FWS asked in Science & MathematicsChemistry · 2 months ago

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

The solubility product constant for calcium fluoride is 1.7 × 10^-10. 

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  • 2 months ago

                      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

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