## Chemistry: The Molecular Science (5th Edition)

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.