solubility - Calculating the pH of a saturated calcium fluoride solution

How can I calculate the pH of a saturated solution of calcium fluoride ($\ce{CaF2}$)? I am given the following values:

$$\begin{align} K_\mathrm{sp}(\ce{CaF2}) &= 3.9 \cdot 10^{-11} \\ K_\mathrm{a}(\ce{HF}) &= 6.8 \cdot 10^{-4} \\ K_\mathrm{w} &= 10^{-14} \end{align}$$

(The $K_\mathrm{sp}$ and $K_\mathrm a$ values are taken from the appendices of Skoog et al. Fundamentals of Analytical Chemistry, 9th edition.)

Answer

Here is my approach. The dissociation of $\ce{CaF2}$ is described by $$\ce{CaF2 <=> Ca^2+ +2F-} \tag{1}$$

Now, if $s$ is the molar solubility of $\ce{CaF2}$ (in $\pu{mol dm^-3}$), then $$\begin{align} K_\mathrm{sp} &= [\ce{Ca^2+}] \cdot [\ce{F-}]^2 \\ &= s \cdot (2s)^2 \\ &= 4s^3 = 3.9 \times 10^{-11} \\ \implies s &= 2.1 \times 10^{-4} \\ \end{align}$$

The fluoride concentration is then equal to $[\ce{F-}] = 4.2 \times 10^{-4} ~\mathrm{M}$. The pH is then governed by the equilibrium

$$\ce{F- + H2O <=> HF + OH-} \tag{2}$$

and since

$$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}]; \quad K_\mathrm{a} = \frac{[\ce{F-}][\ce{H3O+}]}{[\ce{HF}]}$$

we find that the equilibrium constant for reaction $(2)$ is

$$K = \frac{[\ce{HF}][\ce{OH-}]}{[\ce{F-}]} = \frac{K_\mathrm{w}}{K_\mathrm{a}}$$

and hence:

$$[\ce{OH-}][\ce{HF}] = \frac{K_\mathrm{w}}{K_\mathrm{a}}\cdot [\ce{F-}]$$

If we assume now that $[\ce{OH-}] = [\ce{HF}]$ (from the stoichiometry of reaction $(2)$), and that the decrease in $[\ce{F-}]$ due to reaction $(2)$ is negligible, then

$$\begin{align} [\ce{OH-}] &= \sqrt{\frac{K_\mathrm{w}}{K_\mathrm{a}}\cdot [\ce{F-}]} \\ &= \sqrt{\left(\frac{1 \times 10^{-14}}{6.8 \times 10^{-4}}\right) (4.2 \times 10^{-4})} \\ &= 7.9 \times 10^{-8} \end{align}$$

Adding the $1 \times 10^{-7}~\mathrm{M}$ of $\ce{OH-}$ from the autodissociation of water,

$$\begin{align} [\ce{OH-}] &= 1.8 \times 10^{-7} \\ \mathrm{pOH} &= -\log{(1.8 \cdot 10^{-7})} = 6.7\\ \mathrm{pH} &= \mathbf{7.3} \\ \end{align}$$