Answer
$Ka = 3.982\times 10^{- 11}$
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. Now, convert the pH into hydronium ion concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 6.2}$
$[H_3O^+] = 6.3 \times 10^{- 7}$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$
$Ka = \frac{x^2}{[Initial Acid] - x}$
$Ka = \frac{( 6.31\times 10^{- 7})^2}{ 0.01- 6.31\times 10^{- 7}}$
$Ka = \frac{ 3.982\times 10^{- 13}}{ 0.01}$
$Ka = 3.982\times 10^{- 11}$
