According to wikipedia $\ce{Ca(HCO3)2}$ has the following solubility values:

$\pu{16.1 g/100 mL} \,(\pu{0^\circ C)}$
$\pu{16.6 g/100 mL} \,(\pu{20^\circ C)}$
$\pu{18.4 g/100 mL} \,(\pu{100^\circ C)}$

So I assume $\ce{Ca(HCO3)2}$ 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 $\ce{CaCO3}$:

$\ce{Ca(HCO3)2(aq) → CO2(g) + H2O(l) + CaCO3(s)}$

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

  • $\begingroup$ 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. $\endgroup$
    – AChem
    Aug 2, 2019 at 12:58
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    $\begingroup$ Surprisingly, SciFinder is quiet about its solubility and it is only Wikipedia which quotes these values. No reference sadly! $\endgroup$
    – AChem
    Aug 2, 2019 at 13:05
  • $\begingroup$ Yes, all three values are from wikipedia. I edited the text to mention it. But as you write, wikipedia doesn't cite any source... : / $\endgroup$
    – Hans
    Aug 3, 2019 at 1:33
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    $\begingroup$ 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. $\endgroup$
    – AChem
    Aug 3, 2019 at 3:27
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    $\begingroup$ Left on note on Wikipedia calcium bicarbonate page to add citations. $\endgroup$
    – AChem
    Aug 3, 2019 at 3:31

3 Answers 3


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:

$$\ce{Ca(OH)2 + 2CO2 -> Ca^{2+} + 2HCO3^-}$$

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:

$$\ce{Ca^{2+} + 2HCO3^- -> CaCO3 + H2O + CO2}$$

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.


I would assume that when calcium bicarbonate dissolves in water, the bicarbonate reacts to form water and carbon dioxide gas. With the carbon dioxide gas evaporating, you're inevitably left with only calcium and carbonate ions in the solution, so you can't precipitate calcium bicarbonate out of the solution.


The CRC Handbook does not list solubility numbers for $\ce{Ca(HCO3)2}$ or $\ce{Mg(HCO3)2}$. We know that temporary hard water has Mg and/or Ca salts of carbonic acid that will precipitate in hot water heaters at ~$\pu{60 °C}$, which appears to be due to forced evolution of $\ce{CO2}$ rather than an inverse solubility curve.

The solubility of $\ce{CO2}$ ($\pu{0.058 g/100mL}$ at $\pu{60 °C}$) would appear to be insufficient to dissolve ~$\pu{10 g}$ of $\ce{CaCO3}$, so I question the solubility number of $\pu{18 g}$ at $\pu{100 °C}$.

The likely reason for being unable to isolate solid $\ce{Ca(HCO3)2}$ is that $\ce{HCO3-}$ ion is in a labile equilibrium with $\ce{CO2(aq)}$ and $\ce{OH-(aq)}$,

$$\ce{2HCO3- -> H2CO3 + CO3^2-}$$

At higher temperatures, the $\ce{CO2}$ simply escapes by evaporation rather than because of equilibrium processes. Stalactites form at cold temperatures because the $\ce{Ca(HCO3)2}$ solution is exposed to a large surface where $\ce{H2O}$ and $\ce{CO2}$ evaporate leaving $\ce{CaCO3}$ behind.

  • $\begingroup$ "The solubility of CO2 (0.058 g/100mL @ 60 C) would appear to be insufficient to dissolve ~10 g of CaCO3" wrong. One can dissolve way more CO2 in water in presence of CaCO3. Pressure can also be applied and it is good to know at what pressure the solubility is measured. $\endgroup$
    – fraxinus
    May 31, 2020 at 10:08
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    $\begingroup$ @James You are an experienced user. We expect you to learn chemical formatting and typeset so that your posts looks better. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$ May 31, 2020 at 10:14

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