Answer:
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
