Question

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

Answer 1

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