Question

Thinking about considerations of acid/base chemistry and solubility constants. Calcium carbonate (CaCO₃) is added to initially pure water that is exposed to the atmosphere with 400 ppm(v) of carbon dioxide. If solid CaCO₃ remains after a long time – showing dissolution and precipitation are at equilibrium – what is the aqueous concentration of calcium ions? Use the acid constants, Henry’s Law constants and solubility products.

Answer 1

Equilibrium of dissolution of CaCO₃ is given by the equation :

CaCO₃ (s) ⇌ Ca²⁺ (aq)+ CO₃²⁻(aq)

Ksp = 6×10⁻⁹

There are following equilibrium from CO₂ in air to water for this:

1. CO₂ (aq) + H₂O    $\rightleftharpoons$ H₂CO₃ (aq)

2. H₂CO₃ (aq) $\rightleftharpoons$ H⁺ (aq) + HCO₃^- (aq) Ka1 =4.5*10^-7

3. HCO₃^- (aq)  $\rightleftharpoons$ H⁺ (aq) + CO₃^2- (aq) Ka2 =4.7*10^-11

Amount of CO₂ Dissolved in be calculated from Henry 's law ;  

[CO₂ (aq) ] = K_CO2  * P_CO2 (K_CO2  = Henry constant for CO₂ = 3.4 x 10^-2 mol /(L-atm) )

Here we are using 'Henry solubility defined via concentration'.

At 400 ppm , P_CO2 =4*10^-4 atm

[CO₂ (aq) ] =  3.4 x 10^-2 mol /(L-atm) *4*10^-4 atm = 1.36*10^-5 mol/ L

[CO₂ (aq) ] =  [ H₂CO₃ ]

so, from Ka, ; [HCO₃^-  ] = 2.5*10^-6 mol/ L

and [CO₃^2-  ]=1.1*10^-8 mol/ L

we have,   [Ca²⁺  ]= [CO₃^2-  ]= Ksp = 6×10⁻⁹

we have; [Ca²⁺  ]=s

and [CO₃^2-  ]= (s + 1.1*10^-8)

so, s* (s + 1.1*10^-8) = 6×10⁻⁹

[Ca²⁺  ] = s = 7.8×10⁻⁵ M