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: 63c


The solution will not change visibly, because no precipitation is going to happen.

Work Step by Step

This is the equilibrium reaction: $Ca(OH)_2(s) \lt -- \gt Ca^{2+} (aq) + 2OH^-(aq)$ When we add $HCl$ to the solution, it will react with the hydroxide ions: $HCl(aq) + OH^-(aq) -- \gt Cl^-(aq) + H_2O(l)$ Since the $OH^-$ is being consumed, its concentration is going to decrease. Following the Le Chatelier's principle, if we reduce the concentration of one of the products $(OH^-)$, the equilibrium will try to favor that side, to compensate for that. Therefore, more $Ca(OH)_2$ would dissolve, producing more ions. The solution was already saturated (didn't have any precipitate), so that change will have no visual effect.
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