## General Chemistry: Principles and Modern Applications (10th Edition)

(a) $[C_6H_5CH_2{CO_2}^-] = 3.01 \times 10^{-3}M$ (b) pH = 2.614
1. Drawing the ICE table we get: $[H_3O^+] = [C_6H_5CH_2{CO_2}^-] = x$ $[C_6H_5CH_2CO_2H] = [C_6H_5CH_2CO_2H]_{initial} - x$ (a) 2. Now, use the Ka value and equation to find the 'x' value. - Approximating $0.186M - x \approx 0.186M$ we get: $Ka = 4.9 \times 10^{- 5}= \frac{x * x}{ 1.86 \times 10^{- 1}}$ $Ka = 4.9 \times 10^{- 5}= \frac{x^2}{ 1.86 \times 10^{- 1}}$ $9.11 \times 10^{- 6} = x^2$ $x = 3.01 \times 10^{- 3}$ 5% test: $\frac{ 3.01 \times 10^{- 3}}{ 1.86 \times 10^{- 1}} \times 100\% = 1.623\%$ %ionization < 5% : Right approximation. $[C_6H_5CH_2{CO_2}^-] = x = 3.01 \times 10^{-3}M$ ------ (b) 2. Now, use the Ka value and equation to find the 'x' value. - Approximating $0.121M - x \approx 0.121M$ $Ka = 4.9 \times 10^{- 5}= \frac{x * x}{ 1.21 \times 10^{- 1}}$ $Ka = 4.9 \times 10^{- 5}= \frac{x^2}{ 1.21 \times 10^{- 1}}$ $5.92 \times 10^{- 6} = x^2$ $x = 2.43 \times 10^{- 3}$ 5% test: $\frac{ 2.43 \times 10^{- 3}}{ 1.21 \times 10^{- 1}} \times 100\% = 2.012\%$ %ionization < 5% : Right approximation. Therefore: $[H^+] = x = 2.43 \times 10^{-3}$ 3. Find the pH value. $pH = -log[H^+]$ $pH = -log( 2.43 \times 10^{- 3})$ $pH = 2.614$