Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 15 - Additional Aqueous Equilibria - Questions for Review and Thought - Topical Questions - Page 693c: 63a


Some of the $Ca(OH)_2$ would start to precipitate, forming $Ca(OH)_2(s)$, which is a white solid.

Work Step by Step

This is the equilibrium reaction: $Ca(OH)_2(s) \lt -- \gt Ca^{2+} (aq) + 2OH^-(aq)$ When the pH increases, the $[OH^-]$ is increasing too, and the common ion effect will act. So, the reaction will be more reactant-favored. That follows the Le Chatelier's principle. Since we are raising the concentration of one of the products, the equilibrium will try to reduce this concentration, to reduce the change. Since the original solution of $Ca(OH)_2$ is saturated, if the equilibrium moves to the solid's side, the precipitation will occur.
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