## Chemistry 12th Edition

(a) $pH = 2.57$ (b) $pH = 4.44$
(a) 1. We have those concentrations at equilibrium: -$[H_3O^+] = [CH_3COO^-] = x$ -$[CH_3COOH] = [CH_3COOH]_{initial} - x = 0.4 - x$ For approximation, we consider: $[CH_3COOH] = 0.4M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][CH_3COO^-]}{ [CH_3COOH]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.4}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.4}$ $7.2 \times 10^{- 6} = x^2$ $x = 2.7 \times 10^{- 3}$ Percent dissociation: $\frac{ 2.7 \times 10^{- 3}}{ 0.4} \times 100\% = 0.67\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [CH_3COO^-] = x = 2.7 \times 10^{- 3}M$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 2.7 \times 10^{- 3})$ $pH = 2.57$ ---- (b) 1. Drawing the ICE table we get these concentrations at the equilibrium: $CH_3COOH(aq) + H_2O(l) \lt -- \gt CH_3COO^-(aq) + H_3O^+(aq)$ Remember: Reactants at equilibrium = Initial Concentration - x And Products = Initial Concentration + x $[CH_3COOH] = 0.4 M - x$ $[CH_3COO^-] = 0.2M + x$ $[H_3O^+] = 0 + x$ 2. Calculate 'x' using the $K_a$ expression. $1.8\times 10^{- 5} = \frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$ $1.8\times 10^{- 5} = \frac{( 0.2 + x )* x}{ 0.4 - x}$ Considering 'x' has a very small value. $1.8\times 10^{- 5} = \frac{ 0.2 * x}{ 0.4}$ $1.8\times 10^{- 5} = 0.5x$ $\frac{ 1.8\times 10^{- 5}}{ 0.5} = x$ $x = 3.6\times 10^{- 5}$ Percent dissociation: $\frac{ 3.6\times 10^{- 5}}{ 0.4} \times 100\% = 9\times 10^{- 3}\%$ x = $[H_3O^+]$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 3.6 \times 10^{- 5})$ $pH = 4.44$