Question

Given the following equilibrium systems: CaF2(s) <----> Ca2+(aq) + 2F-(aq); Kf = 4.0 x 10-11 HF(aq)...

Given the following equilibrium systems:

CaF2(s) <----> Ca2+(aq) + 2F-(aq); Kf = 4.0 x 10-11

HF(aq) <----> H+(aq) + F-(aq); Ka = 7.2 x 10-4

Determine the molar solubilty of CaF2 in 0.20 M HF solution.

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Answer #1

CaF2(s) <----> Ca2+(aq) + 2F-(aq); Kf = 4.0 x 10-11

Kf = [Ca+2][F-]^2

HF(aq) <----> H+(aq) + F-(aq)

0.20    0    0

0.20 - x    x x

Ka = [H+][F-] / [HF]

7.2 x 10-4 = x^2 / 0.20 - x

x = 0.0116 M

[F-] = 0.0116 M

CaF2(s) <----> Ca2+(aq) + 2F-(aq);

                         S            0.0116

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

4.0 x 10-11 = S x (0.0116)^2

S = 2.97 x 10^-7 M

molar solubility = 3.0 x 10^-7 M

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