For each mole of CaCO3 formed one mole of hydrogen carbonate is converted to carbonic acid, which is in equilibrium with CO2 in air. Thus 100 g of CaCO3 (1 mol) formation result in 44 g of CO2 (also 1 mol): 

2HCO3- + Ca2+ -> CaCO3 + H2CO3.

Thanks Henrik for your support and excellent information 

Henrik,

how do you reconcile that chemical equation with:

Ca2+ + CO32- CaCO3 where the balance depends on the solubility produce Ksp = 2.8x10-9?

Alastair, my process is the total result of calcification. If calcification occur is determined by the equilibrium constant you mentioned.

The reaction described in the equilibrium tells that for each mole of CaCO3 precipated two moles of alkalinity are removed from solution. It is carbonate that is removed by the precipation but this carbonate is in equilibrium with the carbonate buffer system in an open system so in reality removing carbonate results in that it is replaced according to the reaction: 2HCO3- -> H2CO3 + CO32-. 

If you add this process to your calcium precipation reaction you will get reaction I wrote in my first answer.

Henrik,

Much as I hate to say it, hasn't Kenneth got a point? Calcification is a sink for CO2 not a source, despite conventional wisdom.

The equations relating : H2CO3, HCO3-,  and CO32- are

H2O + CO2 H2CO3

H2CO3 + H2O H3O+ + HCO3-

HCO3- + H2O H3O+ + CO32-

Precipitation of calcite will cause the last equation to move to the right and increase [H+] and so decrease pH, but not significantly since it is buffered by carbonic acid and its conjugate base bicarbonate?

Alastair, have no fear, Kenneth is not correct. He just didn't read the question.

Obviously the CO2 that go into CaCO3 is removed but for each mole of CO2 that goes into CaCO3 two moles of alkalinity is removed from the water which means the water becomes less alkaline and thus the total solubility of organic carbon in the water decreases resulting in that (1 mole) CO2 becomes supersaturated.

Read my papers in the field of CO2 release in desalination.