Answer
$Ka = 1.392\times 10^{- 4}$
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 dissociation equation, and find 'x':
%dissociation = $\frac{x}{[Initial Acid]} \times 100$
3= $\frac{x}{ 0.15} \times 100$
0.03= $\frac{x}{ 0.15}$
$ 4.5\times 10^{- 3}= x$
Therefore: $[Conj. Base] and [H_3O^+] = 4.5\times 10^{- 3}$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$
$Ka = \frac{x^2}{[InitialAcid] - x}$
$Ka = \frac{( 4.5\times 10^{- 3})^2}{ 0.15- 4.5\times 10^{- 3}}$
$Ka = \frac{ 2.025\times 10^{- 5}}{ 0.1455}$
$Ka = 1.392\times 10^{- 4}$
