The molar solubility of calcium fluoride in a water solution is

Question

Question

science chemistry

Answer 1

Answer #1

Ksp of CaF2 = 3.9*10^-11

At equilibrium:

CaF2 <----> Ca2+ + 2 F-

s 2s

Ksp = [Ca2+][F-]^2

3.9*10^-11=(s)*(2s)^2

3.9*10^-11= 4(s)^3

s = 2.136*10^-4 M

Answer: 2.1*10^-4 M

Answer 2

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