Answer
$Ka = 3.612\times 10^{- 5}$
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 = $\frac{x}{[Initial Acid]} \times 100$
1.55= $\frac{x}{ 0.148} \times 100$
0.0155= $\frac{x}{ 0.148}$
$ 2.294\times 10^{- 3}= x$
Therefore: $[Conj. Base] and [H_3O^+] = 2.294\times 10^{- 3}$
And, $[Acid] = 0.148 - 2.294 \times 10^{-3} = 0.1457M$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$
$Ka = \frac{x^2}{[InitialAcid] - x}$
$Ka = \frac{( 2.294\times 10^{- 3})^2}{ 0.148- 2.294\times 10^{- 3}}$
$Ka = \frac{ 5.262\times 10^{- 6}}{ 0.1457}$
$Ka = 3.612\times 10^{- 5}$
