Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 16 - Exercises - Page 771: 39


$H_2O$ acting as an acid and $H_2{PO_4}^-$ as a base: $$H_2O(l) + H_2{PO_4}^-(aq) \leftrightharpoons OH^-(aq) + H_3PO_4$$ $H_2O$ is acting as a base and $H_2{PO_4}^-$ as an acid: $$H_2O(l) + H_2P{O_4}^-(aq) \leftrightharpoons H_3O^+(aq) + H{PO_4}^{2-}(aq)$$

Work Step by Step

1. Write the first equation. Since water is acting as an acid, it must donate one proton to the other molecule, and since $H_2{PO_4}^-$ is acting as a base, it must accept one proton from the other substance. 2. For the second one we just invert the roles. Water will receive one proton, and $H_2P{O_4}^-$ will donate one.
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