## Chemistry 10th Edition

(a) 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [CH_3COO^-] = x$ -$[CH_3COOH] = [CH_3COOH]_{initial} - x = 0.15 - x$ For approximation, we consider: $[CH_3COOH] = 0.15M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][CH_3COO^-]}{ [CH_3COOH]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.15}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.15}$ $2.7 \times 10^{- 6} = x^2$ $x = 1.643 \times 10^{- 3}$ Percent ionization: $\frac{ 1.643 \times 10^{- 3}}{ 0.15} \times 100\% = 1.095\%$ %ionization < 5% : Right approximation. (b) 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [CH_3COO^-] = x$ -$[CH_3COOH] = [CH_3COOH]_{initial} - x = 0.015 - x$ For approximation, we consider: $[CH_3COOH] = 0.015M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][CH_3COO^-]}{ [CH_3COOH]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.015}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.015}$ $2.7 \times 10^{- 7} = x^2$ $x = 5.196 \times 10^{- 4}$ Percent ionization: $\frac{ 5.196 \times 10^{- 4}}{ 0.015} \times 100\% = 3.464\%$ %ionization < 5% : Right approximation.