Answer
Percent ionization = $1.7\%$
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.
-$[H_3O^+] = [Conj. Base] = x$
-$[Benzoic Acid] = [Benzoic Acid]_{initial} - x = 0.225 - x$
For approximation, we consider: $[Benzoic Acid] = 0.225M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Benzoic Acid]}$
$Ka = 6.5 \times 10^{- 5}= \frac{x * x}{ 0.225}$
$Ka = 6.5 \times 10^{- 5}= \frac{x^2}{ 0.225}$
$ 1.463 \times 10^{- 5} = x^2$
$x = 3.824 \times 10^{- 3}$
Percent ionization: $\frac{ 3.824 \times 10^{- 3}}{ 0.225} \times 100\% = 1.7\%$
%ionization < 5% : Right approximation.
