Calculate the molar solubility of MgF2 in 0.15 M MgCl2 at 25 °C, given that the Ksp of MgF2 is 6.4 x 10^-9.

• A. 3.2 x 10^-4 M
• B. 1.6 x 10^-4 M
• C. 6.4 x 10^-5 M
• D. 1.0 x 10^-3 M

Answer: (B)

Verified step by step guidance

1. Understand the problem: We need to calculate the molar solubility of MgF2 in a solution that already contains MgCl2. The presence of MgCl2 affects the solubility of MgF2 due to the common ion effect.
2. Write the dissolution equation for MgF2: MgF2(s) ⇌ Mg²⁺(aq) + 2F⁻(aq). The solubility product constant (Ksp) expression for this equilibrium is Ksp = [Mg²⁺][F⁻]².
3. Consider the initial concentrations: The solution already contains 0.15 M Mg²⁺ from MgCl2. Let 's' be the molar solubility of MgF2. Therefore, [Mg²⁺] = 0.15 + s and [F⁻] = 2s.
4. Substitute into the Ksp expression: Ksp = (0.15 + s)(2s)². Given Ksp = 6.4 x 10⁻⁹, set up the equation 6.4 x 10⁻⁹ = (0.15 + s)(4s²).
5. Assume s is small compared to 0.15 M, simplifying the expression to 6.4 x 10⁻⁹ ≈ 0.15 * 4s². Solve for 's' to find the molar solubility of MgF2.