Question

The solubility of CaF2 is 0.00021 mole per liter. What is the solubility product constant for CaF2? A) 7.3 x 10-12 B) 3.7 x 10-11 C) 8.5 x 10-8 D) 4.4 x 10-8 E) 1.9 x 10-11

Answer 1

Answer:

B) Ksp = 3.7 E-11

Explanation:

CaF2 ↔ Ca2+  +  2F-

  S            S          2S,,,,,,,,,,,,,,,in the equilibrium

⇒ Ksp = [ Ca2+ ] * [ F- ]²

⇒ Ksp = S * ( 2S )²

⇒ Ksp = 4S³

⇒ Ksp = 4 * ( 0.00021 )³

⇒ Ksp = 4 * 9.261 E-12

⇒ Ksp = 3.704 E-11