Chemistry - Calculating the pH of a saturated calcium fluoride solution

Solution 1:

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.

Solution 2:

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 $\mathbf{pH = 7.1}$.