Chemistry
brendanjlee5
2016-05-01 21:44:12
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
ANSWERS
gabbykneal
2016-05-01 22:29:37

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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