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](https://rads.stackoverflow.com/amzn/click/0495558281).)




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](https://books.google.com/books?id=CD6kqBW5U_wC&pg=PA74&lpg=PA74&dq=pkb%20calcium%20hydroxide&source=bl&ots=h3F15-O0OW&sig=D82yx8B782Xh0kKFcCan-jPSu-Q&hl=en&sa=X&ei=r56RVZ_bDomyyQTV46Fo&ved=0CB0Q6AEwADgU#v=onepage&q=pkb%20calcium%20hydroxide&f=false)) 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. 


 4


> 
> 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](https://chemistry.stackexchange.com/a/33466/16683) as well as [Linear Christmas's comment](https://chemistry.stackexchange.com/questions/31150/calculating-the-ph-of-a-saturated-calcium-fluoride-solution/82117#comment145313_82117), 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](http://www.wolframalpha.com/input/?i=solve+pq%5E2%3D3.9e-11,+rs%3D1e-14,+qr%2Ft%3D6.8e-4,+2p%2Br%3Dq%2Bs,+2p%3Dq%2Bt), 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 $\mathbf{pH = 7.1}$.


 3

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,


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


 1