Answer
$H_3{PO_4}(aq) + OH^-(aq) \longrightarrow H_2P{O_4}^-(aq) + H_2O(l)$
Work Step by Step
The added base will produce hydroxide ions, which would increase the basicity of the solution, but the buffer will react with these ions to neutralize it.
To neutralize this basicity $(OH^-)$, we need an acid. So, $H_3P{O_4}$ is the other reactant, because it is the acid of the buffer.
$H_3{PO_4}(aq) + OH^-(aq) \longrightarrow$
Since this is an acid-base reaction, $H_3PO_4$ will donate one proton to $OH^-$, producing $H_2P{O_4}^-$ and $H_2O$.
$H_3{PO_4}(aq) + OH^-(aq) \longrightarrow H_2P{O_4}^-(aq) + H_2O(l)$
