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10 January, 22:10

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

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  1. 10 January, 22:37
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    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
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