# Answers to More Chapter 14 Study Questions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1. c)  $$\ce{NaOH}$$ (strong base)