Question 19.6: Determining Ksp from Solubility Problem (a) Lead(II) sulfate......

Plan We are given solubilities in various units and must find K_{sp} values. As always, we write the dissolution equation and ion-product expression for each compound, which show the number of moles of each ion. We use the molar mass to convert the solubility of the compound from the given mass units to molar solubility (molarity), then use that value to find the molarity of each ion, and substitute into the ion-product expression to calculate K_{sp}.

Solution (a) For PbSO_4. Writing the equation and ion-product (K_{sp}) expression:

            PbSO_4(s)   \xrightleftharpoons[]{}  Pb^{2+}(aq)  +  SO_4^{2−}(aq)                K_{sp}  =  [Pb^{2+}][SO_4^{2−}]
Converting solubility to molar solubility:

           \text{Molar solubility of }PbSO_4  =  \frac{0.00425  g  PbSO_4}{100.  \text{mL soln}}  ×  \frac{1000  mL}{1  L}  × \frac{1  \text{mol}  PbSO_4}{303.3  g  PbSO_4}
          =  1.40 \times 10^{-4}  M  PbSO_4

Determining molarities of the ions: Because 1 mol of Pb^{2+} and 1 mol of SO_4^{2−} form when 1 mol of PbSO_4 dissolves, [Pb^{2+}]  =  [SO_4^{2−}]  =  1.40×10^{−4}  M.
Substituting these values into the ion-product expression to calculate K_{sp}:
          K_{sp}  =  [Pb^{2+}][SO_4^{2−}]  =  (1.40×10^{−4})^2  =  1.96×10^{−8}
(b) For PbF_2. Writing the equation and K_{sp} expression:
          PbF_2(s)  \xrightleftharpoons[]{}  Pb^{2+}(aq)  +  2F^−(aq)            K_{sp}  =  [Pb^{2+}][F^−]^2
Converting solubility to molar solubility:

Determining molarities of the ions: Since 1 mol of Pb^{2+} and 2 mol of F^− form when 1 mol of PbF_2 dissolves, we have
           [Pb^{2+}]  =  2.6×10^{−3}  M       and      [F^−]  =  2(2.6×10^{−3}  M)  =  5.2×10^{−3}  M
Substituting these values into the ion-product expression to calculate K_{sp}:
           K_{sp}  =  [Pb^{2+}][F^−]^2  =  (2.6×10^{−3})(5.2×10^{−3})^2  =  7.0×10^{−8}

Check The low solubilities are consistent with K_{sp} values being small. (a) The molar solubility seems about right: ∼  \frac{4×10^{−2} g/L}{3×10^2  \text{g/mol}}  ≈  1.3×10^{−4}  M. Squaring this number gives 1.7×10^{−8}, close to the calculated K_{sp}. (b) Let’s check the math in the final step as follows: ∼(3×10^{−3})(5×10^{−3})^2  =  7.5×10^{−8}, close to the calculated K_{sp}.

Comment 1. In part (b), the formula PbF_2 means that [F^−] is twice [Pb^{2+}]. We follow the ion-product expression exactly and square this value of [F^−].

2. The tabulated K_{sp} values for these compounds (Table 19.2) are lower than our calculated values. For PbF_2, for instance, the tabulated value is 3.6×10^{−8}, but we calculated 7.0×10^{−8} from solubility data. The discrepancy arises because we assumed that PbF_2 in solution dissociates completely to Pb^{2+} and F^−. Here is an example of the complication pointed out earlier. Actually, about a third of the PbF_2 dissolves as PbF^+(aq) and a small amount dissolves as undissociated PbF_2(aq). The solubility given in the problem statement (0.64 g/L) is determined experimentally and includes these other species, which we did not include in our calculation.