Question

Calculate the solubility of laf3 in grams per liter in (a) pure water, (b) 0.010 m kf solution, (c) 0.050 m lacl3 solution

Answer 1

2×10^−3 g/L

2×10^−11 g/L

1×10^−4 g/L

Respectively

Step-by-step explanation:

Other answer's explanation is good, just math is a bit off

Answer 2

Ksp of LaF₃ = 2 x 10⁻¹⁹
a) LaF₃(s) → La³⁺(aq) + 3 F⁻(aq)
Ksp = [La³⁺][F⁻]³
2 x 10⁻¹⁹ = (x)(3x)³
x = 9.28 x 10⁻⁶ M
To obtain solubility in g/L we multiply by molar mass (195.9 g/mol)
Solubility = (9.28 x 10⁻⁶ mol/L) * (195.90 g/mol) = 1.82 x 10⁻³ g/L
b) Solubility in 0.01 M KF is:
LaF₃(s) → La³⁺(aq) + 3 F⁻(aq)
Initial (M) - 0 0.01
Change (M) +x +3 x
Equilibrium (M) x 0.01 + 3x
Ksp = [La³⁺][F⁻]³
2 x 10⁻¹⁹ = (x)(0.01 + 3 x)³
x = 2.0 x 10⁻¹³ M
solubility in g/L = (2.0 x 10⁻¹³) * (195.90) = 3.92 * 10⁻¹¹ g/L
c) In 0.05 M LaCl₃ solution
LaF₃(s) → La³⁺(aq) + 3 F⁻(aq)
Initial (M) - 0.05 0.01
Change (M) +x +3 x
Equilibrium (M) 0.05 + x 3x
Ksp = [La³⁺][F⁻]³
2 x 10⁻¹⁹ = (0.05 + x)(3 x)³
x = 5.29 x 10⁻⁷ M
So solubility in g/L = (5.29 x 10⁻⁷) (195.9) = 1.04 x 10⁻⁴ g/L