# Answers to Chapter 14 Study Questions

1. Fill in the following table:
 $$\ce{[H+]}$$ $$\ce{[OH- ]}$$ pH pOH acid, base or neutral? 1.0 × 10-4 M 1.0 × 10-10 M 4.0 10.0 acid 1.0 × 10-7 M 1.0 × 10-7 M 7.0 7.0 neutral 1.0 × 10-12 M 1.0 × 10-2 M 12.0 2.0 base 1.0 M 1.0 × 10-14 M 0.0 14.0 acid 3.2 × 10-4 M 3.2 × 10-11 M 3.5 10.5 acid 4.6 × 10-3 M 2.2 × 10-12 M 2.3 11.7 acid 1.2 × 10-3 M 8.2 × 10-12 M 2.9 11.1 acid

1. strong, acid, $$\mathrm{HNO_3(aq) \rightarrow H^+(aq) + NO_3^-(aq)}$$
2. weak, acid, $$\mathrm{HF (aq) \rightleftharpoons H^+(aq) + F^-(aq)}$$
3. weak, base, $$\mathrm{F^-(aq) + H_2O \rightleftharpoons HF(aq) + OH^-(aq)}$$
4. amphoteric, acid: $$\mathrm{HSO_3^-(aq) \rightleftharpoons H^+(aq) + SO_3^{2-}(aq)}$$;
base: $$\mathrm{HSO_3^-(aq) + H_2O \rightleftharpoons H_2SO_3 (aq) + OH^-(aq)}$$
5. strong, base, $$\mathrm{KOH(s) \rightarrow K^+(aq) + OH^-(aq)}$$

1. $$\mathrm{HCl = strong\: acid}$$; $$\mathrm{[H^+] = \dfrac{0.010\:mol}{10.0\:L}=1.0\times10^{-13}\:M}$$; $$\mathrm{pH = 3.0}$$
2. NaOH = strong base; $$\mathrm{[NaOH] = \dfrac{6.0\:g\:NaOH}{15.0\:L}\times\dfrac{1\:mol\:NaOH}{40.0\:g\:NaOH}=0.010\: mol/L =}$$

$$\mathrm{[OH^-] = [NaOH] = 1.0 \times 10^{-2}\: M}$$; $$\mathrm{[H^+] = 1.0 \times 10^{-12}\: M}$$; $$\mathrm{pH = 12.0}$$
3. $$\mathrm{HBr = strong\: acid}$$; $$\mathrm{5.0\: mL = 5.0 \times 10^{-3}\: L}$$; $$\mathrm{V_1 \times M_1 = V_2 \times M_2}$$
$$\mathrm{(5.0 \times 10^{-3}\: L) \times (0.40\: M) = (20\: L) \times M_2}$$; $$\mathrm{M_2 = \dfrac{(5.0 \times 10^{-3}\: L) \times (0.40\: M)}{(20\: L)}}$$)
$$\mathrm{M_2 = 1.0 \times 10^{-4}\: M = [H^+]}$$; $$\mathrm{pH = 4.0}$$.
4. $$\mathrm{K_a=\dfrac{[H^+]\times[A^-]}{[HA]}}$$; $$\mathrm{[H^+] = [A^-] = x}$$; $$\mathrm{[HA] \approx 0.10\: M}$$; $$\mathrm{6.4\times10^{-5}=\dfrac{x^2}{0.10\:M}}$$

$$\mathrm{x^2 = (6.4 \times 10^{-5})(0.10) = 6.4 \times 10^{-6}\: M}$$; $$\mathrm{x = (6.4 \times 10^{-6})^{1/2} = 2.5 \times 10^{-3}}$$; $$\mathrm{pH = 2.6}$$
5. $$\mathrm{K_b = \dfrac{[HClO][OH^-]}{[ClO^-]}}$$; [HClO] = [OH-] = x; $$\mathrm{[ClO^-] \approx 0.20\: M}$$;

$$\mathrm{K_a \times K_b = 10^{-14}}$$; $$\mathrm{K_b = \dfrac{10^{-14}}{K_a(HClO)}=\dfrac{10^{-14}}{3.5\times10^{-8}} = 2.9 \times 10^{-7}}$$

$$\mathrm{K_b = \dfrac{[HClO][OH^-]}{[ClO^-]}}$$; $$\mathrm{2.9 \times 10^{-7} = \dfrac{x^2}{0.20\:M}}$$; $$\mathrm{x^2 = (2.9 \times 10^{-7})(0.20\: M) = 5.8 \times 10^{-8}}$$

$$\mathrm{x = [OH^-] = (5.8 \times 10^{-8})^{1/2} = 2.4 \times 10^{-4}\: M}$$; $$\mathrm{pOH = 3.6}$$; $$\mathrm{pH = 14.0 - 3.6 = 10.4}$$
6. 0.10 mol $$\ce{H+}$$ and 0.10 mol $$\mathrm{OH^- \rightarrow}$$ 0.10 mol $$\ce{H2O}$$; 0.10 mol $$\ce{H+}$$ remains in 1.0 L.
$$\mathrm{[H^+] = \dfrac{(0.10\: mol)}{(1.0\: L)} = 1.0 \times 10^{-1}\: M}$$; $$\mathrm{pH = 1.0}$$

1. $$\mathrm{H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)}$$
2. $$\mathrm{K_a = \dfrac{[H^+][HCO_3^-]}{[H_2CO_3]}}$$
3. $$\mathrm{pH = 3.68}$$; $$\mathrm{[H^+] = [HCO_3^-] = 10^{-3.68} = 2.1 \times 10^{-4}\: M}$$; $$\mathrm{[H_2CO_3] \approx 0.10\: M}$$

$$\mathrm{K_a = \dfrac{[H^+][HCO_3^-]}{[H_2CO_3]}}$$; $$\mathrm{K_a = \dfrac{(2.1\times10^{-4})^2}{0.10\:M}=\dfrac{(2.1)^2\times10^{-8}}{1.0\times10^{-1}}=4.4 \times 10^{-7}}$$
4. $$\mathrm{K_a \times K_b = 1.0 \times 10^{-14}}$$; $$\mathrm{K_b = \dfrac{(1.0 \times 10^{-14})}{(4.4 \times 10^{-7})} = 2.3 \times 10^{-8}}$$

1. It can be an $$\ce{H+}$$ acceptor.
2. $$\ce{CH3COOH}$$
3. $$\mathrm{CH_3COO^-(aq) + H_2O \rightleftharpoons CH_3COOH(aq) + OH^-(aq)}$$

1. $$\mathrm{\underset{\Large{acid_1}}{HNO_2(aq)} + \underset{\Large{base_2}}{OH^-(aq)} \rightarrow \underset{\Large{base_1}}{NO_2^-(aq)} + \underset{\Large{acid_2}}{H_2O}}$$

1. weakest acid $$\ce{\rightarrow}$$ strongest acid: $$\mathrm{HCN < HC_2H_3O_2 < HF < HCl}$$

weakest base $$\ce{\rightarrow}$$ strongest base: $$\mathrm{Cl^- < F^- < C_2H_3O_2^- < CN^-}$$

1. basic
2. acidic ($$\ce{NH4+}$$ is an acid, $$\ce{NO3-}$$ is neutral)
3. neutral
4. basic ($$\ce{Na+}$$ is neutral; $$\ce{F-}$$ is a weak base)