Question

Calcium fluoride, CaF2, has a molar solubility of 2.1 x 10^–4 mol L^–1 at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The pKa of HF is 3.17. The answer is 1.83

Answer 1

Let us calculate the solubility of CaF2 at pH = 7

CaF2 ---> Ca+2 + 2F-

Ksp = [Ca⁺²][F-]²

Let solublity = s

Ksp = (s){2s)²

Ksp = 4 X (2.1X10^-4)³ = 3. 7X10^-11

The dissociation equation of HF will be

        HF ---> H⁺ + F-

Ka = antilog(-pKa) = 0.000676 = 6.76 X 10^-4

6.76 X 10^-4 = [H⁺][F^-] / [HF]

Given pH = 3

[H⁺] = 10^-3 M

6.76 X 10^-4 = [10^-3][F^-] / [HF]

[HF] = 1.48[F-]

Ksp = [Ca⁺²][F-]² = 3.7 X 10^-11

The [F-]total = 2Ca⁺²

[F-]total = [F-] + [HF]

The simple calculation will be

[Ca+2] =

[Ca+2] = 3.85 X 10-4

Hence factor = 3.85 X 10-4 / 2.1 X 10-4 = 1.83