Calculate the solubility (in g/L)of calcium fluoride in water at 25°C if the K_sp for CaF₂ is 1.5 × 10^-10. A)9.6 × 10^-4 g/L B)2.6 × 10^-2 g/L C)3.3 × 10^-2 g/L D)4.1 × 10^-2 g/L

Question

Quizplus · Accepted Answer

The expression for the solubility product constant (Ksp) of calcium fluoride (CaF2) is:

Ksp = [Ca2+][F-]2

Assuming that x g/L of CaF2 dissolves in water, we can write:

CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)

At equilibrium, we have:

[Ca2+] = x mol/L
[F-] = 2x mol/L

Substituting these expressions into the Ksp expression, we get:

Ksp = x(2x)2 = 4x3

Given that Ksp = 1.5 × 10-10 at 25°C, we can solve for x:

4x3 = 1.5 × 10-10
x3 = 3.75 × 10-11
x = (3.75 × 10-11)1/3
x = 2.6 × 10-4 g/L

Therefore, the solubility of CaF2 in water at 25°C is 2.6 × 10-4 g/L, which corresponds to answer choice B.

Calculate the Solubility (In G/L)of Calcium Fluoride in Water at 25°C