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

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