Answers to More Chapter 14 Study Questions

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\(\mathrm{NH_4Cl(aq) \rightarrow NH_4^+(aq) + Cl^-(aq)}\); since \(\mathrm{Cl^-}\) is a spectator ion, just consider \(\ce{NH4+}\), which is a weak acid.

\(\mathrm{NH_4^+(aq) \rightleftharpoons H^+(aq) + NH_3(aq)}\); \(\mathrm{K_a = 5.6 \times 10^{-10}}\)

\(\mathrm{K_a=\dfrac{[H^+][NH_3]}{[NH_4^+]}}\) ; \(\mathrm{x = [H^+] = [NH_3]}\); \(\mathrm{[NH_4^+] = 0.030\: M}\); \(\mathrm{5.6\times10^{-10}=\dfrac{x^2}{0.030\:M}}\)

\(\mathrm{x^2 = (0.030)(5.6 \times 10^{-10}) = 1.7 \times 10^{-11}}\); \(\mathrm{x = (1.7 \times 10^{-11})^{1/2} = 4.1 \times 10^{-6}\: M = [H^+]}\)

\(\mathrm{pH = - \log[H^+] = -\log (4.1 \times 10^{-6}\: M)}\); \(\mathrm{pH = 5.4}\)

\(\mathrm{NaF(aq) \rightarrow Na^+(aq) + F^-(aq)}\); since \(\ce{Na+}\) is a spectator ion, just consider \(\ce{F-}\), which is a weak base.

\(\mathrm{F^-(aq) + H_2O \rightleftharpoons HF(aq) + OH^-(aq)}\); \(\mathrm{K_b = \dfrac{10^{-14}}{K_a} = \dfrac{10^{-14}}{(7.2 \times 10^{-4})} = 1.38 \times 10^{-11}}\)

\(\mathrm{K_b=\dfrac{[HF][OH^-]}{[F^-]}}\) ; \(\mathrm{x = [OH^-] = [HF]}\); \(\mathrm{[F^-] = 0.14\: M}\); \(\mathrm{1.38 \times 10^{-11}=\dfrac{x^2}{0.14\:M}}\)

\(\mathrm{x^2 = (0.14)( 1.38 \times 10^{-11}) = 1.9 \times 10^{-12}}\); \(\mathrm{x = (1.9 \times 10^{-12})^{1/2} = 1.4 \times 10^{-6}\: M = [OH^-]}\)

\(\mathrm{pOH = -\log[OH^-] =-\log (1.4 \times 10^{-6}\: M) = 5.9}\); \(\mathrm{pH = 14 - pOH = 14 - 5.9}\); \(\mathrm{pH = 8.1}\)

\(\mathrm{V_1 \times M_1 = V_2 \times M_2}\); \(\ce{HNO3}\) is a strong acid; \(\mathrm{[HNO_3] = [H^+]}\); calculate \(\ce{[HNO3]}\).

\(\mathrm{0.00600\: L \times 3.00\: M = 18.0\: L \times M_2}\); \(\mathrm{M_2=\dfrac{0.00600\times3.00}{18.0}=1.00 \times 10^{-3}\: M = [H^+]}\)

\(\mathrm{pH = -\log[H^+] = -\log (1.00 \times 10^{-3}\: M)}\); \(\mathrm{pH = 3.0}\)

\(\mathrm{NH_4^+(aq) + OH^-(aq) \rightarrow NH_3(aq) + H_2O}\)

conjugate acid base pairs: \(\mathrm{NH_4^+/NH_3}\) and \(\mathrm{H_2O/ OH^-}\)

c) \(\ce{NaOH}\) (strong base)