{"id":997,"date":"2022-06-23T13:14:37","date_gmt":"2022-06-23T17:14:37","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/chapter\/14-4-hydrolysis-of-salts-chemistry-2e\/"},"modified":"2022-07-06T15:36:52","modified_gmt":"2022-07-06T19:36:52","slug":"14-4-hydrolysis-of-salts-chemistry-2e","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chemistry2eengineering\/chapter\/14-4-hydrolysis-of-salts-chemistry-2e\/","title":{"raw":"14.4 Hydrolysis of Salts","rendered":"14.4 Hydrolysis of Salts"},"content":{"raw":"
\r\n\r\n \r\n
\r\n

Learning Objectives<\/strong><\/h3>\r\nBy the end of this section, you will be able to:\r\n
    \r\n \t
  • Calculate the concentrations of the various species in a salt solution<\/li>\r\n \t
  • Describe the acid ionization of hydrated metal ions<\/li>\r\n<\/ul>\r\n<\/div>\r\n
    \r\n

    Salts with Acidic Ions<\/strong><\/h3>\r\n

    Salts are ionic compounds composed of cations and anions, either of which may be capable of undergoing an acid or base ionization reaction with water. Aqueous salt solutions, therefore, may be acidic, basic, or neutral, depending on the relative acid-base strengths of the salt\u2019s constituent ions. For example, dissolving ammonium chloride in water results in its dissociation, as described by the equation<\/p>\r\n\r\n

    NH4<\/sub>Cl(s<\/em>) \u2192 NH4<\/sub>+<\/sup>(aq<\/em>) + Cl\u2013<\/sup>(aq<\/em>)<\/div>\r\n

    The ammonium ion is the conjugate acid of the weak base ammonia, NH3<\/sub>, and so it will undergo acid ionization (or acid hydrolysis<\/em>):<\/p>\r\n\r\n

    NH4<\/sub>+<\/sup>(aq<\/em>) + H2<\/sub>O(l<\/em>) \u21cc H3<\/sub>O+<\/sup>(aq<\/em>) + NH3<\/sub>(aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Ka<\/sub> = Kw<\/sub>\/Kb<\/sub><\/div>\r\n

    Since ammonia is a weak base, K<\/em>b<\/sub> is measurable and K<\/em>a<\/sub> > 0 (ammonium ion is a weak acid).<\/p>\r\n

    The chloride ion is the conjugate base of hydrochloric acid, and so its base ionization (or base hydrolysis<\/em>) reaction is represented by<\/p>\r\n\r\n

    Cl\u2013<\/sup>(aq<\/em>) + H2<\/sub>O(l<\/em>) \u21cc HCl(aq<\/em>) + OH\u2013<\/sup>(aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Kb<\/sub> = Kw<\/sub>\/Ka<\/sub><\/div>\r\n

    Since HCl is a strong acid, K<\/em>a<\/sub> is immeasurably large and K<\/em>b<\/sub> \u2248 0 (chloride ions don\u2019t undergo appreciable hydrolysis).<\/p>\r\n

    Thus, dissolving ammonium chloride in water yields a solution of weak acid cations (NH4<\/sub>+<\/sup>) and inert anions (Cl\u2212<\/sup>), resulting in an acidic solution.<\/p>\r\n\r\n

    \r\n

    Calculating the pH of an Acidic Salt Solution <\/strong><\/p>\r\nAniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, [C6<\/sub>H5<\/sub>NH3<\/sub>]Cl, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M<\/em> solution of anilinium chloride?\r\n

    C6<\/sub>H5<\/sub>NH3<\/sub>+<\/sup>(aq) + H2<\/sub>O(l) \u21cc H3<\/sub>O+<\/sup>(aq<\/em>) + C6<\/sub>H5<\/sub>NH2<\/sub>(aq<\/em>)<\/div>\r\n
    <\/div>\r\n

    Solution:<\/strong><\/p>\r\nThe K<\/em>a<\/sub> for anilinium ion is derived from the K<\/em>b<\/sub> for its conjugate base, aniline (see Appendix H):\r\n

    \"\"<\/div>\r\n

    Using the provided information, an ICE table for this system is prepared:<\/p>\r\n\"This<\/span>\r\n

    Substituting these equilibrium concentration terms into the K<\/em>a<\/sub> expression gives<\/p>\r\n\r\n

    Ka<\/sub> = [C6<\/sub>H5<\/sub>NH2<\/sub>][H3<\/sub>O+<\/sup>]\/[C6<\/sub>H5<\/sub>NH3<\/sub>+<\/sup>]<\/div>\r\n
    2.3 \u00d7 10\u22125<\/sup> = (x<\/em>)(x<\/em>)\/(0.233-x<\/em>)<\/div>\r\n
    <\/div>\r\n

    Assuming x<\/em> < 0.05 \u00d7 0.233M, i.e. x<\/em> < 0.0116 M, the equation is simplified and solved for x<\/em>:<\/p>\r\n\r\n

    2.3 \u00d7 10\u22125<\/sup> = x<\/em>2<\/sup>\/0.233<\/div>\r\n
    x<\/em> = 0.0023 M\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ASSUMPTION VALID<\/div>\r\n
    <\/div>\r\n

    The ICE table defines x<\/em> as the hydronium ion molarity, and so the pH is computed as<\/p>\r\n\r\n

    pH = -log[H3<\/sub>O+<\/sup>] = \u2212log(0.0023 M) = 2.64<\/div>\r\n
    <\/div>\r\n

    Check Your Learning:<\/strong><\/p>\r\nWhat is the hydronium ion concentration in a 0.100-M<\/em> solution of ammonium nitrate, NH4<\/sub>NO3<\/sub>, a salt composed of the ions NH4<\/sub>+<\/sup> and NO3<\/sub>\u2212<\/sup>. Which is the stronger acid, C6<\/sub>H5<\/sub>NH3<\/sub>+<\/sup> or NH4<\/sub>+<\/sup>?\r\n\r\n \r\n

    \r\n
    Answer:<\/strong><\/div>\r\n

    [H3<\/sub>O+<\/sup>] = 7.5 \u00d7 10\u22126<\/sup>M<\/em>; C6<\/sub>H5<\/sub>NH3<\/sub>+<\/sup> is the stronger acid.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n

    Salts with Basic Ions<\/strong><\/h3>\r\n

    As another example, consider dissolving sodium acetate in water:<\/p>\r\n\r\n

    NaCH3<\/sub>CO2<\/sub>(s<\/em>) \u2192 Na+<\/sup>(aq<\/em>) + CH3<\/sub>CO2<\/sub>\u2013<\/sup>(aq<\/em>)<\/div>\r\n

    The sodium ion does not undergo appreciable acid or base ionization and has no effect on the solution pH. This may seem obvious from the ion\u2019s formula, which indicates no hydrogen or oxygen atoms, but some dissolved metal ions function as weak acids, as addressed later in this section.<\/p>\r\n

    The acetate ion, CH3<\/sub>CO2<\/sub>\u2013<\/sup>, is the conjugate base of acetic acid, CH3<\/sub>CO2<\/sub>H, and so its base ionization (or base hydrolysis<\/em>) reaction is represented by<\/p>\r\n\r\n

    CH3<\/sub>CO2<\/sub>\u2013<\/sup>(aq<\/em>) + H2<\/sub>O(l<\/em>) \u21cc CH3<\/sub>CO2<\/sub>H(aq<\/em>) + OH\u2013<\/sup>(aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Kb<\/sub> = Kw<\/sub>\/Ka<\/sub><\/div>\r\n

    Because acetic acid is a weak acid, its K<\/em>a<\/sub> is measurable and K<\/em>b<\/sub> > 0 (acetate ion is a weak base).<\/p>\r\n

    Dissolving sodium acetate in water yields a solution of inert cations (Na+<\/sup>) and weak base anions (CH3<\/sub>CO2<\/sub>\u2013<\/sup>), resulting in a basic solution.<\/p>\r\n\r\n

    \r\n

    Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base <\/strong><\/p>\r\nDetermine the acetic acid concentration in a solution with [CH3<\/sub>CO2<\/sub>\u2013<\/sup>] = 0.050 M and [OH\u2212<\/sup>] = 2.5 \u00d7 10\u22126<\/sup>M<\/em> at equilibrium.\r\n\r\nThe reaction is:\r\n

    CH3<\/sub>CO2<\/sub>\u2013<\/sup>(aq<\/em>) + H2<\/sub>O(l<\/em>) \u21cc CH3<\/sub>CO2<\/sub>H(aq<\/em>) + OH\u2212<\/sup>(aq)<\/div>\r\n
    <\/div>\r\n

    Solution:<\/strong><\/p>\r\nThe provided equilibrium concentrations and a value for the equilibrium constant will permit calculation of the missing equilibrium concentration. The process in question is the base ionization of acetate ion, for which\r\n

    \"\"<\/p>\r\n

    Substituting the available values into the K<\/em>b<\/sub> expression gives<\/p>\r\n\r\n

    \"\"<\/div>\r\n
    \"\"<\/div>\r\n

    Solving the above equation for the acetic acid molarity yields [CH3<\/sub>CO2<\/sub>H] = 1.1 \u00d7 10\u22125<\/sup>M<\/em>.<\/p>\r\n

    Check Your Learning:<\/strong><\/p>\r\nWhat is the pH of a 0.083-M<\/em> solution of NaCN?\r\n\r\n \r\n

    \r\n
    Answer:<\/strong><\/div>\r\n

    11.11<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n

    Key Concepts and Summary<\/strong><\/h3>\r\n

    The ions composing salts may possess acidic or basic character, ionizing when dissolved in water to yield acidic or basic solutions. Acidic cations are typically the conjugate partners of weak bases, and basic anions are the conjugate partners of weak acids.<\/p>\r\n\r\n<\/div>\r\n

    \r\n
    \r\n
    <\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    <\/div>","rendered":"
    \n

     <\/p>\n

    \n

    Learning Objectives<\/strong><\/h3>\n

    By the end of this section, you will be able to:<\/p>\n

      \n
    • Calculate the concentrations of the various species in a salt solution<\/li>\n
    • Describe the acid ionization of hydrated metal ions<\/li>\n<\/ul>\n<\/div>\n
      \n

      Salts with Acidic Ions<\/strong><\/h3>\n

      Salts are ionic compounds composed of cations and anions, either of which may be capable of undergoing an acid or base ionization reaction with water. Aqueous salt solutions, therefore, may be acidic, basic, or neutral, depending on the relative acid-base strengths of the salt\u2019s constituent ions. For example, dissolving ammonium chloride in water results in its dissociation, as described by the equation<\/p>\n

      NH4<\/sub>Cl(s<\/em>) \u2192 NH4<\/sub>+<\/sup>(aq<\/em>) + Cl\u2013<\/sup>(aq<\/em>)<\/div>\n

      The ammonium ion is the conjugate acid of the weak base ammonia, NH3<\/sub>, and so it will undergo acid ionization (or acid hydrolysis<\/em>):<\/p>\n

      NH4<\/sub>+<\/sup>(aq<\/em>) + H2<\/sub>O(l<\/em>) \u21cc H3<\/sub>O+<\/sup>(aq<\/em>) + NH3<\/sub>(aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Ka<\/sub> = Kw<\/sub>\/Kb<\/sub><\/div>\n

      Since ammonia is a weak base, K<\/em>b<\/sub> is measurable and K<\/em>a<\/sub> > 0 (ammonium ion is a weak acid).<\/p>\n

      The chloride ion is the conjugate base of hydrochloric acid, and so its base ionization (or base hydrolysis<\/em>) reaction is represented by<\/p>\n

      Cl\u2013<\/sup>(aq<\/em>) + H2<\/sub>O(l<\/em>) \u21cc HCl(aq<\/em>) + OH\u2013<\/sup>(aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Kb<\/sub> = Kw<\/sub>\/Ka<\/sub><\/div>\n

      Since HCl is a strong acid, K<\/em>a<\/sub> is immeasurably large and K<\/em>b<\/sub> \u2248 0 (chloride ions don\u2019t undergo appreciable hydrolysis).<\/p>\n

      Thus, dissolving ammonium chloride in water yields a solution of weak acid cations (NH4<\/sub>+<\/sup>) and inert anions (Cl\u2212<\/sup>), resulting in an acidic solution.<\/p>\n

      \n

      Calculating the pH of an Acidic Salt Solution <\/strong><\/p>\n

      Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, [C6<\/sub>H5<\/sub>NH3<\/sub>]Cl, a salt prepared by the reaction of the weak base aniline and hydrochloric acid. What is the pH of a 0.233 M<\/em> solution of anilinium chloride?<\/p>\n

      C6<\/sub>H5<\/sub>NH3<\/sub>+<\/sup>(aq) + H2<\/sub>O(l) \u21cc H3<\/sub>O+<\/sup>(aq<\/em>) + C6<\/sub>H5<\/sub>NH2<\/sub>(aq<\/em>)<\/div>\n
      <\/div>\n

      Solution:<\/strong><\/p>\n

      The K<\/em>a<\/sub> for anilinium ion is derived from the K<\/em>b<\/sub> for its conjugate base, aniline (see Appendix H):<\/p>\n

      \"\"<\/div>\n

      Using the provided information, an ICE table for this system is prepared:<\/p>\n

      \"This<\/span><\/p>\n

      Substituting these equilibrium concentration terms into the K<\/em>a<\/sub> expression gives<\/p>\n

      Ka<\/sub> = [C6<\/sub>H5<\/sub>NH2<\/sub>][H3<\/sub>O+<\/sup>]\/[C6<\/sub>H5<\/sub>NH3<\/sub>+<\/sup>]<\/div>\n
      2.3 \u00d7 10\u22125<\/sup> = (x<\/em>)(x<\/em>)\/(0.233-x<\/em>)<\/div>\n
      <\/div>\n

      Assuming x<\/em> < 0.05 \u00d7 0.233M, i.e. x<\/em> < 0.0116 M, the equation is simplified and solved for x<\/em>:<\/p>\n

      2.3 \u00d7 10\u22125<\/sup> = x<\/em>2<\/sup>\/0.233<\/div>\n
      x<\/em> = 0.0023 M\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ASSUMPTION VALID<\/div>\n
      <\/div>\n

      The ICE table defines x<\/em> as the hydronium ion molarity, and so the pH is computed as<\/p>\n

      pH = -log[H3<\/sub>O+<\/sup>] = \u2212log(0.0023 M) = 2.64<\/div>\n
      <\/div>\n

      Check Your Learning:<\/strong><\/p>\n

      What is the hydronium ion concentration in a 0.100-M<\/em> solution of ammonium nitrate, NH4<\/sub>NO3<\/sub>, a salt composed of the ions NH4<\/sub>+<\/sup> and NO3<\/sub>\u2212<\/sup>. Which is the stronger acid, C6<\/sub>H5<\/sub>NH3<\/sub>+<\/sup> or NH4<\/sub>+<\/sup>?<\/p>\n

       <\/p>\n

      \n
      Answer:<\/strong><\/div>\n

      [H3<\/sub>O+<\/sup>] = 7.5 \u00d7 10\u22126<\/sup>M<\/em>; C6<\/sub>H5<\/sub>NH3<\/sub>+<\/sup> is the stronger acid.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      \n

      Salts with Basic Ions<\/strong><\/h3>\n

      As another example, consider dissolving sodium acetate in water:<\/p>\n

      NaCH3<\/sub>CO2<\/sub>(s<\/em>) \u2192 Na+<\/sup>(aq<\/em>) + CH3<\/sub>CO2<\/sub>\u2013<\/sup>(aq<\/em>)<\/div>\n

      The sodium ion does not undergo appreciable acid or base ionization and has no effect on the solution pH. This may seem obvious from the ion\u2019s formula, which indicates no hydrogen or oxygen atoms, but some dissolved metal ions function as weak acids, as addressed later in this section.<\/p>\n

      The acetate ion, CH3<\/sub>CO2<\/sub>\u2013<\/sup>, is the conjugate base of acetic acid, CH3<\/sub>CO2<\/sub>H, and so its base ionization (or base hydrolysis<\/em>) reaction is represented by<\/p>\n

      CH3<\/sub>CO2<\/sub>\u2013<\/sup>(aq<\/em>) + H2<\/sub>O(l<\/em>) \u21cc CH3<\/sub>CO2<\/sub>H(aq<\/em>) + OH\u2013<\/sup>(aq<\/em>)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Kb<\/sub> = Kw<\/sub>\/Ka<\/sub><\/div>\n

      Because acetic acid is a weak acid, its K<\/em>a<\/sub> is measurable and K<\/em>b<\/sub> > 0 (acetate ion is a weak base).<\/p>\n

      Dissolving sodium acetate in water yields a solution of inert cations (Na+<\/sup>) and weak base anions (CH3<\/sub>CO2<\/sub>\u2013<\/sup>), resulting in a basic solution.<\/p>\n

      \n

      Equilibrium in a Solution of a Salt of a Weak Acid and a Strong Base <\/strong><\/p>\n

      Determine the acetic acid concentration in a solution with [CH3<\/sub>CO2<\/sub>\u2013<\/sup>] = 0.050 M and [OH\u2212<\/sup>] = 2.5 \u00d7 10\u22126<\/sup>M<\/em> at equilibrium.<\/p>\n

      The reaction is:<\/p>\n

      CH3<\/sub>CO2<\/sub>\u2013<\/sup>(aq<\/em>) + H2<\/sub>O(l<\/em>) \u21cc CH3<\/sub>CO2<\/sub>H(aq<\/em>) + OH\u2212<\/sup>(aq)<\/div>\n
      <\/div>\n

      Solution:<\/strong><\/p>\n

      The provided equilibrium concentrations and a value for the equilibrium constant will permit calculation of the missing equilibrium concentration. The process in question is the base ionization of acetate ion, for which<\/p>\n

      \"\"<\/p>\n

      Substituting the available values into the K<\/em>b<\/sub> expression gives<\/p>\n

      \"\"<\/div>\n
      \"\"<\/div>\n

      Solving the above equation for the acetic acid molarity yields [CH3<\/sub>CO2<\/sub>H] = 1.1 \u00d7 10\u22125<\/sup>M<\/em>.<\/p>\n

      Check Your Learning:<\/strong><\/p>\n

      What is the pH of a 0.083-M<\/em> solution of NaCN?<\/p>\n

       <\/p>\n

      \n
      Answer:<\/strong><\/div>\n

      11.11<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      \n

      Key Concepts and Summary<\/strong><\/h3>\n

      The ions composing salts may possess acidic or basic character, ionizing when dissolved in water to yield acidic or basic solutions. Acidic cations are typically the conjugate partners of weak bases, and basic anions are the conjugate partners of weak acids.<\/p>\n<\/div>\n

      \n
      \n
      <\/div>\n<\/div>\n<\/div>\n<\/div>\n
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