Answer
$Ka = 5.465\times 10^{- 6}$
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.
-$[H_3O^+] = [Conj. Base] = x$
-$[Acid] = [Acid]_{initial} - x = 3.5 \times 10^{- 5} - x$
2. Write the percent ionization equation, and find 'x':
%ionization = $\frac{x}{[Initial Acid]} \times 100$
0.85= $\frac{x}{ 0.075} \times 100$
0.0085= $\frac{x}{ 0.075}$
$ 6.375\times 10^{- 4}= x$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$
$Ka = \frac{x^2}{[Initial Acid] - x}$
$Ka = \frac{( 6.375\times 10^{- 4})^2}{ 0.075- 6.375\times 10^{- 4}}$
$Ka = \frac{ 4.064\times 10^{- 7}}{ 0.07436}$
$Ka = 5.465\times 10^{- 6}$
