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26 October, 03:55

Calcium fluoride (CaF2) has a solubility constant of 3.45 x 10-11

. What is the molar solubility of CaF2 in water?

4.15 x 10-6 M

5.87 x 10-6 M

2.05 x 10-4 M

3.26 x 10-4 M

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  1. 26 October, 05:40
    0
    Ksp=3.45*10⁻¹¹

    CaF₂ (s) ⇄ Ca²⁺ (aq) + 2F⁻ (aq)

    Ksp=[Ca²⁺][F⁻]²

    [Ca²⁺]=C (CaF₂)

    [F⁻]=2C (CaF₂)

    Ksp=4{C (CaF₂) }³

    C (CaF₂) = ∛ (Ksp/4)

    C (CaF₂) = ∛ (3.45*10⁻¹¹/4) = 2.05*10⁻⁴ mol/L

    2.05*10⁻⁴ M
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