Answer
$Ka = 4.801 \times 10^{-9}$
Work Step by Step
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:
-$[H_3O^+] = [Conj. Base] = x$
-$[Acid] = [Acid]_{initial} - x$
2. Convert the pH into hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 4.52}$
$[H_3O^+] = 3.01 \times 10^{- 5}M$
3.. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$
$Ka = \frac{x^2}{[Initial Acid] - x}$
$Ka = \frac{( 3.02\times 10^{- 5})^2}{ 0.19- 3.02\times 10^{- 5}}$
$Ka = \frac{ 9.12\times 10^{- 10}}{ 0.19}$
$Ka = 4.801\times 10^{- 9}$
