Answer
$$H_2{PO_4}^-(aq) + H_2O(l) \leftrightharpoons HP{O_4}^{2-}(aq) + H_3O^+(aq) (1)$$
$$H_2{PO_4}^-(aq) + H_2O(l) \leftrightharpoons H_3P{O_4}(aq) + OH^-(aq) (2)$$
The acid reaction (1) occurs in a greater extent.
Work Step by Step
$H_2P{O_4}^-$ can act as an acid, forming $H{PO_4}^{2-}$ or as a base, forming $H_3PO_4$.
1. Write the reaction where it donates a proton (acid):
$$H_2{PO_4}^-(aq) + H_2O(l) \leftrightharpoons HP{O_4}^{2-}(aq) + H_3O^+(aq)$$
2. Write the reaction where it receives a proton (base):
$$H_2{PO_4}^-(aq) + H_2O(l) \leftrightharpoons H_3P{O_4}(aq) + OH^-(aq)$$
3. Since pH = 4.7, which is less than 7, the compound forms an acidic solution. Therefore, the acid reaction occurs in larger quantities than the basic reaction. $K_a \gt K_b$
