Calculate the molar solubility of CaF2. (Ksp = 3.9 x 10-11)6.24 x 10-6M2.69 x 10-4M2.14 x 10-4M1.33 x 10-5M

To calculate the molar solubility of CaF2, we first need to understand the dissolution reaction and the expression for the solubility product constant (Ksp).

The dissolution reaction of CaF2 is:

CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)

The expression for the Ksp is:

Ksp = [Ca2+][F-]^2

Given that the Ksp of CaF2 is 3.9 x 10^-11, we can set up the following equation:

3.9 x 10^-11 = [Ca2+][F-]^2

Since the dissolution of one molecule of CaF2 produces one Ca2+ ion and two F- ions, the concentration of F- ions is twice that of Ca2+ ions. Therefore, we can express the concentration of F- ions as 2x and the concentration of Ca2+ ions as x. Substituting these into the Ksp expression gives:

3.9 x 10^-11 = x(2x)^2

Solving this equation for x gives the molar solubility of CaF2.

3.9 x 10^-11 = 4x^3

x^3 = 3.9 x 10^-11 / 4

x^3 = 9.75 x 10^-12

Taking the cube root of both sides gives:

x = 2.14 x 10^-4 M

So, the molar solubility of CaF2 is 2.14 x 10^-4 M.