Answer
$pH = 2.611$
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.
-$[H_3O^+] = [C_6H_5COO^-] = x$
-$[C_6H_5COOH] = [C_6H_5COOH]_{initial} - x = 0.05 - x$
For approximation, we consider: $[C_6H_5COOH] = 0.05M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_6H_5COO^-]}{ [C_6H_5COOH]}$
$Ka = 1.2 \times 10^{- 4}= \frac{x * x}{ 0.05}$
$Ka = 1.2 \times 10^{- 4}= \frac{x^2}{ 0.05}$
$ 6 \times 10^{- 6} = x^2$
$x = 2.449 \times 10^{- 3}$
Therefore: $[H_3O^+] = [C_6H_5COO^-] = x = 2.449 \times 10^{- 3}M $
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 2.449 \times 10^{- 3})$
$pH = 2.611$
