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3 February, 14:10

The ksp of calcium carbonate, caco3, is 3.36 * 10-9 m2. calculate the solubility of this compound in g/l.

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  1. 3 February, 15:38
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    CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,

    CaCO₃ (s) ⇄ Ca²⁺ (aq) + CO₃²⁻ (aq)

    Initial Y - -

    Change - X + X + X

    Equilibrium Y-X X X

    Ksp for the CaCO₃ (s) is 3.36 x 10⁻⁹ M²

    Ksp = [Ca²⁺ (aq) ][CO₃²⁻ (aq) ]

    3.36 x 10⁻⁹ M² = X * X

    3.36 x 10⁻⁹ M² = X²

    X = 5.79 x 10⁻⁵ M

    Hence the solubility of CaCO₃ (s) = 5.79 x 10⁻⁵ M

    = 5.79 x 10⁻⁵ mol/L

    Molar mass of CaCO₃ = 100 g mol⁻¹

    Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹

    = 5.79 x 10⁻³ g/L
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