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Consider the two equilibria

CaF₂(s)  Ca²⁺(aq) + 2 F^-(aq)  K_sp = 4.0*10^-11

F^-(aq) + H₂O(l)  HF(aq) + OH^- (aq) K_b(F^-) = 2.9*10^-11

Write the chemical equation for the overall equilibrium and determine the corresponding equilibrium constant. Determine the solubility of CaF₂ at pH = 3.0.

I multipled the 2nd equation by two and added the two equations to get:

CaF₂(s) + 2H₂O(l)  Ca²⁺(aq) + 2 HF(aq) + 2 OH^- (aq)
K=(2.9*10^-11)²*4.0*10^-11=3.364*10^-32

So I suppose that 3.364*10^-32=[Ca²⁺][HF]²[10^-11]², and then 3.364*10^-10=[Ca²⁺][HF]².

I know we want to find [Ca²⁺], but we don't know [HF], so how do we do that?

[AWK]

Hydrolisis of fluoride you should take into account in water. You have an acidic solution.

[Borek]

You need a mass balance.

Equation you derived is useless here, as at pH 3.00 about 1/3 of fluorine is present as F^-.

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@AWK: I suspect that this has to do with the mass balance that Borek mentioned, but I don't see from where exactly you got the [Ca²⁺]=0.5 ([HF] + [F^-]) from. Could you or someone else elaborate a bit more please?

@Borek: If that is the case, why can't you just let [HF]=3*[F^-] and then since [F^-]=2s from the original K_sp equation, let [HF]=3(2s)=6s and [Ca²⁺]=s and substitute it back into this 3rd equation I derived:

CaF₂(s) + 2H₂O(l)   Ca²⁺(aq) + 2 HF(aq) + 2 OH^- (aq)

Does that work too?

[AWK]

You have an acidic solution. Forget about OH^-.
Without hydrolysis Ca²⁺ = 0.5 [F^-] - this come from CaF₂ formula. After addition of an acid some F^- will become HF. Then - still counting fluorine you have Ca²⁺ = 0.5 ([F^-] + [HF])

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Thanks! That makes more sense now.

And as for forgetting about OH^-, I don't mean to seem like I'm arguing, as I really appreciate the responses, but couldn't you still technically use 10^-8, even though it's acidic?

[Borek]

10^-11, as pH+pOH=pKw.