Answer
$HBrO_2(aq) \lt + H_2O(l)-- \gt H^+(aq) + Br{O_2}^-(aq)$
or
$HBrO_2(aq) \lt + H_2O(l)-- \gt H_3O^+(aq) + Br{O_2}^-(aq)$
----
$K_a = \frac{[H^+][Br{O_2}^-]}{[HBrO_2]}$
or
$K_a = \frac{[H_3O^+][Br{O_2}^-]}{[HBrO_2]}$
Work Step by Step
The ka equation is based on:
$K_a = \frac{[Products]}{[Reactants]}$
- Since the ionization of $HBrO_2$ is represented by the equation:
$HBrO_2(aq) \lt + H_2O(l)-- \gt H^+(aq) + Br{O_2}^-(aq)$
or
$HBrO_2(aq) \lt + H_2O(l)-- \gt H_3O^+(aq) + Br{O_2}^-(aq)$
This is the ka expression for this acid:
$K_a = \frac{[H^+][Br{O_2}^-]}{[HBrO_2]}$
or
$K_a = \frac{[H_3O^+][Br{O_2}^-]}{[HBrO_2]}$
