In hard water containing Ca^2+ at 2.0 × 10^−2 M, what is the maximum fluoride concentration allowed before CaF2(s) begins to precipitate, given Ksp = 4.0 × 10^−11?

• A. 2.0 × 10^−9 M
• B. 2.2 × 10^−5 M
• C. 4.4 × 10^−5 M
• D. 2.0 × 10^−2 M

Answer: (C)

When a sparingly soluble salt begins to precipitate, the dissolution is governed by its solubility product: for calcium fluoride CaF2(s) ⇌ Ca2+ + 2F−, Ksp = [Ca2+][F−]^2. At the point of saturation, the ionic product equals Ksp, so [F−] = sqrt(Ksp/[Ca2+]). With [Ca2+] = 2.0×10^−2 M and Ksp = 4.0×10^−11, [F−] = sqrt(4.0×10^−11 / 2.0×10^−2) = sqrt(2.0×10^−9) ≈ 4.4×10^−5 M. Therefore, the maximum fluoride concentration before CaF2 precipitates is about 4.4×10^−5 M.