Question

A water is in equilibrium with CO2 in the atmosphere (partial pressure is 3.16 * 10^-4 atm) and with CaCO3 (s). The pH is 8.1. Ignoring Ca complexes, what is the solubility (S) of Ca?

Answer 1

CaCO₃ (s) + CO₂ (g) + H₂O (l) = Ca²⁺ + 2HCO₃^-

The electrical neutrality expression for this case is:

2[Ca²⁺] + [H⁺] = [OH^-] + [HCO₃^-] + 2[CO₃^2-]

Ksp = [Ca²⁺][CO₃^2-] = 6x10^-9

[CO₂ (aq)] = K(CO₂) x P(CO₂) K(CO₂) = 2 x10^-3

[Ca²⁺]=Ksp / [CO₃^2-]

[CO₃^2-]=Ka1 x Ka2 [CO₂ (aq)] / [H⁺]²

[Ca²⁺]=Ksp x [H⁺]² / K_a1 x K_a2 [CO₂ (aq)]

[Ca²⁺] = 6x10^-9 x (10^-8.1)² / 2 x10^-3 x 10^-3.6 x 10^-10.3 = 3.79 x 10^-25 / 3.16 x 10^-4 x 2 x 10^-3 x 2.51 x 10^-4 x 5.01 x 10^-11 = 3.79 x 10^-25 / 7.95 x 10^-21 = 4.77 x 10^-5 mol/L

[Ca²⁺] = 4.77 x 10^-5 mol/L