Answer
Percent ionization = $1.897 \%$
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.
-$[H_3O^+] = [Conj. Base] = x$
-$[Acetic Acid] = [Acetic Acid]_{initial} - x = 0.05 - x$
For approximation, we consider: $[Acetic Acid] = 0.05M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acetic Acid]}$
$Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.05}$
$Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.05}$
$ 9 \times 10^{- 7} = x^2$
$x = 9.487 \times 10^{- 4}$
Percent ionization: $\frac{ 9.487 \times 10^{- 4}}{ 0.05} \times 100\% = 1.897\%$
%ionization < 5% : Right approximation.
