If the molar solubility of CaF2 at 35°C is 1.24×10⁻³ mol/L, what is the Ksp at this temperature?

• A. 1.53×10⁻⁴
• B. 1.24×10⁻³
• C. 3.80×10⁻⁹
• D. 6.10×10⁻⁶

Answer: (C)

Verified step by step guidance

1. Identify the dissolution equation for calcium fluoride (CaF₂) in water: CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq).
2. Recognize that the molar solubility of CaF₂ is given as 1.24×10⁻³ mol/L, which represents the concentration of Ca²⁺ ions in the solution at equilibrium.
3. Determine the concentration of fluoride ions (F⁻) in the solution. Since the dissolution of one mole of CaF₂ produces two moles of F⁻, the concentration of F⁻ ions will be 2 × 1.24×10⁻³ mol/L.
4. Write the expression for the solubility product constant (Ksp) for CaF₂: Ksp = [Ca²⁺][F⁻]².
5. Substitute the concentrations of Ca²⁺ and F⁻ into the Ksp expression: Ksp = (1.24×10⁻³) × (2 × 1.24×10⁻³)², and simplify to find the Ksp value.