Answer
$H{CO_3}^-(aq) + H_2O(l) {CO_3}^{2-}(aq) + H^+(aq)$
$H{CO_3}^-(aq) + H_2O(l) {CO_3}^{2-}(aq) + H_3O^+(aq)$
$Ka = \frac{[{CO_3}^{2-}][H^+]}{[HC{O_3}^-]}$
Work Step by Step
An acid will react with water to produce $H^+(H_3O^+)$ and his conjugate base.
$H{CO_3}^-(aq) + H_2O(l) {CO_3}^{2-}(aq) + H^+(aq)$
To find the Ka equation:
$Ka = \frac{[Products]}{[Reactants]}$ or $Ka = \frac{[H^+][A^-]}{[HA]}$
$Ka = \frac{[{CO_3}^{2-}][H^+]}{[HC{O_3}^-]}$
