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
