Question

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

Answer 1

CaCO₃+ H₂O + CO₂ -----------> Ca⁺² + HCO₃^-
In this system the concentrations of H⁺ and Ca⁺² are interconnected to each other as well as P_CO2
The solubility of calcium carbonate and the concentration of Ca²⁺, will vary with changes in
the P_CO2 and [H⁺ ]. To derive K_sp for the [Ca²⁺] as a function of P_CO2, we must first determine the [H⁺ ].
P^H = -log₁₀[H⁺]
8.1 = -log₁₀[H⁺]
10^-8.1 = [H⁺]
[H⁺] = 7.94328 *10^-9 M
K_sp = [Ca⁺²] [CO₃^2-]
[Ca⁺²] = K_sp / [CO₃^2-] --------------------------------- (1)
expressing [Ca⁺²] in terms of P_CO2 and [H⁺]
CO₂ + H₂O <-----------> H⁺ + HCO₃^-
[HCO^{3 -} ] = K_a1 * ( [CO₂] / 7.94328 *10^-9 ) ---------------------- (2)
HCO₃^-  <------------> H⁺ + CO₃^2-
[CO₃^2- ] = Ka₂ * ( [HCO^3- ] / 7.94328 *10^-9 )
[CO₂] = K_w P_CO2 ------------------------- (3)   
making the above substitutions into equation 1
[Ca⁺²] = K_sp [ 7.94328 *10^-9 ]² / (K_a1 K_a2 K_w P_CO2 ) --------- (4)
The charge balance equation
2 [Ca²⁺] + [7.94328 *10^-9] = [HCO^{3 -} ] + 2 [CO₃^2- ] + [OH^- ]
As [Ca²⁺] >> [7.94328 *10^-9 ] & [HCO₃^- ] >> [CO₃^2-] and [OH^- ]
2 [Ca²⁺] = [HCO₃^-]
Finally from charge balance we get equation  [Ca²⁺] = 1/2*[HCO₃^-]
use equations 2 and 3 for the [HCO₃^-] in the terms of [H⁺] and P_CO2
[Ca⁺²] = (K_a1 K_w P_CO2 ) / 2 [7.94328 *10^-9] ------------ (5)
Now just plug in K_a1, K_a2 , Kw , P_CO2 values into equation 4 or equation 5 to get the solubility
of Ca