## Chemistry: A Molecular Approach (3rd Edition)

$H_{2}PO_{4}^{-}(aq) + H_{2}O(l) \rightleftharpoons HPO_{4}^{2-}(aq) + H_{3}O^{+}(aq)$ 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.
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.