# How to predict precipitate upon evaporation from calcium ions and either chloride, carbonate, or nitrate counterions?

An aqueous solution containing $\ce{Ca^2+}$, $\ce{Cl^-}$, $\ce{CO3^2-}$, and $\ce{NO3^-}$ is allowed to evaporate. Which compound will precipitate first?

The correct answer is $\ce{CaCO3}$. I had guessed that, and I know nitrates are soluble, so that wouldn't precipitate, but why the $\ce{CO3^2-}$ and not the $\ce{Cl-}$? Is it because of the equal and opposite charges between them? Does electronegativity play a role?

• Because calcium chloride is fairly well soluble, too. In fact, $\ce{CaCO3}$ would precipitate right away, not waiting for the solution to evaporate. There is no simple rule; it has nothing to do with electronegativity or charges. Jan 27, 2016 at 15:22
• Oh! Duh! Its a halide! I can't believe I missed that Jan 28, 2016 at 15:03

This question is testing your knowledge of general solubility trends, which largely need to be memorized. Here is a fairly comprehensive list, and there are many others scattered through the internet.

You are correct that nearly all nitrates are soluble, so that would be a bad guess.

The halides are generally pretty soluble. There are more exceptions than for nitrates, but as a rule halide salts tend to be soluble.

Carbonates however are frequently insoluble. The Group I cations tend to be an exception, but not the Group II cations, like $\ce{Ca^2+}$.

So, the general trend in solubility for the three anions in your question is $\ce{NO3- > Cl- > CO3^2-}$, so you would expect $\ce{CaCO3}$ to precipitate first.

We can verify this against the known solubilities of each of the salts at room temperature:

$\begin{array}{rr} \pu{\ce{Ca(NO3)2}}: & \pu{1212 g/L }\\ \pu{\ce{CaCl2}}: & \pu{75g/L}\\ \pu{\ce{CaCO3}}: & \pu{0.01g/L} \end{array}$

So, the quantitative solubility data confirms the original prediction and $\ce{CaCO3}$ will clearly precipitate first.

• Frankly I hate problems like this. The problem didn't specify that equal amounts of the anions were present...
– MaxW
Feb 28, 2017 at 19:50
• @MaxW, I guess one aspect of this problem that diminishes that to a fair degree is the ~3-5 orders of magnitude difference in solubility between $\ce{CaCO3}$ and the others. I think the fact that you can only dissolve 10 mg of $\ce{CaCO3}$ in a liter of water makes this a pretty impractical physical experiment though. I mean, you start off with a 100 ml of solution and you're basically just left with dirty glassware! Feb 28, 2017 at 20:03