Answer
$H_2{PO_4}^-(aq) + H_3O^+(aq) \longrightarrow H_3PO_4(aq) + H_2O(l)$
Work Step by Step
The added acid will produce hydronium ions, which would increase the acidity of the solution, but the buffer will react with these ions to neutralize it.
To neutralize this acidity $(H_3O^+)$, we need a base. So, $H_2P{O_4}^-$ is the other reactant, because it is the base of the buffer.
$H_2{PO_4}^-(aq) + H_3O^+(aq) \longrightarrow$
Since this is an acid-base reaction, $H_3O^+$ will donate one proton to $H_2P{O_4}^-$, producing $H_3PO_4$ and $H_2O$.
$H_2{PO_4}^-(aq) + H_3O^+(aq) \longrightarrow H_3PO_4(aq) + H_2O(l)$
