Solubility of LaF3: "S"
First, we need to write the reaction of solubility:
LaF3 (s) <=> La+3 + 3 F-
Initial - 0 +0.011 (F-, from KF)
Change +S +3.S
Equilibrium S 0.011 + 3.S
Now, we know:
Ksp = [La³⁺]x[F⁻]³ = 2×10^−19 = S x (0.011+3.S)³
We work with: 2×10^−19 = S x (0.011+3.S)³, and we clear the "S"
=> S = 1.5x10^-13 mol/L x (195.9 g/mol) = 2.93x10^-11 g/L
The molar mass of LaF3 = 195.9 g/mol
Answer: S = 2.93x10^-11 g/L
