Chemistry: The Central Science (13th Edition)

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

Chapter 17 - Additional Aspects of Aqueous Equilibria - Exercises - Page 768: 17.24c

Answer

This is the net ionic reaction when $KOH$ is added to this buffer: $OH^-(aq) + N{H_4}^+(aq) \lt -- \gt N{H_3}(aq) + H_2O(l)$

Work Step by Step

- Potassium hydroxide: $KOH$. When a base is added to a buffer, its acid is going to try to neutralize the added compound: $KOH(aq) + N{H_4}^+(aq) \lt -- \gt K^+(aq) + N{H_3}(aq) + H_2O(l)$ To write the net ionic equation, identify the totally ionizated/dissociated compounds: $HNO_3$, and break them into ions: $K^+ (aq) + OH^-(aq) + N{H_4}^+(aq) \lt -- \gt K^+(aq) + N{H_3}(aq) + H_2O(l)$ - Remove the spectators ions: $OH^-(aq) + N{H_4}^+(aq) \lt -- \gt N{H_3}(aq) + H_2O(l)$
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