Answer
$Ka = 1.281 \times 10^{-6}$
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 $
2. Write the percent ionization equation, and find 'x':
%ionization = $\frac{x}{[Initial Acid]} \times 100$
0.92= $\frac{x}{ 0.015} \times 100$
0.0092= $\frac{x}{ 0.015}$
$1.38\times 10^{- 4}= x$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$
$Ka = \frac{x^2}{[Initial Acid] - x}$
$Ka = \frac{ (1.38\times 10^{- 4})^2}{ 0.015- 1.38\times 10^{- 4}}$
$Ka = \frac{ 1.904\times 10^{- 8}}{ 1.486\times 10^{- 2}}$
$Ka = 1.281\times 10^{- 6}$
