# Understanding the solubility of Ca(HCO3)2

According to wikipedia Ca(HCO3)2 has the following solubility values:

16.1 g/100 mL (0 °C)
16.6 g/100 mL (20 °C)
18.4 g/100 mL (100 °C)

So I assume Ca(HCO3)2 would precipitate beyond this level. However, the wikipedia page for calcium bicarbonate states that attempts to prepare compounds such as solid calcium bicarbonate by evaporating its solution to dryness invariably yield instead the solid CaCO3:

Ca(HCO3)2(aq) → CO2(g) + H2O(l) + CaCO3(s)

which leaves me wondering if, when the solution is saturated beyond ~16-18 g/100 mL, a precipitate of Ca(HCO3)2 will exist in solution? or if it also dissociating beyond that point while still in solution?

• What is the reference to solubility at 100 degrees? This value could very well be the equilibrium value of dissolved calcium bicarbonate which is in equilibrium with solid calcium carbonate at boiling temperatures. Until and unless we don't know how it was measured, it may be futile to discuss this value. – M. Farooq Aug 2 '19 at 12:58
• Surprisingly, SciFinder is quiet about its solubility and it is only Wikipedia which quotes these values. No reference sadly! – M. Farooq Aug 2 '19 at 13:05
• Yes, all three values are from wikipedia. I edited the text to mention it. But as you write, wikipedia doesn't cite any source... : / – Hans Aug 3 '19 at 1:33
• I found one encyclopedia which quoted the same values without any reference. Most likely copied from Wikipedia without any reference. This is the state of affairs these days. There is German work on the solubility of calcium bicarbonate, however it is in sodium chloride solutions. – M. Farooq Aug 3 '19 at 3:27
• Left on note on Wikipedia calcium bicarbonate page to add citations. – M. Farooq Aug 3 '19 at 3:31

Calcium bicarbonate $$\ce{Ca(HCO3)2}$$simply does not exist. It is impossible to fill up any container with pure $$\ce{Ca(HCO3)2}$$. Nobody has ever been able to produce a powder containing $$\ce{Ca(HCO3)2}$$. It is possible and easy to produce a solution containing the ions $$\ce{Ca^{2+}}$$ and the ions $$\ce{HCO3^-}$$, by bubbling $$\ce{CO2}$$ into a solution of calcium hydroxyde $$\ce{Ca(OH)2}$$. The following reaction will happen : $$\ce{Ca(OH)2} + \ce{2CO2} -> \ce{Ca^{2+} + 2 \ce{HCO3^-}}$$ The result is exactly what could be expected if some $$\ce{Ca(HCO3)2}$$ had been dissolved in water. And of course, it makes sense to admit that by evaporating the solution, the calcium bicarbonate could be obtained dry. Unfortunately it is not the case. When trying to evaporate this solution, $$\ce{Ca(HCO3)2}$$ is always and totally decomposed according to the following equation : $$\ce{Ca^{2+} + 2 \ce{HCO3^-} -> \ce{CaCO3} + H2O + CO2 }$$ And all you obtain by trying to evaporate a solution containing he ions $$\ce{Ca^{2+}}$$ and the ions $$\ce{HCO3^-}$$ is a deposit of calcium carbonate $$\ce{CaCO3}$$, whatever the method used for carrying out this evaporation.
Of course you can calculate the maximum amount of Calcium and of $$\ce{HCO3^-}$$ ions that may exist simultaneously in such a solution. You may even calculate the solubility of $$\ce{Ca(HCO3)2}$$. But this does not prove that $$\ce{Ca(HCO3)2}$$ exists as a pure compound. $$\ce{Ca(HCO3)2}$$ is a compound that only exist in solutions.