In a review article [1] on the solubility of CaCO$_3$(calcite), the solubility is reported to be on the order of 1 gram/litre.

Since CaCO$_3$ has a molar mass of about 100, this solubility is equivalent to $10^{-2}$ M.

The solubility product is the square of this, $K_\mathrm{sp}=10^{-4}$.

However, the solubility product of calcium carbonate (according to various sources, like Wiki) is actually $\sim 10^{-9}$ which is 5 orders of magnitude smaller.

What am I missing?

[1] Miller, John P. "A portion of the system calcium carbonate-carbon dioxide-water, with geological implications." American Journal of Science 250.3 (1952): 161-203.

  • 3
    $\begingroup$ 1 g/l is an insanely high figure for CaCO3; I won't believe it even if I see it appear in glowing red letters on a wall, written by an invisible hand. But there's a catch: what I said only applies to pure water. Add some acid, and you'll easily dissolve this much, and maybe a good deal more. $\endgroup$ May 31, 2019 at 11:39
  • $\begingroup$ Perhaps the influence of carbon dioxide? $\endgroup$
    – Alchimista
    May 31, 2019 at 11:45
  • $\begingroup$ These experiments don't seem to contain any acids that would account for it. According to the literature (Plummer et al, American journal of science 278.2 (1978): 179-216) the range of pCO$_2$ might account for like 1 order of magnitude variation, not 5. I can also cite papers where pH has no more than 1 order of magnitude effect. $\endgroup$
    – lemon
    May 31, 2019 at 12:07
  • 1
    $\begingroup$ Reading the title of the reference gives the hint. $\endgroup$
    – Poutnik
    May 31, 2019 at 16:15
  • $\begingroup$ But what does the "CRC Handbook or Chemistry and Physics" say? Because it is the Bible for chemical and physical constants $\endgroup$
    – SteffX
    Jun 1, 2019 at 17:44


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