Answer
First ionization:
$H_2CO_3(aq) + H_2O(l) \lt -- \gt HC{O_3}^-(aq) + H_3O^+(aq)$
$K_{a1} = \frac{[HC{O_3}^-][H_3O^+]}{[H_2CO_3]}$
First ionization:
$HC{O_3}^-(aq) + H_2O(l) \lt -- \gt C{O_3}^{2-}(aq) + H_3O^+(aq)$
$K_{a2} = \frac{[C{O_3}^{2-}][H_3O^+]}{[HC{O_3}^{-}]}$
Work Step by Step
1. Write the ionization equation, where $H_2CO_3$ donates a proton to the water, producing $HP{O_3}^-$ and $H_3O^+$, and the $K_a$ expression for this reaction:
$K_a = \frac{[Products]}{[Reactants]}$
** $H_2O$ doesn't go in this expression.
2. Repeat the step 1, using the conjugate base of the last reaction as the acid, this will be the next ionization.
