Question

the Ksp of CaF2 = 1.46*10^-10 and Ka of HF = 3.5*10^-4 Calculate the pH of a solution in which the solubility of CaF2 = .0100 moles/liter.

so Ksp =[Ca2+][F-]^2 and
Ka =[H3O+][F-]/[HF]

I'm not sure how to continue...

Answer 1

You need another equation.
2*solubility = (F^-) + (HF)
That + Ksp + Ka
Solve for (H^+).
I get 0.0579 M = (H^+) or pH of 1.237 which rounds to pH = 1.24.
Using 1.24 and going through it from the front end give S = 0.0100. I hope this helps.