Question 17.SE.12: The Ksp for CaF2 is 3.9 × 10^-11 at 25 °C. Assuming equilibr...
The K_{sp} for CaF_2 is 3.9 × 10^{-11} at 25 °C. Assuming equilibrium is established between solid and dissolved CaF_2, and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF_2 in grams per liter.
Analyze We are given K_{sp} for CaF_2 and asked to determine solubility. Recall that the solubility of a substance is the quantity that can dissolve in solvent, whereas the solubility-product constant, K_{sp}, is an equilibrium constant.
Plan To go from K_{sp} to solubility, we follow the steps indicated by the red arrows in Figure 17.16. We first write the chemical equation for the dissolution and set up a table of initial and equilibrium concentrations. We then use the equilibrium expression. In this case we know K_{sp}, and so we solve for the concentrations of the ions in solution. Once we know these concentrations, we use the formula weight to determine solubility in g/L.
Solve
Assume that initially no salt has dissolved, and then allow x mol/L of CaF_2 to dissociate completely when equilibrium is achieved:
\begin{array}{|l|c|c|c|}\hline& CaF _2(s) \rightleftharpoons & Ca ^{2+}(a q) & +2 F ^{-}(a q) \\\hline \text{Initial concentration}(M) & – & 0 & 0 \\\hline \text{Change}(M) & – & +x & +2 x \\\hline \text{Equilibrium concentration}(M) & – & x & 2 x \\\hline\end{array}
The stoichiometry of the equilibrium dictates that 2x mol/L of F^- are produced for each x mol/L of CaF_2 that dissolve. We now use the expression for K_{sp} and substitute the equilibrium concentrations to solve for the value of x:
K_{s p}=\left[ Ca ^{2+}\right]\left[ F ^{-}\right]^2=(x)(2 x)^2=4 x^3=3.9 \times 10^{-11}
(Remember that \sqrt[3]{y}=y^{1 / 3}.) Thus, the molar solubility of CaF_2 is 2.1 × 10^{-4} mol/L.
x=\frac{\sqrt[3]{3.9 \times 10^{-11}}}{4}=2.1 \times 10^{-4}
The mass of CaF_2 that dissolves in water to form 1 L of solution is:
\left(\frac{2.1 \times 10^{-4}\,\cancel{mol \,CaF _2}}{1 \,L \,\text{soln}}\right)\left(\frac{78.1\, g\, CaF _2}{1 \,\cancel{mol \,CaF _2}}\right)=1.6 \times 10^{-2}\,g\, CaF _2 / L \,\text{solution}
Check We expect a small number for the solubility of a slightly soluble salt. If we reverse the calculation, we should be able to recalculate the solubility product: K_{s p}=\left(2.1 \times 10^{-4}\right)\left(4.2 \times 10^{-4}\right)^2=3.7 \times 10^{-11}, close to the value given in the problem statement, 3.9 \times 10^{-11}.
Comment Because F^- is the anion of a weak acid, you might expect hydrolysis of the ion to affect the solubility of CaF_2. The basicity of F^- is so small \left(K_b=1.5 \times 10^{-11}\right), however, that the hydrolysis occurs to only a slight extent and does not significantly influence the solubility. The reported solubility is 0.017 g/L at 25 °C, in good agreement with our calculation.
