Answer
$Ka \approx 0.03333$
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.
-$[H_3O^+] = [Conj. Base] = x$
-$[Acid] = [Acid]_{initial} - x = 0.30M$
$[Acid]_{initial} = 0.30 + x$
2. Write the percent dissociation equation, and find 'x':
%dissociation = $\frac{x}{0.30 + x} \times 100$
25= $\frac{x}{ 0.30 + x} \times 100$
0.25= $\frac{x}{ 0.30 + x}$
$ 0.075 + 0.25x = x$
$0.075 = x - 0.25x = 0.75x$
$\frac{0.075}{0.75} = x$
$x = 0.1M$
Therefore: $[Conj. Base] and [H_3O^+] = 0.1M$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$
$Ka = \frac{x^2}{0.30}$
$Ka = \frac{( 0.1)^2}{ 0.30}$
$Ka = \frac{ 0.01}{ 0.30}$
$Ka = 0.03333$
