Question

Calcium fluoride (CaF2) has a Ksp of 3.45∙10-11. If I have a 0.150 M solution of sodium fluoride, what concentration of calcium acetate will begin precipitation of the fluoride ions?

Answer 1

Given : concentration of sodium fluoride, [NaF] = 0.150 M

Since NaF is an ionic salt, therefore, it will completely dissociate into Na⁺ and F^- ions

concentration of F^- ions, [F^-] = [NaF]

[F^-] = 0.150 M

K_sp CaF₂ = [Ca²⁺][F^-]²

[Ca²⁺] = (K_sp CaF₂) / [F^-]²

[Ca²⁺] = (3.45 x 10^-11) / (0.150 M)²

[Ca²⁺] = 1.53 x 10^-9 M

concentration of calcium acetate = [Ca²⁺]

concentration of calcium acetate = 1.53 x 10^-9 M