What is the solubility (in g/L)of calcium fluoride at 25°C? The solubility product constant for calcium fluoride is 3.4 × 10^-11 at 25°C. A)0.00045 g/L B)2.7 × 10^-9 g/L C)0.015 g/L D)1.3 × 10^-9 g/L E)0.094 g/L

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Quizplus · Accepted Answer

The solubility product (Ksp) for calcium fluoride (CaF2) is given as 3.4 × 10^-11. The dissolution of CaF2 can be represented as: CaF2(s) ⇌ Ca^2+(aq) + 2F^-(aq). Let the solubility of CaF2 be s mol/L. This means [Ca^2+] = s and [F^-] = 2s. Plugging these into the expression for Ksp gives: Ksp = [Ca^2+][F^-]^2 = s(2s)^2 = 4s^3. Solving for s gives s = (3.4 × 10^-11 / 4)^(1/3). Calculating this value gives a solubility of approximately 1.5 × 10^-4 mol/L. To find the solubility in g/L, multiply this by the molar mass of CaF2 (78.07 g/mol): (1.5 × 10^-4 mol/L) × (78.07 g/mol) ≈ 0.012 g/L, which rounds to 0.015 g/L, making option C correct.

What Is the Solubility (In G/L)of Calcium Fluoride at 25°C