What is the molar solubility of MgF2 in a 0.36 M Mg(NO3)2 solution? For MgF2, Ksp = 8.4 × 10^–8

2.4 × 10⁻⁴ M

Explanation:

Step 1: Calculate the concentration of Mg²⁺ coming from Mg(NO₃)₂

Mg(NO₃)₂ is a strong electrolyte and the molar ratio of Mg(NO₃)₂ to Mg²⁺ is 1:1. The initial molar concentration of Mg²⁺ is 1/1 × 0.36 M = 0.36 M.

Step 2: Make an ICE chart for the solution of MgF₂

        MgF₂(s) ⇄ Mg²⁺(aq) + 2 F⁻(aq)

I                           0.36             0

C                           +S             +2S

E                         0.36+S         2S

The solubility product constant is:

Ksp = [Mg²⁺] × [F⁻]² = (0.36+S) × (2S)²

Since S <<< 0.36, 0.36+S ≈ 0.36.

Ksp = 0.36 × 4S² = 8.4 × 10⁻⁸

S = 2.4 × 10⁻⁴ M

the answer is 75.0%.