Answers to Chapter 14 Study Questions

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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. \(\mathrm{CH_3COO^-(aq) + H_2O \rightleftharpoons CH_3COOH(aq) + OH^-(aq)}\)

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

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)