# Answers to More Chapter 14 Study Questions

1.  NH4Cl(aq)  →  NH4+(aq)  + Cl-(aq);  since Cl- is a spectator ion, just consider NH4+, which is a weak acid.

NH4+(aq)  ⇌  H+(aq)  +  NH3(aq);  Ka  =  5.6 x 10-10

$$\mathrm{K_a=\dfrac{[H^+][NH_3]}{[NH_4^+]}}$$ ;  x  =  [H+] = [NH3];  [NH4+] = 0.030 M;  $$\mathrm{5.6\times10^{-10}=\dfrac{x^2}{0.030\:M}}$$

x2  =  (0.030)( 5.6 x 10-10)  =  1.7 x 10-11x = (1.7 x 10-11)1/2  =  4.1 x 10-6 M =  [H+]

pH  =  - log[H+]  =  - log (4.1 x 10-6 M);  pH  = 5.4

2.  NaF(aq)  →   Na+(aq)  +  F-(aq); since Na+ is a spectator ion, just consider F-, which is a weak base.

F-(aq) + H2O ⇌ HF(aq) + OH-(aq);  $$\mathrm{K_b = \dfrac{10^{-14}}{K_a} = \dfrac{10^{-14}}{(7.2 \times 10^{-4})} = 1.38 \times 10^{-11}}$$

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

x2  =  (0.14)( 1.38 x 10-11)  =  1.9 x 10-12x = (1.9 x 10-12)1/2  =  1.4 x 10-6 M =  [OH-]

pOH  =  - log[OH-]  =  - log (1.4 x 10-6 M)  = 5.9;  pH  =  14 - pOH =  14 - 5.9;  pH = 8.1

3.    V1 x M1 =  V2 x M2;  HNO3 is a strong acid;  [HNO3]  =  [H+];  calculate [HNO3].

0.00600 L  x  3.00 M  =  18.0 L  x M2;   $$\mathrm{M_2=\dfrac{0.00600\times3.00}{18.0}=1.00 \times 10^{-3}\: M = [H^+]}$$

pH  =  - log[H+]  =  - log (1.00 x 10-3 M);  pH  = 3.0

4.  NH4+(aq)  +  OH-(aq)   →   NH3(aq)  +  H2O

conjugate acid base pairs: NH4+/NH3  and  H2O/ OH-

5.  c)  NaOH (strong base)