Answer
First ionization:
$H_3PO_4(aq) + H_2O(l) \lt -- \gt H_2P{O_4}(aq) + H_3O^+(aq)$
$K_{a1} = \frac{[H_2{PO_4}^-][H_3O^+]}{[H_3PO_4]}$
Second ionization:
$H_2P{O_4}^-(aq) + H_2O(l) \lt -- \gt HP{O_4}^{2-}(aq) + H_3O^+(aq)$
$K_{a2} = \frac{[H{PO_4}^{2-}][H_3O^+]}{[H_2P{O_4}^-]}$
Third ionization:
$HP{O_4}^{2-}(aq) + H_2O(l) \lt -- \gt P{O_4}^{3-}(aq) + H_3O^+(aq)$
$K_{a3} = \frac{[{PO_4}^{3-}][H_3O^+]}{[HP{O_4}^{2-}]}$
Work Step by Step
Phosphoric acid $(H_3PO_4)$ has 3 ionization steps.
1. Write the ionization equation, where $H_3PO_4$ donates a proton to the water, producing $H_2P{O_4}^-$ 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.
