14
$\begingroup$

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.)

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $\mathbf{pH = 7.1}$.

$\endgroup$
  • $\begingroup$ 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.) $\endgroup$ – Linear Christmas Sep 3 '17 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. $\endgroup$ – orthocresol Sep 3 '17 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 ;} $\endgroup$ – Linear Christmas Sep 3 '17 at 18:49
1
$\begingroup$

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,

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Oscar Lanzi Sep 3 '17 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.) $\endgroup$ – orthocresol Sep 3 '17 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? $\endgroup$ – Tan Yong Boon Sep 5 '17 at 2:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.