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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.)
In homework land you are right, but not in real life.
The $K_\mathrm b$ of $\ce{Ca^2+}$ ($\mathrm pK_\mathrm b = 2.43$ according to some sources) is within 1 log unit of the $K_\mathrm a$ of $\ce{HF}$. In other words, near neutral pH, consideration of the $\ce{Ca^2+/CaOH+}$ equilibrium is almost as important as the $\ce{F-/HF}$ equilibrium.
To begin with, a disclaimer: The approach here only takes into account the equilibria that the question itself cares about, namely $\ce{CaF2}$ dissociation and $\ce{F-/HF}$ acid-base. As pointed out in DavePhD's answer as well as Linear Christmas's comment, for a realistic treatment of the system, extra parameters must be taken into account.
A more accurate calculation would involve setting up five equations for five unknowns and solving them. The first three equations come from the data you provided. All concentrations are in $\pu{mol dm-3}$.
$$\begin{align} [\ce{Ca^2+}][\ce{F-}]^2 &= 3.9 \times 10^{-11} \tag{1} \\ \frac{[\ce{H3O+}][\ce{F-}]}{[\ce{HF}]} &= 6.8 \times 10^{-4} \tag{2} \\ [\ce{H3O+}][\ce{OH-}] &= 1.0 \times 10^{-14} \tag{3} \end{align}$$
The next two equations are the so-called mass balance and charge balance equations:
$$\begin{align} 2[\ce{Ca^2+}] &= [\ce{F-}] + [\ce{HF}] \tag{4} \\ 2[\ce{Ca^2+}] + [\ce{H3O+}] &= [\ce{F-}] + [\ce{OH-}] \tag{5} \end{align}$$
Equation $(4)$ comes from the stoichiometry of $\ce{CaF2}$ dissociation. The total concentration of calcium-containing species, times two, must be equal to the total concentration of fluorine-containing species.
Equation $(5)$ comes from the fact that the system must be electrically neutral, i.e. the positive charges are the same as the negative charges. The concentration of calcium ions is weighted by 2 because it is doubly charged.
In this case, I just plugged it into Wolfram|Alpha, where the concentrations $[\ce{Ca^2+}]$, $[\ce{F-}]$, $[\ce{H3O+}]$, $[\ce{OH-}]$, and $[\ce{HF}]$ are represented by $p$, $q$, $r$, $s$, and $t$ respectively. The only solution that is physically sensible is the one where all the roots are positive and real. From this we find
$$[\ce{H3O+}] = r = 7.8 \times 10^{-8}$$
and hence $\mathbf{pH = 7.1}$.
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,
$$\begin{align} [\ce{OH-}] &= 1.8 \times 10^{-7} \\ \mathrm{pOH} &= -\log{(1.8 \cdot 10^{-7})} = 6.7\\ \mathrm{pH} &= \mathbf{7.3} \\ \end{align}$$
