Given Ksp for CaF2 = 2.1×10^−10 and dissolution CaF2(s) ⇌ Ca2+(aq) + 2F−(aq). Find molar solubility s.

• A. 3.7 × 10^−4 M
• B. 1.0 × 10^−2 M
• C. 2.1 × 10^−10 M
• D. 7.4 × 10^−3 M

Answer: (A)

When a salt dissolves, its ions appear in solution according to the dissolution equation, and the solubility product (Ksp) ties those concentrations together. For calcium fluoride, each dissolved formula unit gives one Ca2+ and two F− ions. If s is the molar solubility, then [Ca2+] = s and [F−] = 2s.

The Ksp expression is [Ca2+][F−]^2. Substituting the concentrations gives Ksp = s(2s)^2 = 4s^3. Set this equal to 2.1×10^−10 and solve for s: 4s^3 = 2.1×10^−10, so s^3 = 5.25×10^−11. Taking the cube root, s ≈ (5.25×10^−11)^(1/3) ≈ 3.7×10^−4 M.

So the molar solubility is about 3.7×10^−4 M.