain one ionizable hydrogen atom in each molecule are called <strong><span data-type=\"term\">monoprotic acids<\/span><\/strong>. Their reactions with water are:<\/p>\r\n\r\n<div id=\"fs-idp136743504\" data-type=\"equation\">[latex]\\begin{array}{c}\\text{HCl}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{\\text{-}}\\left(aq\\right)\\\\ {\\text{HNO}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{NO}}_{3}{}^{\\text{-}}\\left(aq\\right)\\\\ \\text{HCN}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CN}}^{\\text{-}}\\left(aq\\right)\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idp136804288\">Even though it contains four hydrogen atoms, acetic acid, CH<sub>3<\/sub>CO<sub>2<\/sub>H, is also monoprotic because only the hydrogen atom from the carboxyl group (COOH) reacts with bases:<span id=\"fs-idm11559264\" data-type=\"media\" data-alt=\"This image contains two equilibrium reactions. The first shows a C atom bonded to three H atoms and another C atom. The second C atom is double bonded to an O atom and also forms a single bond to another O atom. The second O atom is bonded to an H atom. There is a plus sign and then the molecular formula H subscript 2 O. An equilibrium arrow follows the H subscript 2 O. To the right of the arrow is H subscript 3 O superscript positive sign. There is a plus sign. The final structure shows a C atom bonded the three H atoms and another C atom. This second C atom is double bonded to an O atom and single bonded to another O atom. The entire structure is in brackets and a superscript negative sign appears outside the brackets. The second reaction shows C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ).\">\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213825\/CNX_Chem_14_05_acetic_img.jpg\" alt=\"This image contains two equilibrium reactions. The first shows a C atom bonded to three H atoms and another C atom. The second C atom is double bonded to an O atom and also forms a single bond to another O atom. The second O atom is bonded to an H atom. There is a plus sign and then the molecular formula H subscript 2 O. An equilibrium arrow follows the H subscript 2 O. To the right of the arrow is H subscript 3 O superscript positive sign. There is a plus sign. The final structure shows a C atom bonded the three H atoms and another C atom. This second C atom is double bonded to an O atom and single bonded to another O atom. The entire structure is in brackets and a superscript negative sign appears outside the brackets. The second reaction shows C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ).\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\r\n<p id=\"fs-idp199141696\">Similarly, monoprotic bases are bases that will accept a single proton.<\/p>\r\n<p id=\"fs-idp13686896\"><strong><span data-type=\"term\">Diprotic acids<\/span><\/strong> contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:<\/p>\r\n\r\n<div id=\"fs-idm64885744\" data-type=\"equation\">[latex]\\begin{array}{l}\\\\ \\\\ \\text{First ionization:}{\\text{H}}_{2}{\\text{SO}}_{4}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HSO}}_{4}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}1}=\\text{about}{10}^{\\text{2+}}\\\\ \\text{Second ionization:}{\\text{HSO}}_{4}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{SO}}_{4}{}^{2-}\\left(aq\\right){K}_{\\text{a}2}=1.2\\times {10}^{-2}\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idp137460528\">This <strong><span data-type=\"term\">stepwise ionization<\/span><\/strong> process occurs for all polyprotic acids. When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of acids. Carbonic acid, H<sub>2<\/sub>CO<sub>3<\/sub>, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.<\/p>\r\n\r\n<div id=\"fs-idm41501456\" data-type=\"equation\">[latex]\\begin{array}{l}\\text{First ionization:}\\\\ {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right){K}_{{\\text{H}}_{2}{\\text{CO}}_{3}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=4.3\\times {10}^{-7}\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idp84456096\">The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.<\/p>\r\n\r\n<div id=\"fs-idm146392784\" data-type=\"equation\">[latex]\\begin{array}{l}\\text{Second ionization:}\\\\ {\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right){K}_{{\\text{HCO}}_{3}{}^{\\text{-}}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right]}=4.7\\times {10}^{-11}\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idm74401760\">[latex]{K}_{{\\text{H}}_{2}{\\text{CO}}_{3}}[\/latex] is larger than [latex]{K}_{{\\text{HCO}}_{3}{}^{\\text{-}}}[\/latex] by a factor of 10<sup>4<\/sup>, so H<sub>2<\/sub>CO<sub>3<\/sub> is the dominant producer of hydronium ion in the solution. This means that little of the [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] formed by the ionization of H<sub>2<\/sub>CO<sub>3<\/sub> ionizes to give hydronium ions (and carbonate ions), and the concentrations of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] and [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] are practically equal in a pure aqueous solution of H<sub>2<\/sub>CO<sub>3<\/sub>.<\/p>\r\n<p id=\"fs-idm68059056\">If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This can simplify our work considerably because we can determine the concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] and the conjugate base from the first ionization, then determine the concentration of the conjugate base of the second ionization in a solution with concentrations determined by the first ionization.<\/p>\r\n\r\n<div id=\"fs-idp1592608\" class=\"textbox shaded\" data-type=\"example\">\r\n<h3>Example 1<\/h3>\r\n<h4 id=\"fs-idp100346864\"><span data-type=\"title\">Ionization of a Diprotic Acid<\/span><\/h4>\r\nWhen we buy soda water (carbonated water), we are buying a solution of carbon dioxide in water. The solution is acidic because CO<sub>2<\/sub> reacts with water to form carbonic acid, H<sub>2<\/sub>CO<sub>3<\/sub>. What are [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex], [latex]\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right][\/latex], and [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] in a saturated solution of CO<sub>2<\/sub> with an initial [H<sub>2<\/sub>CO<sub>3<\/sub>] = 0.033 <em data-effect=\"italics\">M<\/em>?\r\n<div id=\"fs-idp126184336\" data-type=\"equation\">[latex]{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}1}=4.3\\times {10}^{-7}[\/latex]<\/div>\r\n<div id=\"fs-idp136502080\" data-type=\"equation\">[latex]{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right){K}_{\\text{a2}}=4.7\\times {10}^{-11}[\/latex]<\/div>\r\n<h4 id=\"fs-idm27159024\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nAs indicated by the ionization constants, H<sub>2<\/sub>CO<sub>3<\/sub> is a much stronger acid than [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex], so H<sub>2<\/sub>CO<sub>3<\/sub> is the dominant producer of hydronium ion in solution. Thus there are two parts in the solution of this problem: (1) Using the customary four steps, we determine the concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] and [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] produced by ionization of H<sub>2<\/sub>CO<sub>3<\/sub>. (2) Then we determine the concentration of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] in a solution with the concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] and [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] determined in (1). To summarize:<span id=\"fs-idm28185792\" data-type=\"media\" data-alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cleft bracket H subscript 2 C O subscript 3 right bracket.\u201d The second is labeled \u201cleft bracket H subscript 3 O superscript plus right bracket and left bracket H C O subscript 3 superscript negative right bracket from H subscript 2 C O subscript 3.\u201d The third is labeled \u201cleft bracket C O subscript 3 superscript 2 negative right bracket from H C O subscript 3 superscript negative.\u201d\">\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213827\/CNX_Chem_14_05_steps1_img.jpg\" alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cleft bracket H subscript 2 C O subscript 3 right bracket.\u201d The second is labeled \u201cleft bracket H subscript 3 O superscript plus right bracket and left bracket H C O subscript 3 superscript negative right bracket from H subscript 2 C O subscript 3.\u201d The third is labeled \u201cleft bracket C O subscript 3 superscript 2 negative right bracket from H C O subscript 3 superscript negative.\u201d\" width=\"881\" height=\"156\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<ol id=\"fs-idp108659776\" class=\"stepwise\" data-number-style=\"arabic\">\r\n\t<li><em data-effect=\"italics\">Determine the concentrations of<\/em> [latex]{H}_{3}{O}^{\\text{+}}[\/latex] <em data-effect=\"italics\">and<\/em> [latex]HC{O}_{3}{}^{\\text{-}}[\/latex].\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idm22716864\" data-type=\"equation\">[latex]{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}1}=4.3\\times {10}^{-7}[\/latex]<\/div>\r\nAs for the ionization of any other weak acid:<span id=\"fs-idp99810176\" data-type=\"media\" data-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\">\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213828\/CNX_Chem_14_05_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=\"842\" height=\"149\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\nAn abbreviated table of changes and concentrations shows:<span id=\"fs-idm37059488\" data-type=\"media\" data-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 constant ( M ). The second column has the header of \u201cH subscript 2 C O subscript 3 plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus sign H C O subscript 3 superscript negative sign.\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.033, negative sign x, 0.033 minus sign x. The second column has the following: approximately 0, x, x. The third column has the following: 0, x, x.\">\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213830\/CNX_Chem_14_05_ICETable1_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 constant ( M ). The second column has the header of \u201cH subscript 2 C O subscript 3 plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus sign H C O subscript 3 superscript negative sign.\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.033, negative sign x, 0.033 minus sign x. The second column has the following: approximately 0, x, x. The third column has the following: 0, x, x.\" width=\"838\" height=\"227\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\nSubstituting the equilibrium concentrations into the equilibrium gives us:\r\n<div id=\"fs-idm26667456\" data-type=\"equation\">[latex]{K}_{{\\text{H}}_{2}{\\text{CO}}_{3}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{0.033-x}=4.3\\times {10}^{-7}[\/latex]<\/div>\r\nSolving the preceding equation making our standard assumptions gives:\r\n<div id=\"fs-idm73884976\" data-type=\"equation\">[latex]x=1.2\\times {10}^{-4}[\/latex]<\/div>\r\nThus:\r\n<div id=\"fs-idm57413264\" data-type=\"equation\">[latex]\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]=0.033M[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idp20713136\" data-type=\"equation\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right]=1.2\\times {10}^{-4}M[\/latex]<\/div><\/li>\r\n\t<li><em data-effect=\"italics\">Determine the concentration of [latex]C{O}_{3}{}^{2-}[\/latex] in a solution at equilibrium with<\/em> [latex]\\left[{H}_{3}{O}^{\\text{+}}\\right][\/latex] <em data-effect=\"italics\">and<\/em> [latex]\\left[HC{O}_{3}{}^{\\text{-}}\\right][\/latex] <em data-effect=\"italics\">both equal to 1.2<\/em> [latex]\\times [\/latex] <em data-effect=\"italics\">10<sup>\u22124<\/sup> M<\/em>.\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idp103313952\" data-type=\"equation\">[latex]{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idm27760608\" data-type=\"equation\">[latex]{K}_{{\\text{HCO}}_{3}{}^{\\text{-}}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right]}=\\frac{\\left(1.2\\times {10}^{-4}\\right)\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{1.2\\times {10}^{-4}}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div id=\"fs-idm37931200\" data-type=\"equation\">[latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right]=\\frac{\\left(4.7\\times {10}^{-11}\\right)\\left(1.2\\times {10}^{-4}\\right)}{1.2\\times {10}^{-4}}=4.7\\times {10}^{-11}M[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<p id=\"fs-idp167647504\">To summarize: In part 1 of this example, we found that the H<sub>2<\/sub>CO<sub>3<\/sub> in a 0.033-<em data-effect=\"italics\">M<\/em> solution ionizes slightly and at equilibrium [H<sub>2<\/sub>CO<sub>3<\/sub>] = 0.033 <em data-effect=\"italics\">M<\/em>; [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 1.2 [latex]\\times [\/latex] 10<sup>\u22124<\/sup>; and [latex]\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right]=1.2\\times {10}^{-4}M[\/latex]. In part 2, we determined that [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right]=4.7\\times {10}^{-11}M[\/latex].<\/p>\r\n\r\n<h4 id=\"fs-idp3314688\"><span data-type=\"title\">Check Your Learning<\/span><\/h4>\r\nThe concentration of H<sub>2<\/sub>S in a saturated aqueous solution at room temperature is approximately 0.1 <em data-effect=\"italics\">M<\/em>. Calculate [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex], [HS<sup>\u2212<\/sup>], and [S<sup>2\u2212<\/sup>] in the solution:\r\n<div id=\"fs-idp170235264\" data-type=\"equation\">[latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HS}}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}1}=1.0\\times {10}^{-7}[\/latex]<\/div>\r\n<div id=\"fs-idp22220080\" data-type=\"equation\">[latex]{\\text{HS}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right){K}_{\\text{a}2}=1.0\\times {10}^{-19}[\/latex]<\/div>\r\n<div id=\"fs-idm58434032\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:\u00a0<\/strong>[H<sub>2<\/sub>S] = 0.1 <em data-effect=\"italics\">M<\/em>; [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = [HS<sup>\u2212<\/sup>] = 0.0001 <em data-effect=\"italics\">M<\/em>; [S<sup>2\u2212<\/sup>] = 1 [latex]\\times [\/latex] 10<sup>\u221219<\/sup><em data-effect=\"italics\">M<\/em><\/div>\r\n&nbsp;\r\n\r\nWe note that the concentration of the sulfide ion is the same as <em data-effect=\"italics\">K<\/em><sub>a2<\/sub>. This is due to the fact that each subsequent dissociation occurs to a lesser degree (as acid gets weaker).\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idm49318672\">A <b>triprotic acid<\/b> is an acid that has three dissociable protons that undergo stepwise ionization: Phosphoric acid is a typical example:<\/p>\r\n\r\n<div id=\"fs-idm72609920\" data-type=\"equation\">[latex]\\begin{array}{l}\\text{First ionization:}{\\text{H}}_{3}{\\text{PO}}_{4}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}{\\text{PO}}_{4}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}1}=7.5\\times {10}^{-3}\\\\ \\text{Second ionization:}{\\text{H}}_{2}{\\text{PO}}_{4}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HPO}}_{4}{}^{2-}\\left(aq\\right){K}_{\\text{a}2}=6.3\\times {10}^{-8}\\\\ \\text{Third ionization:}{\\text{HPO}}_{4}{}^{2-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{PO}}_{4}{}^{3-}\\left(aq\\right){K}_{\\text{a}3}=3.6\\times {10}^{-13}\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-idp15743456\">As with the diprotic acids, the differences in the ionization constants of these reactions tell us that in each successive step the degree of ionization is significantly weaker. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 10<sup>5<\/sup> to 10<sup>6<\/sup>.<\/p>\r\n<p id=\"fs-idp108834896\">This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H<sub>3<\/sub>PO<sub>4<\/sub> complicated. However, because the successive ionization constants differ by a factor of 10<sup>5<\/sup> to 10<sup>6<\/sup>, the calculations can be broken down into a series of parts similar to those for diprotic acids.<\/p>\r\n<p id=\"fs-idp178452064\">Polyprotic bases can accept more than one hydrogen ion in solution. The carbonate ion is an example of a <b>diprotic base<\/b>, since it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions:<\/p>\r\n\r\n<div id=\"fs-idm16615264\" data-type=\"equation\">[latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)\\rightleftharpoons {\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right)\\text{and}{\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)\\rightleftharpoons {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right)[\/latex]<\/div>\r\n<section id=\"fs-idm40783936\" class=\"summary\" data-depth=\"1\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h2>Key Concepts and Summary<\/h2>\r\n<p id=\"fs-idm21286928\">An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude. As long as the difference between the successive values of <em data-effect=\"italics\">K<\/em><sub>a<\/sub> of the acid is greater than about a factor of 20, it is appropriate to break down the calculations of the concentrations of the ions in solution into a series of steps.<\/p>\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-idm45487824\" class=\"exercises\" data-depth=\"1\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Chemistry End of Chapter Exercises<\/h3>\r\n<div id=\"fs-idp94973072\" data-type=\"exercise\">\r\n<div id=\"fs-idp149185520\" data-type=\"problem\">\r\n<ol>\r\n\t<li id=\"fs-idm49321712\">Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134-<em data-effect=\"italics\">M<\/em> solution of H<sub>2<\/sub>CO<sub>3<\/sub>, a diprotic acid: [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex], [OH<sup>\u2212<\/sup>], [H<sub>2<\/sub>CO<sub>3<\/sub>], [latex]\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right][\/latex], [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right]?[\/latex] No calculations are needed to answer this question.<\/li>\r\n\t<li>Calculate the concentration of each species present in a 0.050-<em data-effect=\"italics\">M<\/em> solution of H<sub>2<\/sub>S.<\/li>\r\n\t<li>Calculate the concentration of each species present in a 0.010-<em data-effect=\"italics\">M<\/em> solution of phthalic acid, C<sub>6<\/sub>H<sub>4<\/sub>(CO<sub>2<\/sub>H)<sub>2<\/sub>.<\/li>\r\n\t<li>Salicylic acid, HOC<sub>6<\/sub>H<sub>4<\/sub>CO<sub>2<\/sub>H, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.\r\n<ol>\r\n\t<li>(a) Both functional groups of salicylic acid ionize in water, with <em data-effect=\"italics\">K<\/em><sub>a<\/sub> = 1.0 [latex]\\times [\/latex] 10<sup>\u22123<\/sup> for the\u2014CO<sub>2<\/sub>H group and 4.2 [latex]\\times [\/latex] 10<sup>\u221213<\/sup> for the \u2212OH group. What is the pH of a saturated solution of the acid (solubility = 1.8 g\/L).<\/li>\r\n\t<li>(b) Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, CH<sub>3<\/sub>CO<sub>2<\/sub>C<sub>6<\/sub>H<sub>4<\/sub>CO<sub>2<\/sub>H. The \u2212CO<sub>2<\/sub>H functional group is still present, but its acidity is reduced, <em data-effect=\"italics\">K<\/em><sub>a<\/sub> = 3.0 [latex]\\times [\/latex] 10<sup>\u22124<\/sup>. What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).<\/li>\r\n\t<li>(c) Under some conditions, aspirin reacts with water and forms a solution of salicylic acid and acetic acid:\r\n[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{\\text{C}}_{6}{\\text{H}}_{4}{\\text{CO}}_{2}\\text{H(}aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{HOC}}_{6}{\\text{H}}_{4}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)[\/latex]\r\n<ol>\r\n\t<li>i. Which of the acids salicylic acid or acetic acid produces more hydronium ions in solution such a solution?<\/li>\r\n\t<li>ii. What are the concentrations of molecules and ions in a solution produced by the hydrolysis of 0.50 g of aspirin dissolved in enough water to give 75 mL of solution?<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>The ion HTe<sup>\u2212<\/sup> is an amphiprotic species; it can act as either an acid or a base.\r\n<ol>\r\n\t<li>a) What is <em data-effect=\"italics\">K<\/em><sub>a<\/sub> for the acid reaction of HTe<sup>\u2212<\/sup> with H<sub>2<\/sub>O?<\/li>\r\n\t<li>(b) What is <em data-effect=\"italics\">K<\/em><sub>b<\/sub> for the reaction in which HTe<sup>\u2212<\/sup> functions as a base in water?<\/li>\r\n\t<li>(c) Demonstrate whether or not the second ionization of H<sub>2<\/sub>Te can be neglected in the calculation of [HTe<sup>\u2212<\/sup>] in a 0.10 M solution of H<sub>2<\/sub>Te.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp153083664\" data-type=\"exercise\"><\/div>\r\n<\/div>\r\n<\/section>\r\n<div class=\"bcc-box bcc-info\">\r\n<h4>Selected Answers<\/h4>\r\n1.\u00a0[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] and [latex]\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right][\/latex] are equal in a 0.134-<em data-effect=\"italics\">M<\/em> solution of H<sub>2<\/sub>CO<sub>3<\/sub>. <em data-effect=\"italics\">K<\/em><sub>a<\/sub> of H<sub>2<\/sub>CO<sub>3<\/sub> is significantly larger than <em data-effect=\"italics\">K<\/em><sub>a<\/sub> for [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex]. Therefore, very little of [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] ionizes to give hydronium ions and [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] ions, and the concentrations of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] and [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] are practically equal in an aqueous solution of H<sub>2<\/sub>CO<sub>3<\/sub>.\r\n\r\n3.\u00a0[latex]\\begin{array}{l}{\\text{C}}_{6}{\\text{H}}_{4}{\\left({\\text{CO}}_{2}\\text{H}\\right)}_{2}\\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}}_{4}\\left({\\text{CO}}_{2}\\text{H}\\right){\\left({\\text{CO}}_{2}\\right)}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}}=1.1\\times {10}^{-3}\\\\ {\\text{C}}_{6}{\\text{H}}_{4}\\left({\\text{CO}}_{2}\\text{H}\\right)\\left({\\text{CO}}_{2}\\right)\\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}}_{4}{\\left({\\text{CO}}_{2}\\right)}_{2}{}^{2-}\\left(aq\\right){K}_{\\text{a}}=3.9\\times {10}^{-6}\\end{array}[\/latex]\r\n<p id=\"fs-idp126090960\">[C<sub>6<\/sub>H<sub>4<\/sub>(CO<sub>2<\/sub>H)<sub>2<\/sub>] 7.2 [latex]\\times [\/latex] 10<sup>\u22123<\/sup><em data-effect=\"italics\">M<\/em>, [C<sub>6<\/sub>H<sub>4<\/sub>(CO<sub>2<\/sub>H)(CO<sub>2<\/sub>)<sup>\u2212<\/sup>] = [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] 2.8 [latex]\\times [\/latex] 10<sup>\u22123<\/sup><em data-effect=\"italics\">M<\/em>, [latex]\\left[{\\text{C}}_{6}{\\text{H}}_{4}{\\left({\\text{CO}}_{2}\\right)}_{2}{}^{2-}\\right][\/latex] 3.9 [latex]\\times [\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>, [OH<sup>\u2212<\/sup>] 3.6 [latex]\\times [\/latex] 10<sup>\u221212<\/sup><em data-effect=\"italics\">M<\/em><\/p>\r\n5. \u00a0(a) as an acid,\r\n\r\n[latex]\\begin{array}{l}\\\\ {\\text{HTe}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{Te}}^{2-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)\\\\ {K}_{\\text{a2}}=\\frac{\\left[{\\text{Te}}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{HTe}}^{\\text{-}}\\right]}=1\\times {10}^{-5}\\end{array}[\/latex];\r\n\r\n(b) as a base,\r\n\r\n[latex]\\begin{array}{l}\\\\ {\\text{HTe}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{2}\\text{Te}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right)\\\\ \\\\ {K}_{\\text{b}}=\\frac{\\left[{\\text{H}}_{2}\\text{Te}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{HTe}}^{\\text{-}}\\right]}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a1}}}=\\frac{1.0\\times {10}^{-14}}{2.3\\times {10}^{-3}}=4.3\\times {10}^{-12}\\end{array}[\/latex];\r\n\r\n(c) The reactions and ionization constants are:\r\n\r\n[latex]\\begin{array}{l}\\\\ {\\text{H}}_{2}\\text{Te}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{HTe}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right){K}_{\\text{a1}}=2.3\\times {10}^{-3}\\\\ {\\text{HTe}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{Te}}^{2-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right){K}_{\\text{a2}}=1\\times {10}^{-5}\\end{array}[\/latex]\r\n\r\nAs a general rule, if the first ionization constant is larger than the second by a factor of at least 20, then the second ionization can be neglected. Since [latex]{K}_{\\text{a1}}[\/latex] is 230-times larger than [latex]{K}_{\\text{a2}}[\/latex], the assumption should hold true for HTe<sup>\u2212<\/sup>. To test the assumptions, find [HTe<sup>\u2212<\/sup>] from the first reaction. The equilibrium expression for this reaction is [latex]{K}_{\\text{a1}}=\\frac{\\left[{\\text{HTe}}^{\\text{-}}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{H}}_{2}\\text{Te}\\right]}=2.3\\text{\\times }{10}^{-3}[\/latex].\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm16795072\" data-type=\"media\" data-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 \u201c[ H subscript 2 T e ] [ H T E superscript negative sign ] [ H subscript 3 O superscript positive sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.10, negative sign x, 0.10 minus sign x. The second column has the following: 0, x, x. The third column has the following: 0, x, x.\">\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213836\/CNX_Chem_14_05_ICETable6_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 \u201c[ H subscript 2 T e ] [ H T E superscript negative sign ] [ H subscript 3 O superscript positive sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.10, negative sign x, 0.10 minus sign x. The second column has the following: 0, x, x. The third column has the following: 0, x, x.\" width=\"879\" height=\"238\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (0.10 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.10, gives [latex]\\frac{\\left[{\\text{HTe}}^{\\text{-}}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{H}}_{2}\\text{Te}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.10-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.10}=2.3\\times {10}^{-3}[\/latex].\r\n\r\nSolving for <em data-effect=\"italics\">x<\/em> gives 0.0152 <em data-effect=\"italics\">M<\/em>. Because this value is 15% of 0.10 <em data-effect=\"italics\">M<\/em>, our assumption is incorrect. Therefore, use the quadratic formula. Using the data gives the quadratic equation:\r\n\r\n<em data-effect=\"italics\">x<\/em><sup>2<\/sup> + 2.3 [latex]\\times [\/latex] 10<sup>\u22123<\/sup><em data-effect=\"italics\">x<\/em> \u2212 2.3 [latex]\\times [\/latex] 10<sup>\u22124<\/sup> = 0\r\n\r\nUsing the quadratic formula gives (<em data-effect=\"italics\">a<\/em> = 1, <em data-effect=\"italics\">b<\/em> = 2.3 [latex]\\times [\/latex] 10<sup>\u22123<\/sup>, and <em data-effect=\"italics\">c<\/em> = \u22122.3 [latex]\\times [\/latex] 10<sup>\u22124<\/sup>)\r\n\r\n[latex]\\begin{array}{ll}\\hfill x&amp; =\\frac{-b\\pm \\sqrt{{b}^{\\text{2+}}-4ac}}{2a}=\\frac{-\\left(2.3\\times {10}^{-3}\\right)\\pm \\sqrt{{\\left(2.3\\times {10}^{-3}\\right)}^{\\text{2+}}-4\\left(1\\right)\\left(\\text{-2.3}\\times {10}^{-4}\\right)}}{2\\left(1\\right)}\\hfill \\\\ &amp; =\\frac{-\\left(2.3\\times {10}^{-3}\\right)\\pm \\left(0.0304\\right)}{2}=0.0141M\\text{(positive root)}\\hfill \\end{array}[\/latex]\r\n\r\nThus [HTe<sup>\u2212<\/sup>] = 0.014 <em data-effect=\"italics\">M<\/em>. For the second ionization, [latex]{K}_{\\text{a2}}=\\frac{\\left[{\\text{Te}}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{HTe}}^{\\text{-}}\\right]}=1\\text{\\times }{10}^{-5}[\/latex].\r\n\r\nThe initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idp59348832\" data-type=\"media\" data-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 \u201c[ H T e superscript negative sign ] [ T e to the second power superscript negative sign ] [ H subscript 3 O superscript positive sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.0141, negative sign x, 0.0141 minus sign x. The second column has the following: 0, x, x. The third column has the following: 0.0141, x, 0.0141 plus sign x.\">\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213837\/CNX_Chem_14_05_ICETable7_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 \u201c[ H T e superscript negative sign ] [ T e to the second power superscript negative sign ] [ H subscript 3 O superscript positive sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.0141, negative sign x, 0.0141 minus sign x. The second column has the following: 0, x, x. The third column has the following: 0.0141, x, 0.0141 plus sign x.\" width=\"875\" height=\"237\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n\r\nSubstituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.0140 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 and (0.0141 + <em data-effect=\"italics\">x<\/em>) \u2248 0.0141, gives:\r\n\r\n[latex]\\frac{\\left[{\\text{Te}}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{HTe}}^{\\text{-}}\\right]}=\\frac{\\left(x\\right)\\left(0.0141+x\\right)}{\\left(0.0141-x\\right)}\\approx \\frac{\\left(x\\right)\\left(0.0141\\right)}{0.0141}=1\\text{\\times }{10}^{-5}[\/latex]\r\n\r\nSolving for <em data-effect=\"italics\">x<\/em> gives 1 [latex]\\times [\/latex] 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em>. Therefore, compared with 0.014 <em data-effect=\"italics\">M<\/em>, this value is negligible (0.071%).\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Glossary<\/h3>\r\n<div id=\"fs-idp88439392\" data-type=\"definition\">\r\n\r\n<span data-type=\"term\"><strong>diprotic acid<\/strong>\r\n<\/span>acid containing two ionizable hydrogen atoms per molecule. A diprotic acid ionizes in two steps\r\n\r\n<\/div>\r\n<div id=\"fs-idm124847888\" data-type=\"definition\">\r\n\r\n<span data-type=\"term\"><strong>diprotic base<\/strong>\r\n<\/span>base capable of accepting two protons. The protons are accepted in two steps\r\n\r\n<\/div>\r\n<div id=\"fs-idp98784944\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">monoprotic acid\r\n<\/span><\/strong>acid containing one ionizable hydrogen atom per molecule\r\n\r\n<\/div>\r\n<div id=\"fs-idm63142976\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">stepwise ionization\r\n<\/span><\/strong>process in which an acid is ionized by losing protons sequentially\r\n\r\n<\/div>\r\n<div id=\"fs-idm25629392\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">triprotic acid\r\n<\/span><\/strong>acid that contains three ionizable hydrogen atoms per molecule; ionization of triprotic acids occurs in three steps\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>LEARNING OBJECTIVES<\/h3>\n<p>By the end of this module, you will be able to:<\/p>\n<ul>\n<li>Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idp98649872\">We can classify acids by the number of protons per molecule that they can give up in a reaction. Acids such as HCl, HNO<sub>3<\/sub>, and HCN that contain one ionizable hydrogen atom in each molecule are called <strong><span data-type=\"term\">monoprotic acids<\/span><\/strong>. Their reactions with water are:<\/p>\n<div id=\"fs-idp136743504\" data-type=\"equation\">[latex]\\begin{array}{c}\\text{HCl}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{\\text{-}}\\left(aq\\right)\\\\ {\\text{HNO}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{NO}}_{3}{}^{\\text{-}}\\left(aq\\right)\\\\ \\text{HCN}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CN}}^{\\text{-}}\\left(aq\\right)\\end{array}[\/latex]<\/div>\n<p id=\"fs-idp136804288\">Even though it contains four hydrogen atoms, acetic acid, CH<sub>3<\/sub>CO<sub>2<\/sub>H, is also monoprotic because only the hydrogen atom from the carboxyl group (COOH) reacts with bases:<span id=\"fs-idm11559264\" data-type=\"media\" data-alt=\"This image contains two equilibrium reactions. The first shows a C atom bonded to three H atoms and another C atom. The second C atom is double bonded to an O atom and also forms a single bond to another O atom. The second O atom is bonded to an H atom. There is a plus sign and then the molecular formula H subscript 2 O. An equilibrium arrow follows the H subscript 2 O. To the right of the arrow is H subscript 3 O superscript positive sign. There is a plus sign. The final structure shows a C atom bonded the three H atoms and another C atom. This second C atom is double bonded to an O atom and single bonded to another O atom. The entire structure is in brackets and a superscript negative sign appears outside the brackets. The second reaction shows C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ).\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23213825\/CNX_Chem_14_05_acetic_img.jpg\" alt=\"This image contains two equilibrium reactions. The first shows a C atom bonded to three H atoms and another C atom. The second C atom is double bonded to an O atom and also forms a single bond to another O atom. The second O atom is bonded to an H atom. There is a plus sign and then the molecular formula H subscript 2 O. An equilibrium arrow follows the H subscript 2 O. To the right of the arrow is H subscript 3 O superscript positive sign. There is a plus sign. The final structure shows a C atom bonded the three H atoms and another C atom. This second C atom is double bonded to an O atom and single bonded to another O atom. The entire structure is in brackets and a superscript negative sign appears outside the brackets. The second reaction shows C H subscript 3 C O O H ( a q ) plus H subscript 2 O ( l ) equilibrium arrow H subscript 3 O ( a q ) plus C H subscript 3 C O O superscript negative sign ( a q ).\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p id=\"fs-idp199141696\">Similarly, monoprotic bases are bases that will accept a single proton.<\/p>\n<p id=\"fs-idp13686896\"><strong><span data-type=\"term\">Diprotic acids<\/span><\/strong> contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:<\/p>\n<div id=\"fs-idm64885744\" data-type=\"equation\">[latex]\\begin{array}{l}\\\\ \\\\ \\text{First ionization:}{\\text{H}}_{2}{\\text{SO}}_{4}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HSO}}_{4}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}1}=\\text{about}{10}^{\\text{2+}}\\\\ \\text{Second ionization:}{\\text{HSO}}_{4}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{SO}}_{4}{}^{2-}\\left(aq\\right){K}_{\\text{a}2}=1.2\\times {10}^{-2}\\end{array}[\/latex]<\/div>\n<p id=\"fs-idp137460528\">This <strong><span data-type=\"term\">stepwise ionization<\/span><\/strong> process occurs for all polyprotic acids. When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of acids. Carbonic acid, H<sub>2<\/sub>CO<sub>3<\/sub>, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.<\/p>\n<div id=\"fs-idm41501456\" data-type=\"equation\">[latex]\\begin{array}{l}\\text{First ionization:}\\\\ {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right){K}_{{\\text{H}}_{2}{\\text{CO}}_{3}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=4.3\\times {10}^{-7}\\end{array}[\/latex]<\/div>\n<p id=\"fs-idp84456096\">The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.<\/p>\n<div id=\"fs-idm146392784\" data-type=\"equation\">[latex]\\begin{array}{l}\\text{Second ionization:}\\\\ {\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right){K}_{{\\text{HCO}}_{3}{}^{\\text{-}}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right]}=4.7\\times {10}^{-11}\\end{array}[\/latex]<\/div>\n<p id=\"fs-idm74401760\">[latex]{K}_{{\\text{H}}_{2}{\\text{CO}}_{3}}[\/latex] is larger than [latex]{K}_{{\\text{HCO}}_{3}{}^{\\text{-}}}[\/latex] by a factor of 10<sup>4<\/sup>, so H<sub>2<\/sub>CO<sub>3<\/sub> is the dominant producer of hydronium ion in the solution. This means that little of the [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] formed by the ionization of H<sub>2<\/sub>CO<sub>3<\/sub> ionizes to give hydronium ions (and carbonate ions), and the concentrations of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] and [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] are practically equal in a pure aqueous solution of H<sub>2<\/sub>CO<sub>3<\/sub>.<\/p>\n<p id=\"fs-idm68059056\">If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This can simplify our work considerably because we can determine the concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] and the conjugate base from the first ionization, then determine the concentration of the conjugate base of the second ionization in a solution with concentrations determined by the first ionization.<\/p>\n<div id=\"fs-idp1592608\" class=\"textbox shaded\" data-type=\"example\">\n<h3>Example 1<\/h3>\n<h4 id=\"fs-idp100346864\"><span data-type=\"title\">Ionization of a Diprotic Acid<\/span><\/h4>\n<p>When we buy soda water (carbonated water), we are buying a solution of carbon dioxide in water. The solution is acidic because CO<sub>2<\/sub> reacts with water to form carbonic acid, H<sub>2<\/sub>CO<sub>3<\/sub>. What are [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex], [latex]\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right][\/latex], and [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] in a saturated solution of CO<sub>2<\/sub> with an initial [H<sub>2<\/sub>CO<sub>3<\/sub>] = 0.033 <em data-effect=\"italics\">M<\/em>?<\/p>\n<div id=\"fs-idp126184336\" data-type=\"equation\">[latex]{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}1}=4.3\\times {10}^{-7}[\/latex]<\/div>\n<div id=\"fs-idp136502080\" data-type=\"equation\">[latex]{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right){K}_{\\text{a2}}=4.7\\times {10}^{-11}[\/latex]<\/div>\n<h4 id=\"fs-idm27159024\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>As indicated by the ionization constants, H<sub>2<\/sub>CO<sub>3<\/sub> is a much stronger acid than [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex], so H<sub>2<\/sub>CO<sub>3<\/sub> is the dominant producer of hydronium ion in solution. Thus there are two parts in the solution of this problem: (1) Using the customary four steps, we determine the concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] and [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] produced by ionization of H<sub>2<\/sub>CO<sub>3<\/sub>. (2) Then we determine the concentration of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] in a solution with the concentration of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] and [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] determined in (1). To summarize:<span id=\"fs-idm28185792\" data-type=\"media\" data-alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cleft bracket H subscript 2 C O subscript 3 right bracket.\u201d The second is labeled \u201cleft bracket H subscript 3 O superscript plus right bracket and left bracket H C O subscript 3 superscript negative right bracket from H subscript 2 C O subscript 3.\u201d The third is labeled \u201cleft bracket C O subscript 3 superscript 2 negative right bracket from H C O subscript 3 superscript negative.\u201d\"><br \/>\n<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\/23213827\/CNX_Chem_14_05_steps1_img.jpg\" alt=\"Four tan rectangles are shown that are connected with right pointing arrows. The first is labeled \u201cleft bracket H subscript 2 C O subscript 3 right bracket.\u201d The second is labeled \u201cleft bracket H subscript 3 O superscript plus right bracket and left bracket H C O subscript 3 superscript negative right bracket from H subscript 2 C O subscript 3.\u201d The third is labeled \u201cleft bracket C O subscript 3 superscript 2 negative right bracket from H C O subscript 3 superscript negative.\u201d\" width=\"881\" height=\"156\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<ol id=\"fs-idp108659776\" class=\"stepwise\" data-number-style=\"arabic\">\n<li><em data-effect=\"italics\">Determine the concentrations of<\/em> [latex]{H}_{3}{O}^{\\text{+}}[\/latex] <em data-effect=\"italics\">and<\/em> [latex]HC{O}_{3}{}^{\\text{-}}[\/latex].\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idm22716864\" data-type=\"equation\">[latex]{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}1}=4.3\\times {10}^{-7}[\/latex]<\/div>\n<p>As for the ionization of any other weak acid:<span id=\"fs-idp99810176\" data-type=\"media\" data-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\"><br \/>\n<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\/23213828\/CNX_Chem_14_05_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=\"842\" height=\"149\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p>An abbreviated table of changes and concentrations shows:<span id=\"fs-idm37059488\" data-type=\"media\" data-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 constant ( M ). The second column has the header of \u201cH subscript 2 C O subscript 3 plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus sign H C O subscript 3 superscript negative sign.\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.033, negative sign x, 0.033 minus sign x. The second column has the following: approximately 0, x, x. The third column has the following: 0, x, x.\"><br \/>\n<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\/23213830\/CNX_Chem_14_05_ICETable1_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 constant ( M ). The second column has the header of \u201cH subscript 2 C O subscript 3 plus sign H subscript 2 O equilibrium arrow H subscript 3 O superscript positive sign plus sign H C O subscript 3 superscript negative sign.\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.033, negative sign x, 0.033 minus sign x. The second column has the following: approximately 0, x, x. The third column has the following: 0, x, x.\" width=\"838\" height=\"227\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p>Substituting the equilibrium concentrations into the equilibrium gives us:<\/p>\n<div id=\"fs-idm26667456\" data-type=\"equation\">[latex]{K}_{{\\text{H}}_{2}{\\text{CO}}_{3}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{0.033-x}=4.3\\times {10}^{-7}[\/latex]<\/div>\n<p>Solving the preceding equation making our standard assumptions gives:<\/p>\n<div id=\"fs-idm73884976\" data-type=\"equation\">[latex]x=1.2\\times {10}^{-4}[\/latex]<\/div>\n<p>Thus:<\/p>\n<div id=\"fs-idm57413264\" data-type=\"equation\">[latex]\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]=0.033M[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idp20713136\" data-type=\"equation\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right]=1.2\\times {10}^{-4}M[\/latex]<\/div>\n<\/li>\n<li><em data-effect=\"italics\">Determine the concentration of [latex]C{O}_{3}{}^{2-}[\/latex] in a solution at equilibrium with<\/em> [latex]\\left[{H}_{3}{O}^{\\text{+}}\\right][\/latex] <em data-effect=\"italics\">and<\/em> [latex]\\left[HC{O}_{3}{}^{\\text{-}}\\right][\/latex] <em data-effect=\"italics\">both equal to 1.2<\/em> [latex]\\times[\/latex] <em data-effect=\"italics\">10<sup>\u22124<\/sup> M<\/em>.\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idp103313952\" data-type=\"equation\">[latex]{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idm27760608\" data-type=\"equation\">[latex]{K}_{{\\text{HCO}}_{3}{}^{\\text{-}}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right]}=\\frac{\\left(1.2\\times {10}^{-4}\\right)\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{1.2\\times {10}^{-4}}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div id=\"fs-idm37931200\" data-type=\"equation\">[latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right]=\\frac{\\left(4.7\\times {10}^{-11}\\right)\\left(1.2\\times {10}^{-4}\\right)}{1.2\\times {10}^{-4}}=4.7\\times {10}^{-11}M[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p id=\"fs-idp167647504\">To summarize: In part 1 of this example, we found that the H<sub>2<\/sub>CO<sub>3<\/sub> in a 0.033-<em data-effect=\"italics\">M<\/em> solution ionizes slightly and at equilibrium [H<sub>2<\/sub>CO<sub>3<\/sub>] = 0.033 <em data-effect=\"italics\">M<\/em>; [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = 1.2 [latex]\\times[\/latex] 10<sup>\u22124<\/sup>; and [latex]\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right]=1.2\\times {10}^{-4}M[\/latex]. In part 2, we determined that [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right]=4.7\\times {10}^{-11}M[\/latex].<\/p>\n<h4 id=\"fs-idp3314688\"><span data-type=\"title\">Check Your Learning<\/span><\/h4>\n<p>The concentration of H<sub>2<\/sub>S in a saturated aqueous solution at room temperature is approximately 0.1 <em data-effect=\"italics\">M<\/em>. Calculate [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex], [HS<sup>\u2212<\/sup>], and [S<sup>2\u2212<\/sup>] in the solution:<\/p>\n<div id=\"fs-idp170235264\" data-type=\"equation\">[latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HS}}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}1}=1.0\\times {10}^{-7}[\/latex]<\/div>\n<div id=\"fs-idp22220080\" data-type=\"equation\">[latex]{\\text{HS}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right){K}_{\\text{a}2}=1.0\\times {10}^{-19}[\/latex]<\/div>\n<div id=\"fs-idm58434032\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer:\u00a0<\/strong>[H<sub>2<\/sub>S] = 0.1 <em data-effect=\"italics\">M<\/em>; [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] = [HS<sup>\u2212<\/sup>] = 0.0001 <em data-effect=\"italics\">M<\/em>; [S<sup>2\u2212<\/sup>] = 1 [latex]\\times[\/latex] 10<sup>\u221219<\/sup><em data-effect=\"italics\">M<\/em><\/div>\n<p>&nbsp;<\/p>\n<p>We note that the concentration of the sulfide ion is the same as <em data-effect=\"italics\">K<\/em><sub>a2<\/sub>. This is due to the fact that each subsequent dissociation occurs to a lesser degree (as acid gets weaker).<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-idm49318672\">A <b>triprotic acid<\/b> is an acid that has three dissociable protons that undergo stepwise ionization: Phosphoric acid is a typical example:<\/p>\n<div id=\"fs-idm72609920\" data-type=\"equation\">[latex]\\begin{array}{l}\\text{First ionization:}{\\text{H}}_{3}{\\text{PO}}_{4}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}{\\text{PO}}_{4}{}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}1}=7.5\\times {10}^{-3}\\\\ \\text{Second ionization:}{\\text{H}}_{2}{\\text{PO}}_{4}{}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{HPO}}_{4}{}^{2-}\\left(aq\\right){K}_{\\text{a}2}=6.3\\times {10}^{-8}\\\\ \\text{Third ionization:}{\\text{HPO}}_{4}{}^{2-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{PO}}_{4}{}^{3-}\\left(aq\\right){K}_{\\text{a}3}=3.6\\times {10}^{-13}\\end{array}[\/latex]<\/div>\n<p id=\"fs-idp15743456\">As with the diprotic acids, the differences in the ionization constants of these reactions tell us that in each successive step the degree of ionization is significantly weaker. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 10<sup>5<\/sup> to 10<sup>6<\/sup>.<\/p>\n<p id=\"fs-idp108834896\">This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H<sub>3<\/sub>PO<sub>4<\/sub> complicated. However, because the successive ionization constants differ by a factor of 10<sup>5<\/sup> to 10<sup>6<\/sup>, the calculations can be broken down into a series of parts similar to those for diprotic acids.<\/p>\n<p id=\"fs-idp178452064\">Polyprotic bases can accept more than one hydrogen ion in solution. The carbonate ion is an example of a <b>diprotic base<\/b>, since it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions:<\/p>\n<div id=\"fs-idm16615264\" data-type=\"equation\">[latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)\\rightleftharpoons {\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right)\\text{and}{\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\text{HCO}}_{3}{}^{\\text{-}}\\left(aq\\right)\\rightleftharpoons {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right)[\/latex]<\/div>\n<section id=\"fs-idm40783936\" class=\"summary\" data-depth=\"1\">\n<div class=\"bcc-box bcc-success\">\n<h2>Key Concepts and Summary<\/h2>\n<p id=\"fs-idm21286928\">An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude. As long as the difference between the successive values of <em data-effect=\"italics\">K<\/em><sub>a<\/sub> of the acid is greater than about a factor of 20, it is appropriate to break down the calculations of the concentrations of the ions in solution into a series of steps.<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idm45487824\" class=\"exercises\" data-depth=\"1\">\n<div class=\"bcc-box bcc-info\">\n<h3>Chemistry End of Chapter Exercises<\/h3>\n<div id=\"fs-idp94973072\" data-type=\"exercise\">\n<div id=\"fs-idp149185520\" data-type=\"problem\">\n<ol>\n<li id=\"fs-idm49321712\">Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134-<em data-effect=\"italics\">M<\/em> solution of H<sub>2<\/sub>CO<sub>3<\/sub>, a diprotic acid: [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex], [OH<sup>\u2212<\/sup>], [H<sub>2<\/sub>CO<sub>3<\/sub>], [latex]\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right][\/latex], [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right]?[\/latex] No calculations are needed to answer this question.<\/li>\n<li>Calculate the concentration of each species present in a 0.050-<em data-effect=\"italics\">M<\/em> solution of H<sub>2<\/sub>S.<\/li>\n<li>Calculate the concentration of each species present in a 0.010-<em data-effect=\"italics\">M<\/em> solution of phthalic acid, C<sub>6<\/sub>H<sub>4<\/sub>(CO<sub>2<\/sub>H)<sub>2<\/sub>.<\/li>\n<li>Salicylic acid, HOC<sub>6<\/sub>H<sub>4<\/sub>CO<sub>2<\/sub>H, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.\n<ol>\n<li>(a) Both functional groups of salicylic acid ionize in water, with <em data-effect=\"italics\">K<\/em><sub>a<\/sub> = 1.0 [latex]\\times[\/latex] 10<sup>\u22123<\/sup> for the\u2014CO<sub>2<\/sub>H group and 4.2 [latex]\\times[\/latex] 10<sup>\u221213<\/sup> for the \u2212OH group. What is the pH of a saturated solution of the acid (solubility = 1.8 g\/L).<\/li>\n<li>(b) Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, CH<sub>3<\/sub>CO<sub>2<\/sub>C<sub>6<\/sub>H<sub>4<\/sub>CO<sub>2<\/sub>H. The \u2212CO<sub>2<\/sub>H functional group is still present, but its acidity is reduced, <em data-effect=\"italics\">K<\/em><sub>a<\/sub> = 3.0 [latex]\\times[\/latex] 10<sup>\u22124<\/sup>. What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).<\/li>\n<li>(c) Under some conditions, aspirin reacts with water and forms a solution of salicylic acid and acetic acid:<br \/>\n[latex]{\\text{CH}}_{3}{\\text{CO}}_{2}{\\text{C}}_{6}{\\text{H}}_{4}{\\text{CO}}_{2}\\text{H(}aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{HOC}}_{6}{\\text{H}}_{4}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)[\/latex]<\/p>\n<ol>\n<li>i. Which of the acids salicylic acid or acetic acid produces more hydronium ions in solution such a solution?<\/li>\n<li>ii. What are the concentrations of molecules and ions in a solution produced by the hydrolysis of 0.50 g of aspirin dissolved in enough water to give 75 mL of solution?<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<li>The ion HTe<sup>\u2212<\/sup> is an amphiprotic species; it can act as either an acid or a base.\n<ol>\n<li>a) What is <em data-effect=\"italics\">K<\/em><sub>a<\/sub> for the acid reaction of HTe<sup>\u2212<\/sup> with H<sub>2<\/sub>O?<\/li>\n<li>(b) What is <em data-effect=\"italics\">K<\/em><sub>b<\/sub> for the reaction in which HTe<sup>\u2212<\/sup> functions as a base in water?<\/li>\n<li>(c) Demonstrate whether or not the second ionization of H<sub>2<\/sub>Te can be neglected in the calculation of [HTe<sup>\u2212<\/sup>] in a 0.10 M solution of H<sub>2<\/sub>Te.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"fs-idp153083664\" data-type=\"exercise\"><\/div>\n<\/div>\n<\/section>\n<div class=\"bcc-box bcc-info\">\n<h4>Selected Answers<\/h4>\n<p>1.\u00a0[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] and [latex]\\left[{\\text{HCO}}_{3}{}^{\\text{-}}\\right][\/latex] are equal in a 0.134-<em data-effect=\"italics\">M<\/em> solution of H<sub>2<\/sub>CO<sub>3<\/sub>. <em data-effect=\"italics\">K<\/em><sub>a<\/sub> of H<sub>2<\/sub>CO<sub>3<\/sub> is significantly larger than <em data-effect=\"italics\">K<\/em><sub>a<\/sub> for [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex]. Therefore, very little of [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] ionizes to give hydronium ions and [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] ions, and the concentrations of [latex]{\\text{H}}_{3}{\\text{O}}^{\\text{+}}[\/latex] and [latex]{\\text{HCO}}_{3}{}^{\\text{-}}[\/latex] are practically equal in an aqueous solution of H<sub>2<\/sub>CO<sub>3<\/sub>.<\/p>\n<p>3.\u00a0[latex]\\begin{array}{l}{\\text{C}}_{6}{\\text{H}}_{4}{\\left({\\text{CO}}_{2}\\text{H}\\right)}_{2}\\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}}_{4}\\left({\\text{CO}}_{2}\\text{H}\\right){\\left({\\text{CO}}_{2}\\right)}^{\\text{-}}\\left(aq\\right){K}_{\\text{a}}=1.1\\times {10}^{-3}\\\\ {\\text{C}}_{6}{\\text{H}}_{4}\\left({\\text{CO}}_{2}\\text{H}\\right)\\left({\\text{CO}}_{2}\\right)\\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}}_{4}{\\left({\\text{CO}}_{2}\\right)}_{2}{}^{2-}\\left(aq\\right){K}_{\\text{a}}=3.9\\times {10}^{-6}\\end{array}[\/latex]<\/p>\n<p id=\"fs-idp126090960\">[C<sub>6<\/sub>H<sub>4<\/sub>(CO<sub>2<\/sub>H)<sub>2<\/sub>] 7.2 [latex]\\times[\/latex] 10<sup>\u22123<\/sup><em data-effect=\"italics\">M<\/em>, [C<sub>6<\/sub>H<sub>4<\/sub>(CO<sub>2<\/sub>H)(CO<sub>2<\/sub>)<sup>\u2212<\/sup>] = [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex] 2.8 [latex]\\times[\/latex] 10<sup>\u22123<\/sup><em data-effect=\"italics\">M<\/em>, [latex]\\left[{\\text{C}}_{6}{\\text{H}}_{4}{\\left({\\text{CO}}_{2}\\right)}_{2}{}^{2-}\\right][\/latex] 3.9 [latex]\\times[\/latex] 10<sup>\u22126<\/sup><em data-effect=\"italics\">M<\/em>, [OH<sup>\u2212<\/sup>] 3.6 [latex]\\times[\/latex] 10<sup>\u221212<\/sup><em data-effect=\"italics\">M<\/em><\/p>\n<p>5. \u00a0(a) as an acid,<\/p>\n<p>[latex]\\begin{array}{l}\\\\ {\\text{HTe}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{Te}}^{2-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)\\\\ {K}_{\\text{a2}}=\\frac{\\left[{\\text{Te}}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{HTe}}^{\\text{-}}\\right]}=1\\times {10}^{-5}\\end{array}[\/latex];<\/p>\n<p>(b) as a base,<\/p>\n<p>[latex]\\begin{array}{l}\\\\ {\\text{HTe}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{2}\\text{Te}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right)\\\\ \\\\ {K}_{\\text{b}}=\\frac{\\left[{\\text{H}}_{2}\\text{Te}\\right]\\left[{\\text{OH}}^{\\text{-}}\\right]}{\\left[{\\text{HTe}}^{\\text{-}}\\right]}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a1}}}=\\frac{1.0\\times {10}^{-14}}{2.3\\times {10}^{-3}}=4.3\\times {10}^{-12}\\end{array}[\/latex];<\/p>\n<p>(c) The reactions and ionization constants are:<\/p>\n<p>[latex]\\begin{array}{l}\\\\ {\\text{H}}_{2}\\text{Te}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{HTe}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right){K}_{\\text{a1}}=2.3\\times {10}^{-3}\\\\ {\\text{HTe}}^{\\text{-}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{Te}}^{2-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right){K}_{\\text{a2}}=1\\times {10}^{-5}\\end{array}[\/latex]<\/p>\n<p>As a general rule, if the first ionization constant is larger than the second by a factor of at least 20, then the second ionization can be neglected. Since [latex]{K}_{\\text{a1}}[\/latex] is 230-times larger than [latex]{K}_{\\text{a2}}[\/latex], the assumption should hold true for HTe<sup>\u2212<\/sup>. To test the assumptions, find [HTe<sup>\u2212<\/sup>] from the first reaction. The equilibrium expression for this reaction is [latex]{K}_{\\text{a1}}=\\frac{\\left[{\\text{HTe}}^{\\text{-}}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{H}}_{2}\\text{Te}\\right]}=2.3\\text{\\times }{10}^{-3}[\/latex].<\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idm16795072\" data-type=\"media\" data-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 \u201c[ H subscript 2 T e ] [ H T E superscript negative sign ] [ H subscript 3 O superscript positive sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.10, negative sign x, 0.10 minus sign x. The second column has the following: 0, x, x. The third column has the following: 0, x, x.\"><br \/>\n<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\/23213836\/CNX_Chem_14_05_ICETable6_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 \u201c[ H subscript 2 T e ] [ H T E superscript negative sign ] [ H subscript 3 O superscript positive sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.10, negative sign x, 0.10 minus sign x. The second column has the following: 0, x, x. The third column has the following: 0, x, x.\" width=\"879\" height=\"238\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (0.10 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 0.10, gives [latex]\\frac{\\left[{\\text{HTe}}^{\\text{-}}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{H}}_{2}\\text{Te}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{\\left(0.10-x\\right)}\\approx \\frac{\\left(x\\right)\\left(x\\right)}{0.10}=2.3\\times {10}^{-3}[\/latex].<\/p>\n<p>Solving for <em data-effect=\"italics\">x<\/em> gives 0.0152 <em data-effect=\"italics\">M<\/em>. Because this value is 15% of 0.10 <em data-effect=\"italics\">M<\/em>, our assumption is incorrect. Therefore, use the quadratic formula. Using the data gives the quadratic equation:<\/p>\n<p><em data-effect=\"italics\">x<\/em><sup>2<\/sup> + 2.3 [latex]\\times[\/latex] 10<sup>\u22123<\/sup><em data-effect=\"italics\">x<\/em> \u2212 2.3 [latex]\\times[\/latex] 10<sup>\u22124<\/sup> = 0<\/p>\n<p>Using the quadratic formula gives (<em data-effect=\"italics\">a<\/em> = 1, <em data-effect=\"italics\">b<\/em> = 2.3 [latex]\\times[\/latex] 10<sup>\u22123<\/sup>, and <em data-effect=\"italics\">c<\/em> = \u22122.3 [latex]\\times[\/latex] 10<sup>\u22124<\/sup>)<\/p>\n<p>[latex]\\begin{array}{ll}\\hfill x& =\\frac{-b\\pm \\sqrt{{b}^{\\text{2+}}-4ac}}{2a}=\\frac{-\\left(2.3\\times {10}^{-3}\\right)\\pm \\sqrt{{\\left(2.3\\times {10}^{-3}\\right)}^{\\text{2+}}-4\\left(1\\right)\\left(\\text{-2.3}\\times {10}^{-4}\\right)}}{2\\left(1\\right)}\\hfill \\\\ & =\\frac{-\\left(2.3\\times {10}^{-3}\\right)\\pm \\left(0.0304\\right)}{2}=0.0141M\\text{(positive root)}\\hfill \\end{array}[\/latex]<\/p>\n<p>Thus [HTe<sup>\u2212<\/sup>] = 0.014 <em data-effect=\"italics\">M<\/em>. For the second ionization, [latex]{K}_{\\text{a2}}=\\frac{\\left[{\\text{Te}}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{HTe}}^{\\text{-}}\\right]}=1\\text{\\times }{10}^{-5}[\/latex].<\/p>\n<p>The initial and equilibrium concentrations for this system can be written as follows:<span id=\"fs-idp59348832\" data-type=\"media\" data-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 \u201c[ H T e superscript negative sign ] [ T e to the second power superscript negative sign ] [ H subscript 3 O superscript positive sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.0141, negative sign x, 0.0141 minus sign x. The second column has the following: 0, x, x. The third column has the following: 0.0141, x, 0.0141 plus sign x.\"><br \/>\n<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\/23213837\/CNX_Chem_14_05_ICETable7_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 \u201c[ H T e superscript negative sign ] [ T e to the second power superscript negative sign ] [ H subscript 3 O superscript positive sign ].\u201d Under the second column is a subgroup of three columns and three rows. The first column has the following: 0.0141, negative sign x, 0.0141 minus sign x. The second column has the following: 0, x, x. The third column has the following: 0.0141, x, 0.0141 plus sign x.\" width=\"875\" height=\"237\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<p>Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.0140 \u2212 <em data-effect=\"italics\">x<\/em>) \u2248 and (0.0141 + <em data-effect=\"italics\">x<\/em>) \u2248 0.0141, gives:<\/p>\n<p>[latex]\\frac{\\left[{\\text{Te}}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}{\\left[{\\text{HTe}}^{\\text{-}}\\right]}=\\frac{\\left(x\\right)\\left(0.0141+x\\right)}{\\left(0.0141-x\\right)}\\approx \\frac{\\left(x\\right)\\left(0.0141\\right)}{0.0141}=1\\text{\\times }{10}^{-5}[\/latex]<\/p>\n<p>Solving for <em data-effect=\"italics\">x<\/em> gives 1 [latex]\\times[\/latex] 10<sup>\u22125<\/sup><em data-effect=\"italics\">M<\/em>. Therefore, compared with 0.014 <em data-effect=\"italics\">M<\/em>, this value is negligible (0.071%).<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Glossary<\/h3>\n<div id=\"fs-idp88439392\" data-type=\"definition\">\n<p><span data-type=\"term\"><strong>diprotic acid<\/strong><br \/>\n<\/span>acid containing two ionizable hydrogen atoms per molecule. A diprotic acid ionizes in two steps<\/p>\n<\/div>\n<div id=\"fs-idm124847888\" data-type=\"definition\">\n<p><span data-type=\"term\"><strong>diprotic base<\/strong><br \/>\n<\/span>base capable of accepting two protons. The protons are accepted in two steps<\/p>\n<\/div>\n<div id=\"fs-idp98784944\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">monoprotic acid<br \/>\n<\/span><\/strong>acid containing one ionizable hydrogen atom per molecule<\/p>\n<\/div>\n<div id=\"fs-idm63142976\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">stepwise ionization<br \/>\n<\/span><\/strong>process in which an acid is ionized by losing protons sequentially<\/p>\n<\/div>\n<div id=\"fs-idm25629392\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">triprotic acid<br \/>\n<\/span><\/strong>acid that contains three ionizable hydrogen atoms per molecule; ionization of triprotic acids occurs in three steps<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/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-3496\">\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":5,"menu_order":7,"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-3496","chapter","type-chapter","status-publish","hentry"],"part":2988,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/3496","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/users\/5"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/3496\/revisions"}],"predecessor-version":[{"id":4975,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/3496\/revisions\/4975"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/parts\/2988"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/3496\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/media?parent=3496"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapter-type?post=3496"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/contributor?post=3496"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/license?post=3496"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}