## Chemistry (4th Edition)

1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [C_6H_5O^-] = x$ -$[C_6H_5OH] = [C_6H_5OH]_{initial} - x$ For approximation, we consider: $[C_6H_5OH] = [C_6H_5OH]_{initial}$ (a) 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_6H_5O^-]}{ [C_6H_5OH]}$ $Ka = 1.3 \times 10^{- 10}= \frac{x * x}{ 5.6 \times 10^{- 1}}$ $Ka = 1.3 \times 10^{- 10}= \frac{x^2}{ 5.6 \times 10^{- 1}}$ $7.27 \times 10^{- 11} = x^2$ $x = 8.53 \times 10^{- 6}$ 5% test: $\frac{ 8.53 \times 10^{- 6}}{ 5.6 \times 10^{- 1}} \times 100\% = 0.00152\%$ %ionization < 5% : Right approximation. (b) 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_6H_5O^-]}{ [C_6H_5OH]}$ $Ka = 1.3 \times 10^{- 10}= \frac{x * x}{ 2.5 \times 10^{- 1}}$ $Ka = 1.3 \times 10^{- 10}= \frac{x^2}{ 2.5 \times 10^{- 1}}$ $3.25 \times 10^{- 11} = x^2$ $x = 5.7 \times 10^{- 6}$ 5% test: $\frac{ 5.7 \times 10^{- 6}}{ 2.5 \times 10^{- 1}} \times 100\% = 0.00228\%$ %ionization < 5% : Right approximation. (c) 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_6H_5O^-]}{ [C_6H_5OH]}$ $Ka = 1.3 \times 10^{- 10}= \frac{x * x}{ 1.8 \times 10^{- 6}}$ $Ka = 1.3 \times 10^{- 10}= \frac{x^2}{ 1.8 \times 10^{- 6}}$ $2.33 \times 10^{- 16} = x^2$ $x = 1.52 \times 10^{- 8}$ 5% test: $\frac{ 1.52 \times 10^{- 8}}{ 1.8 \times 10^{- 6}} \times 100\% = 0.85\%$ %ionization < 5% : Right approximation.