Question

WHat is the pH of MgF2 in water if the Ksp for MgF2 is 6.4*10^-9 for HF is 7.2 *10^4.

The answer is 7.26, but I am not 100% sure how to geth there, thank you!

Answer 1

MgF₂ Mg⁺² & 2 F^-

Let the solubility be x

[Mg⁺²] = x and [F^-] = 2x

Ksp = [Mg⁺²] [F^-]²

6.4 x 10^-9 = (x) (2x)²

4x³ = 6.4 x 10^-9

x³ = 1.6 x 10^-9

x = 1.17 x 10^-3

[Mg²⁺] = 1.17 x 10^-3 M, [F^-] = 2.34 x 10^-3 M

Now,

F^- + H₂O HF + OH^-

Kb = [HF] [OH^-] / [F^-]

Here Ka of HF = 7.2 x 10⁴

Ka x Kb = Kw = 1 x 10^-14

Kb = (1 x 10^-14) / ( 7.2 x 10⁴) = 1.39 x 10^-19

Kb = [HF] [OH^-] / [F^-]

1.39 x 10^-19= x² / (2.34 x 10^-3- x )

1.39 x 10^-19 = x² /  2.34 x 10^-3 (x << 2.34 x 10^-3, so neglecting in denominator)

x² = 3.25 x 10^-22

x = 1.80 x 10^-11

[OH^-] = 1.80 x 10^-11

pOH = -log [OH^-] = -log(1.80 x 10^-11) = 10.74

pH + pOH =14

pH = 14 - pOH = 14 - 10.74 = 3.26