Answer
$[H_3O^+] = 1.775 \times 10^{- 3}M $
$[C_6H_5COO^-] = 1.775 \times 10^{- 3}M $
$[C_6H_5COOH] \approx 0.05M$
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.
$Ka = \frac{[H_3O^+][C_6H_5COO^-]}{ [C_6H_5COOH]}$
$Ka = 6.3 \times 10^{- 5}= \frac{x * x}{ 0.05}$
$Ka = 6.3 \times 10^{- 5}= \frac{x^2}{ 0.05}$
$ 3.15 \times 10^{- 6} = x^2$
$x = 1.775 \times 10^{- 3}$
Percent ionization: $\frac{ 1.775 \times 10^{- 3}}{ 0.05} \times 100\% = 3.55\%$
%ionization < 5% : Right approximation.
Therefore: $[H_3O^+] = [C_6H_5COO^-] = x = 1.775 \times 10^{- 3}M $
And, since 'x' has a very small value (compared to the initial concentration): $[C_6H_5COOH] \approx 0.05M$
