Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 16 - Acid-Base Equilibria - Exercises - Page 719: 16.70b


${HPO_4}^{2-} + H_2O -> {H_2PO_4}^{-} + OH^-$ $Kb = \frac{[{H_2PO_4}^-][OH^-]}{[{HPO_4}^{2-}]}$

Work Step by Step

1. Write the reaction: ${HPO_4}^{2-} + H_2O -> {H_2PO_4}^{-} + OH^-$ 2. Now, write the Kb formula: $Kb = \frac{[Conjugate.Acid]*[OH^-]}{[Base]}$ $Kb = \frac{[{H_2PO_4}^-][OH^-]}{[{HPO_4}^{2-}]}$
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