Question 17.SE.13: Calculate the molar solubility of CaF2 at 25 °C in a solutio...
Calculate the molar solubility of CaF_2 at 25 °C in a solution that is (a) 0.010 M in Ca(NO_3)_2 and (b) 0.010 M in NaF.
Analyze We are asked to determine the solubility of CaF_2 in the presence of two strong electrolytes, each containing an ion common to CaF_2. In (a) the common ion is Ca^{2 +} , and NO_3^- is a spectator ion. In (b) the common ion is F^-, and Na^+ is a spectator ion.
Plan Because the slightly soluble compound is CaF_2, we need to use K_{sp} for this compound, which Appendix D gives as 3.9 × 10^{-11}. The value of K_{sp} is unchanged by the presence of additional solutes. Because of the common-ion effect, however, the solubility of the salt decreases in the presence of common ions. We use our standard equilibrium techniques of starting with the equation for CaF_2 dissolution, setting up a table of initial and equilibrium concentrations, and using the K_{sp} expression to determine the concentration of the ion that comes only from CaF_2.
Solve
(a) The initial concentration of Ca^{2 +} is 0.010 M because of the dissolved Ca(NO_3)_2:
\begin{array}{|l|c|c|c|}\hline & CaF _2(s) \rightleftharpoons& Ca ^{2+}(a q)+& 2 F ^{-}(a q) \\\hline \text{Initial Concentration}(M) & – & 0.010 & 0 \\\hline \text{Change}(M) & – & +x & +2 x \\\hline \text{Equilibrium Concentration}(M) & – & (0.010+x) & 2 x \\\hline\end{array}
Substituting into the solubility-product expression gives:
K_{s p}=3.9 \times 10^{-11}=\left[ Ca ^{2+}\right]\left[ F ^{-}\right]^2=(0.010+x)(2 x)^2
If we assume that x is small compared to 0.010, we have:
3.9 \times 10^{-11}=(0.010)(2 x)^2
This very small value for x validates the simplifying assumption we made. Our calculation indicates that 3.1 × 10^{-5} mol of solid CaF_2 dissolves per liter of 0.010 M Ca(NO_3)_2 solution.
\begin{aligned} x^2 &=\frac{3.9 \times 10^{-11}}{4(0.010)}=9.8 \times 10^{-10}\\x &=\sqrt{9.8 \times 10^{-10}}=3.1 \times 10^{-5}\,M\end{aligned}
(b) The common ion is F^-, and at equilibrium we have:
\left[ Ca ^{2+}\right]=x\,\, \text{and}\,\,\left[ F ^{-}\right]=0.010+2 x
Assuming that 2x is much smaller than 0.010 M (that is, 0.010 + 2x \simeq 0.010), we have:
3.9 \times 10^{-11}=(x)(0.010+2 x)^2 \simeq x(0.010)^2
Thus, 3.9 × 10^{-7} mol of solid CaF_2 should dissolve per liter of 0.010 M NaF solution.
x=\frac{3.9 \times 10^{-11}}{(0.010)^2}=3.9 \times 10^{-7}\,M
Comment The molar solubility of CaF_2 in water is 2.1 × 10^{-4} M (Sample Exercise 17.12). By comparison, our calculations here give a CaF_2 solubility of 3.1 \times 10^{-5} M in the presence of 0.010 M Ca ^{2+} and 3.9 × 10^{-7} M in the presence of 0.010 M F^- ion. Thus, the addition of either Ca ^{2+} or F^- to a solution of CaF_2 decreases the solubility. However, the effect of F^- on the solubility is more pronounced than that of Ca ^{2+} because \left[ F ^{-}\right] appears to the second power in the K_{sp} expression for CaF_2, whereas \left[ Ca ^{2+}\right] appears to the first power.