Question 19.8: Calculating the Effect of a Common Ion on Solubility Problem......

Plan Addition of Ca^{2+}, the common ion, should lower the solubility. We write the equation and ion-product expression and set up a reaction table, with [Ca^{2+}]_{\text{init}} reflecting the 0.10 M Ca(NO_3)_2 and S equal to [Ca^{2+}]_{\text{from }Ca(OH)_2}. To simplify the math, we assume that, because K_{sp} is low, S is so small relative to [Ca^{2+}]_{\text{init}} that it can be neglected. Then we solve for S and check the assumption.

Solution Ca(NO_3)_2 is a soluble salt that dissociates completely in water; in a solution of 0.10 M Ca(NO_3)_2,  [Ca^{2+}]  =  0.10  M, and NO_3^− is a spectator ion that does not affect the solubility of Ca(OH)_2.

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

Making the assumption: K_{sp} is small, so S << 0.10 M; thus, 0.10 M + S ≈ 0.10 M.
Substituting into the ion-product expression and solving for S:
             K_{sp}  =  [Ca^{2+}][OH^−]^2  =  6.5×10^{−6}  ≈  (0.10)(2S)^2
Therefore,           4S^2  ≈  \frac{6.5×10^{−6}}{0.10}\text{     so      }S  ≈  \sqrt{\frac{6.5×10^{−5}}{4}}  =  4.0×10^{−3}  M
Checking the assumption by comparing the magnitude of S to 0.10 M:
             \frac{4.0×10^{−3}  M}{0.10  M}  ×  100  =  4.0\%  <  5\%
Check In Sample Problem 19.7, the solubility of Ca(OH)_2 was 0.012 M; here it is 0.0040 M, one-third as much. As expected, the solubility decreased in the presence of added Ca^{2+}, the common ion.