Chemistry: The Molecular Science (5th Edition)

The pH of a $0.15M$ $CH_3COOH$ solution is equal to $2.78$.
1. We have these concentrations at equilibrium: -$[H_3O^+] = [CH_3COO^-] = x$ -$[CH_3COOH] = [CH_3COOH]_{initial} - x = 0.15 - x$ 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 dissociation: $\frac{ 1.643 \times 10^{- 3}}{ 0.15} \times 100\% = 1.095\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [CH_3COO^-] = x = 1.643 \times 10^{- 3}M$ And, since 'x' has a very small value (compared to the initial concentration): $[CH_3COOH] \approx 0.15M$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.643 \times 10^{- 3})$ $pH = 2.784$