Calculate the molar solubility of calcium fluoride (CaF2) in g/L if its solubility product constant (Ksp) is 5.3×10^-11.

• A. 0.0016 g/L
• B. 0.16 g/L
• C. 0.016 g/L
• D. 0.001 g/L

Answer: (C)

Verified step by step guidance

1. Start by writing the balanced chemical equation for the dissolution of calcium fluoride (CaF2) in water: CaF2(s) ⇌ Ca²⁺(aq) + 2F⁻(aq).
2. Express the solubility product constant (Ksp) for CaF2. The Ksp expression is: $K_{\mathrm{sp}}=\; [Ca²⁺][F⁻]²$.
3. Let the molar solubility of CaF2 be 's'. Therefore, the concentration of Ca²⁺ ions will be 's' and the concentration of F⁻ ions will be '2s'. Substitute these into the Ksp expression: $K_{\mathrm{sp}}=\; (s)(2s)²\; =\; 4s³$.
4. Set the Ksp value equal to the expression derived: $4s³\; =\; 5.3\; imes\; 10^\{-11\}$. Solve for 's' to find the molar solubility in mol/L.
5. Convert the molar solubility from mol/L to g/L using the molar mass of CaF2. The molar mass of CaF2 is approximately 78.08 g/mol. Use the formula: $s\; imes\; 78.08\; ext\{\; g/mol\}$ to find the solubility in g/L.