## Chemistry (4th Edition)

(a) 1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [N{H_4}^+] = x$ -$[NH_3] = [NH_3]_{initial} - x = 0.1 - x$ For approximation, we consider: $[NH_3] = 0.1M$ 2. Now, use the Kb value and equation to find the 'x' value. $Ka = \frac{[OH^-][N{H_4}^+]}{ [NH_3]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.1}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{0.1}$ $1.8 \times 10^{- 6} = x^2$ $x = 1.342 \times 10^{- 3}$ Percent ionization: $\frac{ 1.342 \times 10^{- 3}}{ 1 \times 10^{- 1}} \times 100\% = 1.342\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = x = 1.342 \times 10^{- 3}$ 3. Convert $[OH^-]$ into pH. $pOH = -log[OH^-]$ $pOH = -log( 1.342 \times 10^{- 3})$ $pOH = 2.872$ $pH + pOH = 14$ $pH + 2.872 = 14$ $pH = 11.128$ ----- (b) 1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [C_5H_5NH^+] = x$ -$[C_5H_5N] = [C_5H_5N]_{initial} - x = 0.05 - x$ For approximation, we consider: $[C_5H_5N] = 0.05M$ 2. Now, use the Kb value and equation to find the 'x' value. $Ka = \frac{[OH^-][C_5H_5NH^+]}{ [C_5H_5N]}$ $Ka = 1.7 \times 10^{- 9}= \frac{x * x}{ 5 \times 10^{- 2}}$ $Ka = 1.7 \times 10^{- 9}= \frac{x^2}{ 5 \times 10^{- 2}}$ $8.5 \times 10^{- 11} = x^2$ $x = 9.22 \times 10^{- 6}$ Percent ionization: $\frac{ 9.22 \times 10^{- 6}}{ 5 \times 10^{- 2}} \times 100\% = 0.01844\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = x = 9.22 \times 10^{- 6}$ 3. Convert $[OH^-]$ into pH. $pOH = -log[OH^-]$ $pOH = -log( 9.22 \times 10^{- 6})$ $pOH = 5.035$ $pH + pOH = 14$ $pH + 5.035 = 14$ $pH = 8.965$