Answer
$$K_a = 5.3\times 10^{- 5}$$
Work Step by Step
1. Calculate the $[H_3O^+]$
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 2.68}$
$[H_3O^+] = 2.1 \times 10^{-3}$
Therefore : $x = 2.1\times 10^{- 3}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
$C_6H_5CH_2COOH(aq) + H_2O(l) \lt -- \gt C_6H_5CH_2COOH^-(aq) + H_3O^+(aq) $
\begin{matrix} & [C_6H_5CH_2COOH] & [H_3O^+] & [C_6H_5CH_2COOH^-] \\ Initial & 0.085 & 0 & 0 \\ Change & -x & +x & +x\\ Equil & 0.085 -x & x & x \end{matrix}
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][C_6H_5CH_2COOH^-]}{ [C_6H_5CH_2COOH]}$
$Ka = \frac{x^2}{0.085 - x}$
$Ka = \frac{( 2.1\times 10^{- 3})^2}{ 0.085- 2.1\times 10^{- 3}}$
$Ka = 5.3\times 10^{- 5}$