Asked by Natash
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...
so Ksp =[Ca2+][F-]^2 and
Ka =[H3O+][F-]/[HF]
I'm not sure how to continue...
Answers
Answered by
DrBob222
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.
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.
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