## Chemistry: The Central Science (13th Edition)

$HBrO_2(aq) \lt + H_2O(l)-- \gt H^+(aq) + Br{O_2}^-(aq)$ 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 $K_a = \frac{[H_3O^+][Br{O_2}^-]}{[HBrO_2]}$
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 $K_a = \frac{[H_3O^+][Br{O_2}^-]}{[HBrO_2]}$