[MangoPlant]

Find the solubility of CaF₂ in a solution that is buffered at pH = 3.

My approach:

CaF₂  Ca²⁺ + 2F^-    K = K_sp

2F^- + 2H⁺  2HF    K = (1/K_{a HF})²

Adding the two above equations, we get:

CaF₂ + 2H⁺  Ca²⁺ + 2HF    K = K_sp(1/K_{a HF})²


K_sp *(1/K_{a HF})² = x²/[H⁺]²

and at pH of 3, we have [H⁺] = 10^-3 and using the values, K_sp = 3.1 x 10^-11 and K_{a HF} = 6.8 x 10^-4 we have,

3.1 x 10^-11 * (1/6.8 x 10^-4)² = x²/(10^-3)²

Solving for x gives x = 2.58 x 10^-5, which is wrong. What was the problem?

[Borek]

You are assuming all F^- is protonated - it is not.

[MangoPlant]

Doesn't using the K_a of HF account for the fact that not all of the F^- is protonated?

[Borek]

No.

[Ca²⁺] = [F^-] + [HF]

and you approach is equivalent to assuming

[Ca²⁺] = [HF]

[MangoPlant]

Thanks.