First, we write our reaction:
We call solubility to "S"
LaF₃(s) → La³⁺(aq) + 3 F⁻(aq)
Initial - 0.055M (LaCl3) 0
Change +S +3.S
Eq. 0.055+S 3.S
Ksp = 2×10^−19 = [La³⁺]x[F⁻]³ = (0.055+S)x(3.S)³
2×10^−19 = (0.055+S)x(3.S)³ => S = 5.12x10^-7 mol/L x (195.9 g/mol) =>
S = 1.00x10^-4 g/L
(195.9 g/mol = The molar mass of LaF3)
Answer: S = 1.00x10^-4 g/L
