Calculate the solubility of calcium fluoride in a 0.020 M fluoride solution.?

                  CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq)      Ksp = 1.7 × 10⁻¹⁰
Initial:                            0 M        0.020 M
Change:                       +s M         +2s M
Eqm:    　　　              s M     (0.020+2s) M
                                                ≈ 0.020 M

At equilibrium:
Ksp = [Ca²⁺] [F⁻]²
1.7 × 10⁻¹⁰ = s (0.020)²
s = (1.7 × 10⁻¹⁰ ) / (0.020)²
s = 4.3 × 10⁻⁷ (to 2 sig. fig.)

Solubility of CaF₂ in 0.020 M fluoride solution = 4.3 × 10⁻⁷ M