This is Q36 on the 2018 Chemistry Olympiad:

Calcium fluoride, $\ce{CaF2}$, has a molar solubility of $\pu{2.1e–4 mol L–1}$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The pKa of $\ce{HF}$ is 3.17 (source).

Here are my steps:

  1. Calculate the solubility product of calcium fluoride from the data at pH=7.
  2. Combine the dissolution and the acid/base reaction to $$\ce{CaF2(s) + 2 H+ <=> Ca^2+(aq) + 2HF(aq)}$$
  3. Calculate the equilibrium constant of the combined reaction from the known constants of the separate reactions.
  4. Calculate the equilibrium concentration of $\ce{Ca^{2+}}$ at pH = 3.00
  5. Calculate the molar solubility and compared it to the one given for ph = 7.

The arithmetic is shown on the picture, resulting in a $\ce{Ca^{2+}}$ concentration of $\pu{4.325e-4 mol L-1}$.

enter image description here

I don't understand what I did wrong as the answer is B, 1.83.

Basically I combined the equilibrium of $\ce{CaF2}$ dissociating and the equilibrium of the formation of $\ce{HF}$. I then using the equilibrium product found out the concentration of $\ce{Ca^{2+}}$. It would be great if you could show me where I went wrong.

  • 1
    $\begingroup$ Just to clarify, but did you just take the reciprocal of the Ka as the Kb? Because that’s definitely not how Ka/Kb works. $\endgroup$ – ANZGC FlyingFalcon Mar 9 '19 at 9:52
  • $\begingroup$ There is no reaction stated that requires Kb (no hydroxide). Here, the acid dissociation of HF is just reversed and multiplied by two, giving you K = 1/square(Ka). $\endgroup$ – Karsten Theis Mar 9 '19 at 15:03
  • $\begingroup$ chemistry.stackexchange.com/questions/111041/… $\endgroup$ – Adnan AL-Amleh Mar 16 '19 at 22:22

I don't understand what I did wrong as the answer is B, 1.83.

You assumed that $\ce{[Ca^2+]}$ is equal to $\ce{[HF]}$ (you both call them "x") but forgot that an appreciable fraction of fluorine exists in the deprotonated form, i.e. fluoride ion.

It would be great if you could dictate the steps on paper.

Sorry, I can't do that. I can, however, sketch the path to an answer. At any given pH, some of the fluoride generated from dissolving calcium fluoride will end up as fluoride ion (deprotonated form) and some will be in the form of hydrogen fluoride (protonated form). The deprotonated fraction will be relevant for the dissolution equilibrium while the protonated fraction won't ("hidden" from the equilibrium reaction).

By what factor does its molar solubility increase in a solution with pH = 3.00?

You can use Henderson-Hasselbalch to figure out which fraction of total soluble fluorine is in the deprotonated state. At pH = 7, it will be almost 100% (in your answer, you neglected the presence of HF at pH = 7, which is probably fine). At pH = 3.00, there is more hydrogen fluoride than fluoride ion (pH is lower than pKa), but fluoride ion is not negligible (this is the conceptual error of your solution attempt). In fact, about 2/5 are fluoride ion and 3/5 are hydrogen fluoride.

You can either use your combined reaction or the original reaction to figure out the rest.

  • $\begingroup$ Here is an approach with ICE tables to solve this problem. $\endgroup$ – Karsten Theis Mar 18 '19 at 6:28

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