Question 17.10: Calculating Molar Solubility in the Presence of a Common Ion...
Calculating Molar Solubility in the Presence of a Common Ion
What is the molar solubility of CaF_2 in a solution containing 0.100 M NaF?
Begin by writing the reaction by which solid CaF_2 dissolves into its constituent aqueous ions. Write the corresponding expression for K_{sp}.
CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2 \ F^-(aq)
K_{sp} = [Ca^{2+}][F^-]
Use the stoichiometry of the reaction to prepare an ICE table showing the initial concentration of the common ion. Fill in the equilibrium concentrations of Ca^{2+} and F^- relative to S, the amount of CaF_2 that dissolves.
CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2 \ F^-(aq)
| [Ca^{2+}] | [F^−] | |
| Initial | 0.00 | 0.00 |
| Change | +S | +2S |
| Equil | S | 0.100 +2S |
Substitute the equilibrium expressions for [Ca^{2+}] and [F^-] from the previous step into the expression for Ksp. Since K_{sp} is small, you can make the approximation that 2S is much less than 0.100 and will therefore be insignificant when added to 0.100 (this is similar to the x is small approximation in equilibrium problems).
K_{sp} = [Ca^{2+}][F^-]^2
= S(0.100 + \cancel{2S})^2 (S is small)
= S(0.100)^2
Solve for S and substitute the numerical value of K_{sp}(from Table 17.2) to calculate S.
Note that the calculated value of S is indeed small compared to 0.100; your approximation is valid.
K_{sp}=S(0.100)^2
S=\frac{K_{sp}}{0.0100}=\frac{1.46 \times 10^{-10}}{0.0100}=1.46 \times 10^{-8} MFor comparison, the molar solubility of CaF_2 in pure water is 3.32 \times 10^{-4} M, which means CaF_2 is over 20,000 times more soluble in water than in the NaF solution. (Confirm this for yourself by calculating its solubility in pure water from the value of K_{sp}).
| TABLE 17. 2 Selected Solubility-Product Constants (K_{sp}) at 25 °C | |||||
| Compound | Formula | K_{sp} | Compound | Formula | K_{sp} |
| Barium fluoride | BaF_2 | 2.45 × 10^{-5} | Lead(II) chloride | PbCl_2 | 1.17 × 10^{-5} |
| Barium sulfate | BaSO_4 | 1.07× 10^{-10} | Lead(II) bromide | PbBr_2 | 4.67 × 10^{-6} |
| Calcium carbonate | CaCO_3 | 4.96 × 10^{-9} | Lead(II) sulfate | PbSO_4 | 1.82 × 10^{-8} |
| Calcium fluoride | CaF_2 | 1.46 × 10^{-10} | Lead(II) sulfide* | PbS | 9.04 × 10^{-29} |
| Calcium hydroxide | Ca(OH)_2 | 4.68 × 10^{-6} | Magnesium carbonate | MgCO_3 | 6.82 × 10^{-6} |
| Calcium sulfate | CaSO_4 | 7.10 × 10^{-5} | Magnesium hydroxide | Mg(OH)_2 | 2.06 × 10^{-13} |
| Copper(II) sulfide* | CuS | 1.27 × 10^{-36} | Silver chloride | AgCl | 1.77 × 10^{-10} |
| Iron(II) carbonate | FeCO_3 | 3.07 × 10^{-11} | Silver chromate | Ag_2CrO_4 | 1.12 × 10^{-12} |
| Iron(II) hydroxide | Fe(OH)_2 | 4.87 × 10^{-17} | Silver bromide | AgBr | 5.36 × 10^{-13} |
| Iron(II) sulfide* | FeS | 3.72× 10^{-19} | Silver iodide | AgI | 8.51 × 10^{-17} |
*Sulfide equilibrium is of the type: MS(s) + H_2O(l) \rightleftharpoons M_{2+}(aq) + HS^-(aq) + OH^-(aq)