The ksp of copper (ii) carbonate, cuco3, is 1.4 * 10-10. calculate the molar solubility of this compound.

As,

CuCO₃ ⇆ Cu²⁺ + CO₃²⁻

So,

Kc = [Cu²⁺] [CO₃²⁻] / CuCO₃

Or,

Kc (CuCO₃) = [Cu²⁺] [CO₃²⁻]

Or,

Ksp = [Cu²⁺] [CO₃²⁻]

As,

Ksp = 1.4 * 10⁻¹⁰

So,

1.4 * 10⁻¹⁰ = [x] [x]

Or,

x² = 1.4 * 10⁻¹⁰

Or,

x = 1.18 * 10⁻⁵ mol/L

To cahnge ito g/L,

x = 1.18 * 10⁻⁵ mol/L * 123.526 g/mol

x = 1.45 * 10⁻³ g/L