Question

5.0 mg of limestone is dispersed in 1.0 L of water containing 2.5 mg/L CO3. How much solid will remain after the system reaches equilibrium?

Answer 1

I will suppose you have 2.5 mg CO₃^2-/L not 2.5 mg CO₂/L.

Ksp CaCO₃ = 4.5x10^-9 (verify this value)

5.0 x10^-3g / 100 g CaCO₃/mol = 5.0x10^-5 mol CaCO₃

[CO₃^2-]_initial = 2.5x10^-3 g / 60 gCO₃^2-/mol = 4.16x10^-5 mol CO₃^2- / mol

After limestone dissolution:

[CO₃^2-] = [CO₃^2-]_initial + [Ca²⁺] and

[Ca²⁺] [CO₃^2-] = Ksp

  [Ca²⁺]= x

x (x + 4.16·10^-5 ) = 3.2x10^-9

100 000 x² + 4.16x = 0.0004

X= 0.00004 mol/L

The molar solubility of CaCO₃ = [Ca²⁺]= x = 0.00005 mol/L

Or   0.00005 mol/L · 100 g/mol = 0.005 g/L

No solid in solution (complete dissolution) if you use this value of Ksp.