Answer
$C_6H_5COOH(aq) + H_2O(l) C_6H_5COO^-(aq) + H^+(aq)$
$C_6H_5COOH(aq) + H_2O(l) C_6H_5COO^-(aq) + H_3O^+(aq)$
$Ka = \frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}$
Work Step by Step
An acid will react with water to produce $H^+(H_3O^+)$ and his conjugate base:
$C_6H_5COOH(aq) + H_2O(l) C_6H_5COO^-(aq) + H_3O^+(aq)$
To find the Ka equation:
$Ka = \frac{[Products]}{[Reactants]}$ or $Ka = \frac{[A^-][H^+]}{[AH]}$
$Ka = \frac{[C_6H_5COO^-][H^+]}{[C_6H_5COOH]}$
