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

1 answer:

3 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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