Explanation:
[tex]HPO_4^{2-}[/tex] is a conjugate base of [tex]H_2PO_4^{-}[/tex] but it has one acidic hydrogen which dissociates in the solution and exists in the equilibrium as shown below:
[tex]HPO_4^{2-}+H_2O\rightleftharpoons H_3O^++PO_4^{3-}[/tex]
The expression for the dissociation constant, [tex]K_a[/tex] for [tex]HPO_4^{2-}[/tex] is shown below as:
[tex]K_a=\frac {[H_3O^+][PO_4^{3-}]}{[HPO_4^{2-}]}[/tex]
