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
