Unit 4 Acid-Base and Solubility Equilibria

4.2 Exercises

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Section 4.2 Exercises

  1. Explain why a sample of pure water at 40 °C is neutral even though [H3O+] = 1.7 × 10−7 M. Kw is 2.9 × 10−14 at 40 °C.
  2. The ionization constant for water (Kw) is 9.311 × 10−14 at 60 °C. Calculate [H3O+], [OH−], pH, and pOH for pure water at 60 °C.
  3. Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:

    (a) 0.000259 M HClO4
    (b) 0.21 M NaOH
    (c) 0.000071 M Ba(OH)2
    (d) 2.5 M KOH

  4. What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
  5. Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. The pH of lime juice is approximately 2.00.
  6. The hydroxide ion concentration in household ammonia is 3.2 × 10−3 M at 25 °C. What is the concentration of hydronium ions in the solution?
  7. Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.
  8. The odour of vinegar is due to the presence of acetic acid, CH3CO2H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1-M aqueous solution of this acid.
  9. Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.
  10. What is the ionization constant at 25 °C for the weak acid CH3NH3+, the conjugate acid of the weak base CH3NH2, Kb = 4.4 × 10−4.
  11. Using the answer from Exercise 10, which base, CH3NH2 or (CH3)2NH, is the stronger base? Which conjugate acid, CH3NH3+ or (CH3)2NH2+, is the stronger acid?  The Kb of (CH3)2NH is 5.9 × 10−4.
  12. Explain why equilibrium calculations are not necessary to determine ionic concentrations in solutions of certain strong electrolytes such as NaOH and HCl. Under what conditions are equilibrium calculations necessary as part of the determination of the concentrations of all ions of some other strong electrolytes in solution?
  13. What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid?
  14. Which of the following will increase the percent of NH3 that is converted to the ammonium ion in water (Hint: Use Le Châtelier’s principle.)?

    (a) addition of NaOH
    (b) addition of HCl
    (c) addition of NH4Cl

  15. What is the effect on the concentrations of NO2−, HNO2, and OH− when the following are added to a solution of KNO2 in water:

    (a) HCl
    (b) HNO2
    (c) NaOH
    (d) NaCl
    (e) KNO2
    The equation for the equilibrium is:
    [latex]\text{NO}_2^{\;\;-}(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{HNO}_2(aq)\;+\;\text{OH}^{-}(aq)[/latex]

  16. Why is the hydronium ion concentration in a solution that is 0.10 M in HCl and 0.10 M in HCOOH determined by the concentration of HCl?
  17. From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.

    (a) NH3: [OH−] = 3.1 × 10−3 M;
    [NH4+] = 3.1 × 10−3 M;
    [NH3] = 0.533 M;
    (b) HNO2: [H3O+] = 0.011 M;
    [NO2−] = 0.0438 M;
    [HNO2] = 1.07 M;
    (c) (CH3)3N: [(CH3)3N] = 0.25 M;
    [(CH3)3NH+] = 4.3 × 10−3 M;
    [OH−] = 4.3 × 10−3 M;
    (d) [NH4+]: [NH4+] = 0.100 M;
    [NH3] = 7.5 × 10−6 M;
    [H3O+] = 7.5 × 10−6 M

  18. Determine Ka for hydrogen sulfate ion, HSO4−. In a 0.10-M solution the acid is 29% ionized.
  19. Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

    (a) [latex]\text{HTe}^{-}[/latex] (as a base)
    (b) [latex](\text{CH}_3)_3\text{NH}^{+}[/latex]
    (c) [latex]\text{HAsO}_4^{\;\;3-}[/latex] (as a base)
    (d) [latex]\text{HO}_2^{\;\;-}[/latex] (as a base)
    (e) [latex]\text{C}_6\text{H}_5\text{NH}_3^{\;\;+}[/latex]
    (f) [latex]\text{HSO}_3^{\;\;-}[/latex] (as a base)

  20. Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 M in HNO2 and 0.120 M in HBrO.  Hint: The stronger of the two acids is essentially the sole source of H3O+.
  21. Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 M in NH3 and 0.100 M in C6H5NH2.  Hint: The stronger of the two bases is essentially the sole source of OH−.
  22. Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants are provided for weak acids and weak bases; source: OpenStax Chemistry 2e

    (a) 0.0092 M HClO, a weak acid
    (b) 0.0784 M C6H5NH2, a weak base
    (c) 0.0810 M HCN, a weak acid
    (d) 0.11 M (CH3)3N, a weak base
    (e) 0.120 M Fe(H2O)62+ a weak acid, Ka = 1.6 × 10−7.  The ionization equation is
    [latex]\text{Fe}(\text{H}_2\text{O})_6^{2+}(aq)\;+\;\text{H}_2\text{O}(l)\;\rightleftharpoons\;\text{Fe}(\text{H}_2\text{O})_5(\text{OH})^+(aq)\;+\;\text{H}_3\text{O}^+(aq)[/latex]

  23. White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm3, what is the pH?  Start by calculating the molar concentration of acetic acid.  Note that 1 cm3 = 1 mL = 10−3 L.
  24. The pH of a 0.15-M solution of HSO4− is 1.43. Determine Ka for HSO4− from these data.
  25. The pH of a solution of household ammonia, a 0.950 M solution of NH3, is 11.612. Determine Kb for NH3 from these data.

Solutions

  1. In a neutral solution [H3O+] = [OH−]. At 40 °C, [H3O+] = [OH−] = (2.910−14)1/2 = 1.7 × 10−7 M.
  2. x = 3.051 × 10−7 M = [H3O+] = [OH−]
    pH = −log3.051 × 10−7 = −(−6.5156) = 6.5156
    pOH = pH = 6.5156
  3. (a) pH = 3.587; pOH = 10.41
    (b) pH = 0.68; pOH = 13.32
    (c) pOH = 3.85; pH = 10.15
    (d) pH = −0.40; pOH = 14.40
  4. [H3O+] = 3.0 × 10−7 M; [OH−] = 3.3 × 10−8 M
  5. approximately [H3O+] = 1 .0 × 10−2 M; approximately [OH−] = 1.0 × 10−12 M
  6. [OH−] = 3.1 × 10−12 M
  7. The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH−, which causes the solution to be basic.
    [latex]\text{HA}(aq)\;+\;\text{OH}^-(aq)\;\longrightarrow\;\text{H}_2\text{O}(l)\;+\;\text{A}^-(aq)[/latex]
    [latex]\text{A}^-(aq)\;+\;\text{H}_2\text{O}(l)\;\rightleftharpoons\;\text{HA}(aq)\;+\;\text{OH}^-(aq)[/latex]
  8. [latex][\text{H}_2\text{O}]\;>\;[\text{CH}_3\text{CO}_2\text{H}]\;>\;[\text{H}_3\text{O}^{+}]\;{\approx}\;[\text{CH}_3\text{CO}_2^{\;\;-}]\;>\;[\text{OH}^{-}][/latex]
  9. [latex]\begin{array}{ccccccccc} \text{Mg(OH)}_2(s) & + & 2\text{HCl}(aq) & {\longrightarrow} & \text{Mg}^{2+}(aq) & + & 2\text{Cl}^{-}(aq) & + & 2\text{H}_2\text{O}(l) \\[0.5em] \text{BB} & & \text{BA} & & \text{CB} & & \text{CA} & & \end{array}[/latex]
  10. [latex]K_{\text{a}} = 2.3\;\times\;10^{-11}[/latex]
  11. The stronger base or stronger acid is the one with the larger Kb or Ka, respectively. In these two examples, they are (CH3)2NH (stronger base) and CH3NH3+ (stronger acid).
  12. Strong electrolytes are 100% ionized, and, as long as the component ions are neither weak acids nor weak bases, the ionic species present result from the dissociation of the strong electrolyte. Equilibrium calculations are necessary when one (or more) of the ions is a weak acid or a weak base.
  13. 1. Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration. 2. Assume we can neglect the contribution of water to the equilibrium concentration of H3O+.
  14. (b) The addition of HCl.
  15. (a) Adding HCl will add H3O+ ions, which will then react with the OH− ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO2, and decreasing the concentration of NO2− ions.
    (b) Adding HNO2 increases the concentration of HNO2 and shifts the equilibrium to the left, increasing the concentration of NO2− ions and decreasing the concentration of OH− ions.
    (c) Adding NaOH adds OH− ions, which shifts the equilibrium to the left, increasing the concentration of NO2− ions and decreasing the concentrations of HNO2.
    (d) Adding NaCl has no effect on the concentrations of the ions.
    (e) Adding KNO2 adds NO2− ions and shifts the equilibrium to the right, increasing the HNO2 and OH− ion concentrations.
  16. This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO2H exists primarily as HCO2H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO2H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [H3O+] produced by the stronger acid.
  17. (a) [latex]K_{\text{b}} = 1.8\;\times\;10^{-5}[/latex];
    (b) [latex]K_{\text{a}} = 4.5\;\times\;10^{-4}[/latex];
    (c) [latex]K_{\text{b}} = 7.4\;\times\;10^{-5}[/latex];
    (d) [latex]K_{\text{a}} = 5.6\;\times\;10^{-10}[/latex]
  18. [latex]K_{\text{a}} = 1.2\;\times\;10^{-2}[/latex]
  19. (a) [latex]K_{\text{b}} = 4.3\;\times\;10^{-12}[/latex];
    (b) [latex]K_{\text{a}} = 1.6\;\times\;10^{-8}[/latex];
    (c) [latex]K_{\text{b}} = 5.9\;\times\;10^{-7}[/latex];
    (d) [latex]K_{\text{b}} = 4.2\;\times\;10^{-3}[/latex];
    (e) [latex]K_{\text{b}} = 2.3\;\times\;10^{-3}[/latex];
    (f) [latex]K_{\text{b}} = 6.3\;\times\;10^{-13}[/latex]
  20. Use the quadratic formula since % ionization of HNO2 exceeded 5% (it is 6%).
    [H3O+] = 7.4 × 10−3 M
    [HNO2] = 0.127 M
    [OH−] = 1.3 × 10−12 M
    [BrO−] = 4.5 × 10−8 M
    [HBrO] = 0.120 M
  21. [OH−] = [NH4+] = 0.0014 M
    [NH3] = 0.114 M
    [H3O+] = 7.1 × 10−12 M
    [C6H5NH3+] = 3.1 × 10−8 M
    [C6H5NH2] = 0.100 M
  22. (a) [latex]\frac{[\text{H}_3\text{O}^{+}][\text{ClO}^{-}]}{[\text{HClO}]} = \frac{(x)(x)}{(0.0092\;-\;x)}\;{\approx}\;\frac{(x)(x)}{0.0092} = 2.9\;\times\;10^{-8}[/latex]
    Solving for x gives 1.63 × 10−5 M. This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
    [H3O+] = [ClO] = 5.8 × 10−5 M
    [HClO] = 0.00092 M
    [OH−] = 6.1 × 10−10 M;
    (b) [latex]\frac{[\text{C}_6\text{H}_5\text{NH}_3^{\;\;+}][\text{OH}^{-}]}{[\text{C}_6\text{H}_5\text{NH}_2]} = \frac{(x)(x)}{(0.0784\;-\;x)}\;{\approx}\;\frac{(x)(x)}{0.0784} = 4.3\;\times\;10^{-10}[/latex]
    Solving for x gives 5.81 × 10−6 M. This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
    [C6H5NH3+] = [OH−] = 5.8 × 10−6 M
    [C6H5NH2] = 0.00784 M
    [H3O+] = 1.7× 10−9 M;
    (c) [latex]\frac{[\text{H}_3\text{O}^{+}][\text{CN}^{-}]}{[\text{HCN}]} = \frac{(x)(x)}{(0.0810\;-\;x)}\;{\approx}\;\frac{(x)(x)}{0.0810} = 4.9\;\times\;10^{-10}[/latex]
    Solving for x gives 6.30 × 10−6 M. This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
    [H3O+] = [CN−] = 6.3 × 10−6 M
    [HCN] = 0.0810 M
    [OH−] = 1.6 × 10−9 M;
    (d) [latex]\frac{[(\text{CH}_3)_3\text{NH}^{+}][\text{OH}^{-}]}{[(\text{CH}_3)_3\text{N}]} = \frac{(x)(x)}{(0.11\;-\;x)}\;{\approx}\;\frac{(x)(x)}{0.11} = 6.3\;\times\;10^{-5}[/latex]
    Solving for x gives 2.63 × 10−3 M. This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
    [(CH3)3NH+] = [OH−] = 2.6 × 10−3 M
    [(CH3)3N] = 0.11 M
    [H3O+] = 3.8 × 10−12 M;
    (e) [latex]\frac{[\text{Fe}(\text{H}_2\text{O})_5(\text{OH})^{+}][\text{H}_3\text{O}^{+}]}{[\text{Fe}(\text{H}_2\text{O})_6^{\;\;2+}]} = \frac{(x)(x)}{(0.120\;-\;x)}\;{\approx}\;\frac{(x)(x)}{0.120} = 1.6\;\times\;10^{-7}[/latex]
    Solving for x gives 1.39 × 10−4 M. This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
    [Fe(H2O)5(OH)+] = [H3O+] = 1.4 × 10−4 M
    [Fe(H2O)62+] = 0.120 M
    [OH−] = 7.2 × 10−11 M
  23. The initial (nonionized) concentration of acetic acid is 0.83 M.  pH = 2.41
  24. [latex]K_{\text{a}} = 1.2\;\times\;10^{-2}[/latex]
  25. [latex]K_{\text{b}} = 1.8\;\times\;10^{-5}[/latex]

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