If the dissolution of CaF2 has Ksp = 4.0 × 10^−11 and the equilibrium [Ca^2+] = 1.0 × 10^−2 M, what is [F−] at equilibrium?

• A. 2.0 × 10^−4 M
• B. 2.0 × 10^−3 M
• C. 4.0 × 10^−5 M
• D. 1.0 × 10^−9 M

Answer: (A)

Dissolving CaF2 follows CaF2(s) ⇌ Ca2+ + 2F−, so the solubility product is Ksp = [Ca2+][F−]^2. With [Ca2+] = 1.0×10−2 M, solve for [F−]:

[F−]^2 = Ksp / [Ca2+] = (4.0×10−11) / (1.0×10−2) = 4.0×10−9

[F−] = sqrt(4.0×10−9) ≈ 6.3×10−5 M.

Thus the equilibrium fluoride concentration is about 6.3×10−5 M. None of the options match exactly, though the closest magnitude among them is 4.0×10−5 M. The key idea is using the Ksp expression and the 1:2 ratio between Ca2+ and F− in the dissolution.