Ksp = 2 × 10^(-19) for LaF3. Calculate the solubility of LaF3 in grams per liter in a solution that is 0.013 M in KF.

• A. 3.4 × 10^(-5) g/L
• B. 7.8 × 10^(-2) g/L
• C. 1.2 × 10^(-7) g/L
• D. 5.6 × 10^(-3) g/L

Answer: (C)

Verified step by step guidance

1. Identify the dissolution reaction for LaF3: LaF3(s) ⇌ La³⁺(aq) + 3F⁻(aq).
2. Write the expression for the solubility product constant (Ksp): Ksp = [La³⁺][F⁻]³.
3. Since the solution is 0.013 M in KF, assume that the concentration of F⁻ from KF is 0.013 M. Let the solubility of LaF3 be 's' mol/L, then [La³⁺] = s and [F⁻] = 0.013 + 3s.
4. Substitute the concentrations into the Ksp expression: 2 × 10^(-19) = s(0.013 + 3s)³.
5. Assume 3s is negligible compared to 0.013, simplify the expression to: 2 × 10^(-19) ≈ s(0.013)³, and solve for 's'. Convert the solubility from mol/L to g/L using the molar mass of LaF3.