This question already has an answer here:

Consider the following reaction.

$\ce{2 HF(aq) + Ca(OH)2(aq) -> CaF2(s) + 2 H2O(l)}$
The reaction produced a fairly insoluble salt.

Assume that this reaction proceeds entirely to completion as the acid and bases were initially in equimolar quantities with respect to their stoichiometry.

What is the resulting pH of the solution at equilibrium?

After the entire neutralization reaction has taken place we have the following reactions taking place in the solution.

$\ce{CaF2(s) -> Ca^2+(aq) + 2F- (aq)}$ with some $K_\mathrm{sp}$ value
and this reaction
$\ce{F- (aq) + H2O(aq) -> HF(aq) + OH- (aq)}$ with some $K_\mathrm{b}$ value

How would you go about calculating the values of $\ce{F- (aq)}$? I ask because every time some $\ce{CaF2}$ breaks off those free ions are used in the base reaction with water which according to Le Chatelier would result in more $\ce{CaF2}$ dissolving.

Does the chemically determined $K_\mathrm{sp}$ value in textbooks take into consideration the reaction of fluoride ions deprotonating water?


marked as duplicate by Loong, M.A.R. ಠ_ಠ, user15489, ron, jerepierre Jul 17 '15 at 15:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Technically fluoride doesn't deprotonate water much but binds hydronium. And yeah I'm rather nitpicky now ;) $\endgroup$ – Mithoron Jul 16 '15 at 23:38
  • $\begingroup$ @Mithoron 'deprotonating' was my edit from 'hydrolyzing'. If you like 'binds hydronium' better, please edit. $\endgroup$ – jerepierre Jul 17 '15 at 15:10