Question

What is the solubility (in g/L) of calcium fluoride at 25°C? The solubility product constant for calcium fluoride is 3.4 × 10–11 at 25°C.

Answer 1

The salt dissolves as:

CaF2 <----> Ca2+ + 2 F-

   s 2s

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

3.4*10^-11=(s)*(2s)^2

3.4*10^-11= 4(s)^3

s = 2.041*10^-4 M

Molar mass of CaF2 = 1*MM(Ca) + 2*MM(F)

= 1*40.08 + 2*19.0

= 78.08 g/mol

Molar mass of CaF2= 78.08 g/mol

s = 2.041*10^-4 mol/L

To covert it to g/L, multiply it by molar mass

s = 2.041*10^-4 mol/L * 78.08 g/mol

s = 1.59*10^-2 g/L

Answer: 1.59*10^-2 g/L