Answer
$H_{2}PO_{4}^{-}(aq) + H_{2}O(l) \rightleftharpoons HPO_{4}^{2-}(aq) + H_{3}O^{+}(aq)$
Here, $H_{2}PO_{4}^{-}(aq)$ is the acid and $H_{2}O(l)$ is the base.
$H_{2}PO_{4}^{-}(aq) + H_{2}O(l) \rightleftharpoons H_{3}PO_{4}(aq) + OH^{-}(aq)$
Here, $H_{2}PO_{4}^{-}(aq)$ is the base and $H_{2}O(l)$ is the acid.
Work Step by Step
According to the definition,
Acid: proton ($H^{+}$ ion) donor
Base: proton ($H^{+}$ ion) acceptor
In the first reaction, $H_{2}PO_{4}^{-}$ is an acid because, in solution, it donates a proton to water.
$H_{2}O$ is a base because it accepts the proton.
In the second reaction, $H_{2}O$ is an acid because, in solution, it donates a proton.
$H_{2}PO_{4}^{-}$ is a base because it accepts the proton.
