the solubility product of lead fluoride is 3.6 x 10-8. what is its solubility in 0.10M NaF solution, in grams per liter

8.8 * 10⁻³ g/L

Explanation:

NaF is a strong electrolyte that ionizes according to the following reaction.

NaF (aq) → Na⁺ (aq) + F⁻ (aq)

Then, the concentration of F⁻ will also be 0.10 M.

In order to find the solubility of PbF₂ (S), we will use an ICE Chart.

PbF₂ (s) ⇄ Pb²⁺ (aq) + 2 F⁻ (aq)

I 0 0.10

C + S + 2S

E S 0.10 + 2S

The solubility product (Kps) is:

Kps = 3.6 * 10⁻⁸ = [Pb²⁺].[F⁻]² = S. (0.10 + 2S) ²

In the term 0.10 + 2S, 2S is negligible in comparison with 0.10 and we can omit it to simplify calculations.

Kps = 3.6 * 10⁻⁸ = S. (0.10) ²

S = 3.6 * 10⁻⁵ M

The molar mass of PbF₂ is 245.20 g/mol. The solubility of PbF₂ in g/L is:

3.6 * 10⁻⁵ mol/L * 245.20 g/mol = 8.8 * 10⁻³ g/L