Question

Part B) Calculate the solubility of LaF3 in grams per liter in 1.2×10⁻² M KF solution.

Part C) Calculate the solubility of LaF3 in grams per liter in 5.0×10⁻² M LaCl3 solution.

Answer 1

PART B : -

LiF₃ (s) <-> La³⁺(aq) + 3 F^-(aq)

So the ionic molarities in saturated solution satisfy
Ksp = [La³⁺] * [F^-]³

Ksp = solubility product, -------it has to be given , i think it is given in part A

In a 1.2 * 10^-2 = 0.012 KF solution there are already 0.012M fluoride ions. Thus,

[La³⁺] = x
[F^-]= 0.012 + 3*x
Since x is small compared to 0.012, you can approximate:

Ksp = [La³⁺] *[F⁻]³
=>Ksp = x∙(0.012)³= 1.728*10^-6 x
=> x = Ksp / 1.728*10^-6 mol / L    (substitute the value of Ksp and calcualte x )

PART C :-

In a 0.05M LaCl₃ solution there are already 0.05M lanthanum ions. Thus,
[La³⁺] = 0.05 + x
[F^-]= 3x
Since x is small compared to 0.05, you can approximate:

=> Ksp = [La³⁺]*[F^-]³

By substituting the value of Ksp we can find the value of the solubility.