Question

Calculate the molar solubility of SrF2 in water in a solution buffered at pH 2. Ka of HF is 7.2 x 10^-4

Answer 1

Ksp for SrF2 = [Sr2+][F-]^2 = 7.9 x 10^-10

For every Sr2+ we will have 2 F- in solution

let x be the molar solubility for Sr2+, then 2x will be the molar solubility of F-

Fluoride in solution will be present as both F- and HF at this pH

[F-] + [HF] = 2x

From HF we can write Ka expression,

Ka = [H+][F-]/[HF] = 7.2 x 10^-4

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

pH = -log[H+] = 2

[H+] = 0.01 M

[F-] + ([H+][F-]/Ka) = 2x

[F-] = 2x/(1+[H+]/Ka)

Feed in Ksp equation,

Ksp = (x)(2x/(1+[H+]/Ka))^2

7.9 x 10^-10 = 4x^3/(1+(0.01)/(7.2 x 10^-4))^2

Solving for x we get,

x = cube root (3.83 x 10^-8) = 3.37 x 10^-3 M is the molar solubility of SrF2