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.)


3 Answers 3


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 $\mathbf{pH = 7.1}$.

  • $\begingroup$ How significantly would the answer change if we included the equilibrium $$\ce {F- + HF <=> HF2-}? \tag {p$K_\mathrm {a}$ = 0.58}$$ I have always steered clear of simplified calculations for $\ce {HF} $ due to this reason, and have advocated for others to do the same. Since this is a saturated solution, I would fear the effect is even more noticable. Hydrolysis of calcium ions might also be important. (I'm not really sure because of the saturation.) $\endgroup$ Commented Sep 3, 2017 at 18:18
  • $\begingroup$ @LinearChristmas I agree that it's undoubtedly more complicated than this. Should have added a disclaimer, but was in a rush... :/ I'll add one now. Obviously it is possible to simply set up more equations and solve them, but one would need the data. $\endgroup$ Commented Sep 3, 2017 at 18:34
  • $\begingroup$ If you ever have the time and find resources/data, feel free to include it (difluoride and hydrolysis). I guess it would be worth a 200 bounty from me. 400, if you also include more complicated equations to take into account departure from ideality (generalised Hückel eq-s, or sth similar). An additional 100 if you go and measure the pH and include uncertainty. (Maybe it's better to post a new question?) Anyways, the offer stands. Have fun ;} $\endgroup$ Commented Sep 3, 2017 at 18:49

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,

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

  • $\begingroup$ Not $10^{-7}$ mol hydroxide from autodissociation. When you add an acid or base to water the dissociation has to be less to keep the $K_w$ product. $\endgroup$ Commented Sep 3, 2017 at 9:46
  • $\begingroup$ @OscarLanzi actually, that was from me. I edited it from the original; the original used incorrect values and obtained [OH-] = 6.32e-7. However I was not keen on keeping the incorrect values up, hence the editing. I inserted that extra assumption (1e-7 hydroxide from autodissociation) because I wanted to keep the approach simplistic, in the spirit of the original answer. (As you can see this approach also assumes the F-/HF equilibrium does not disturb the CaF2 equilibrium.) $\endgroup$ Commented Sep 3, 2017 at 11:30
  • $\begingroup$ Hmmm... But why didn't you factor the hydroxide ion concentration from the auto-ionisation of water in the intermediate steps but only in the final step? You could have factored it in by letting the final hydroxide ion concentration be (X+10^(-7)) M and the HF concentration to be X M. Then, solve for X quadratically, using the Kb expression to form an equation like you just did. Also, how do you justify your approximation? $\endgroup$ Commented Sep 5, 2017 at 2:29

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