Answer
$C_2H_5COOH(aq) + H_2O(l) \lt-- \gt H^+(aq) + C_2H_5COO^-(aq)$
or
$C_2H_5COOH(aq) + H_2O(l) \lt-- \gt H_3O^+(aq) + C_2H_5COO^-(aq)$
------
$Ka = \frac{[C_2H_5COO^-][H^+]}{[C_2H_5COOH]}$
or
$Ka = \frac{[C_2H_5COO^-][H_3O^+]}{[C_2H_5COOH]}$
Work Step by Step
1. Write the equations:
$C_2H_5COOH(aq) + H_2O(l) \lt-- \gt H^+(aq) + C_2H_5COO^-(aq)$
or
$C_2H_5COOH(aq) + H_2O(l) \lt-- \gt H_3O^+(aq) + C_2H_5COO^-(aq)$
2. Since the ka expression is based on:
$Ka = \frac{[Products]}{[Reactants]}$
The expressions will be:
$Ka = \frac{[C_2H_5COO^-][H^+]}{[C_2H_5COOH]}$
or
$Ka = \frac{[C_2H_5COO^-][H_3O^+]}{[C_2H_5COOH]}$
