Calcium fluoride, CaF2, has a Ksp of 1.5 × 10–10 at a certain temperature. What is the solubility in mol/L of calcium fluoride?

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  • 1 month ago

              CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq)    Ksp = 1.5 × 10⁻¹⁰

    Initial:                    0 M            0 M

    Change:              +s M          +2s M

    Eqm:                s M            2s M

    At equilibrium:

    Ksp = [Ca²⁺] [F⁻]²

    1.5 × 10⁻¹⁰ = s (2s)²

    4s³ = 5.2 × 10⁻¹¹

    s = ³√(1.5 × 10⁻¹⁰/ 4)

    s = 3.35 × 10⁻⁴

    Solubility of CaF₂ = 3.35 × 10⁻⁴ M

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