2
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Ivan Neretin Jan 27 '16 at 15:22
  • $\begingroup$ Oh! Duh! Its a halide! I can't believe I missed that $\endgroup$ – zogfotpik Jan 28 '16 at 15:03
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Frankly I hate problems like this. The problem didn't specify that equal amounts of the anions were present... $\endgroup$ – MaxW Feb 28 '17 at 19:50
  • 1
    $\begingroup$ @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! $\endgroup$ – airhuff Feb 28 '17 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.