Unit 4 Acid-Base and Solubility Equilibria

4.3 Exercises

OpenStax

Section 4.3 Exercises

  1. Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134-M solution of H2CO3, a diprotic acid: [H3O+], [OH−], [H2CO3], [HCO3−], [CO32−]? No calculations are needed to answer this question.
  2. Calculate the concentration of each species present in a 0.010-M solution of phthalic acid, C6H4(CO2H)2.
    [latex]\text{C}_6\text{H}_4(\text{CO}_2\text{H})_2(aq)\;+\;\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^{+}(aq)\;+\;\text{C}_6\text{H}_4(\text{CO}_2\text{H})(\text{CO}_2)^{-}(aq)  \\K_{\text{a}} = 1.1\;\times\;10^{-3} \\ \\\text{C}_6\text{H}_4(\text{CO}_2\text{H})(\text{CO}_2)^-(aq)\;+\;\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^{+}(aq)\;+\;\text{C}_6\text{H}_4(\text{CO}_2)_2^{\;\;2-}(aq) \\K_{\text{a}} = 3.9\;\times\;10^{-6}[/latex]
  3. The ion HTe− is an amphiprotic species; it can act as either an acid or a base.  Consult the table of ionization constants of weak acids (Source: OpenStax Chemistry 2e).

    (a) What is Ka for the acid reaction of HTe− with H2O?
    (b) What is Kb for the reaction in which HTe− functions as a base in water?
    (c) Demonstrate whether or not the second ionization of H2Te can be neglected in the calculation of [HTe−] in a 0.10 M solution of H2Te.

  4. The active ingredient formed by aspirin in the body is salicylic acid, HOC6H4COOH. The carboxyl group (−COOH) acts as a weak acid. The phenol group (an −OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001-M aqueous solution of HOC6H4COOH.

Solutions

  1. [H3O+] and [HCO3−] are practically equal
  2. [C6H4(CO2H)2] = 7.2 × 10−3 M, [C6H4(CO2H)(CO2)−] = [H3O+] = 2.8 × 10−3 M, [C6H4(CO2)22−] = 3.9 × 10−6 M, [OH−] = 3.6 × 10−12 M
  3. (a) Ka2 = 1.6 × 10−11
    (b) Kb = 4.3 × 10−12
    (c) [latex]K_{\text{a2}}=\frac{[\text{Te}^{2-}][\text{H}_3\text{O}^{+}]}{[\text{HTe}^{-}]} = \frac{(x)(0.015\;+\;x)}{(0.015\;-\;x)}\;{\approx}\;\frac{(x)(0.015)}{0.015} = 1.6\;\times\;10^{-11}[/latex]
    Solving for x gives 1.6 × 10−11 M. Therefore, compared with 0.015 M, this value is negligible (1.1 × 10−7%).
  4. [latex][\text{H}_2\text{O}]\;>\;[\text{HOC}_6\text{H}_4\text{COO}^-]\;\approx\;[\text{H}_3\text{O}^{+}]\;\gg\;[\text{OH}^{-}]\;>\;[\text{OC}_6\text{H}_4\text{COO}^{2-}][/latex]

License

Icon for the Creative Commons Attribution 4.0 International License

4.3 Exercises Copyright © by OpenStax is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Share This Book