Question 19.7: Determining Solubility from Ksp Problem Calcium hydroxide (s......

Plan We write the dissolution equation and the ion-product expression. We know K_{sp}  (6.5×10^{−6}), so to find molar solubility (S), we set up a reaction table that expresses [Ca^{2+}] and [OH^−] in terms of S, substitute into the ion-product expression, and solve for S.

Solution Writing the equation and ion-product expression:
             Ca(OH)_2(s)  \xrightleftharpoons[]{}  Ca^{2+}(aq)  +  2OH^−(aq)           K_{sp}  =  [Ca^{2+}][OH^−]^2  =  6.5×10^{−6}
Setting up a reaction table (Table 1), with S = molar solubility:

Substituting into the ion-product expression and solving for S:
             K_{sp}  =  [Ca^{2+}][OH^−]^2  = (S)(2S)^2  =  (S)(4S^2)  =  4S^3  =  6.5×10^{−6}
             S  =  \sqrt[3]{\frac{6.5×10^{−6}}{4}}  =  1.2×10^{−2}  M
Check We expect a low solubility from a slightly soluble salt. If we reverse the calculation, we should obtain the given K_{sp}:  4(1.2×10^{−2})^3  =  6.9×10^{−6}, which is close to 6.5×10^{−6}.

Comment 1. Note that we did not double and then square [OH^−]. 2S is the [OH^−], so we just squared it, as the ion-product expression required.
2. Once again, we assumed that the solid dissociates completely. Actually, the solubility is increased to about 2.0×10^{−2}  M by the presence of CaOH^+(aq) formed in the reaction Ca(OH)_2(s)  \xrightleftharpoons[]{}  CaOH^+(aq)  +  OH^−(aq). Our calculated answer is only approximate because we did not take this other species into account.