Answer
pH = 5.18
Work Step by Step
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [C_6H_5O^-] = x$
-$[C_6H_5OH] = [C_6H_5OH]_{initial} - x = 0.34 - x$
For approximation, we consider: $[C_6H_5OH] = 0.34M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_6H_5O^-]}{ [C_6H_5OH]}$
$Ka = 1.3 \times 10^{- 10}= \frac{x * x}{0.34}$
$Ka = 1.3 \times 10^{- 10}= \frac{x^2}{ 0.34}$
$ 4.42 \times 10^{- 11} = x^2$
$x = 6.64 \times 10^{- 6}$
5% test: $\frac{ 6.64 \times 10^{- 6}}{ 0.34} \times 100\% = 0.00196\%$
%ionization < 5% : Right approximation.
Therefore: $[H_3O^+] = x = 6.64 \times 10^{- 6} $
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 6.64 \times 10^{- 6})$
$pH = 5.18$
