Answer
$Ka = 2.297\times 10^{- 6}$
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.
-$[H_3O^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x$
2.Calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 3.29}$
$[H_3O^+] = 5.129 \times 10^{- 4}$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][A^-]}{ [HA]}$
$Ka = \frac{x^2}{[InitialHA] - x}$
$Ka = \frac{( 5.129\times 10^{- 4})^2}{ 0.115- 5.129\times 10^{- 4}}$
$Ka = \frac{ 2.63\times 10^{- 7}}{ 0.1145}$
$Ka = 2.297\times 10^{- 6}$
