Answer
The pH of a $0.15M$ $CH_3COOH$ solution is equal to $2.78$.
Work Step by Step
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
$pH = -log[H_3O^+]$
$pH = -log( 1.643 \times 10^{- 3})$
$pH = 2.784$
