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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

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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.

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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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