Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 15 - Sections 15.1-15.12 - Exercises - Problems by Topic: 39

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.
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