I have two mixtures: $\ce{CaCO3}$ in NaCl and $\ce{CaCO3}$ in deionized water (DI). Is it possible to get more $\ce{Ca^2+}$ ions in NaCl solution than DI?

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    $\begingroup$ NaCl? You mean salt water? $\endgroup$
    – Karl
    Commented Sep 25, 2016 at 21:05

2 Answers 2


You claim to have a solid solution of Calcium carbonate in solid (or molten? you didn't specify temperature!) sodium chloride. I doubt it. Because $\ce{CaCl2}$ is more soluble in water than $\ce{CaCO3}$, addition of $\ce{Cl^-}$ ions would be expected to increase the solubility of $\ce{CaCO3}$ in water. In other words, "yes, it is possible." One crude way to think about it is that the "average" environment for the $\ce{Ca^{2+}}$ ions is more "friendly" in a mix of $\ce{CO3^{2-}}$ and $\ce{Cl^-}$ ions than in a mixture of $\ce{CO3^{2-}}$ions alone. The effect is called the "Common Ion" effect. Look it up on Wikipedia.


It depends... I'm going to assume no carbonate in either the pure water or the NaCl solution.

Calcium carbonate is fairly insoluble. The $\rm K_{sp}$ from Wikipedia is $3.3×10^{−9}$.

Now we typically just use the concentrations of $\ce{Ca^{2+}}$ and $\ce{CO3^{2-}}$ when using the formula, but it is really the activities that should be used not the concentrations.

In a weak solution of NaCl then the $\ce{Na^{+}}$ cations and $\ce{Cl^{-}}$ anions won't effect the activity of the $\ce{CO3^{2-}}$ so more calcium carbonate won't dissolve.

In a strong NaCl solution, just because of the ionic strength the activities of $\ce{Ca^{2+}}$ and $\ce{CO3^{2-}}$ will be lowered. In addition there will be clusters/complexes such as $\ce{NaCO3^{-}}$ forming which lower the activity of the $\ce{CO3^{2-}}$ species even more. This will cause a little more calcium carbonate to dissolve than it would in pure water.

I'll digress and point out that the solubility product itself really isn't a constant but depends on the ionic strength of the solution.

But, off the top of my head, I can't remember if there is a way to calculate [$\ce{Ca^{2+}}$] as a function of [NaCl]. Such a calculation would have to consider both the activities of $\ce{Ca^{2+}}$ and $\ce{CO3^{2-}}$ and the change in the solubility product itself.


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