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I have a solution with only $\ce{CO_3^{2-}}$ ions and water.

What would happen with the carbonate ions if I were to evaporate the water of that solution (in a vacuum)?

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    $\begingroup$ I'm pretty sure that you also have some counter ion in there $\endgroup$ – Mad Scientist Nov 14 '13 at 16:35
  • $\begingroup$ but what if the ion was selectively let through by a semi penetrable membrane into this solution? $\endgroup$ – user2117 Nov 14 '13 at 16:37
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    $\begingroup$ You can't selectively remove one type of charge and sustain the difference indefinitely. Because the electrostatic force constant is so huge, without work being done in the form of an opposing electric field, significant net charges would quickly be neutralized by tearing out the opposite charge from whatever they can, far before getting to macroscopically measurable amount of excess ions. Leaving the system in a vacuum is not enough as the charge would jump through empty space. $\endgroup$ – Nicolau Saker Neto Nov 14 '13 at 22:14
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    $\begingroup$ There is nothing that says solutions must be exactly electrically neutral, just that they can't have too much excess charge. It's a problem of quantifying the limit. After some research, it seems that you theoretically could prepare a solution with excess $\ce{CO^{2-}_3}$, though the excess would have to be quite small (apparently well under $1\ \mu M$). I'm not exactly sure what would happen if you removed the water entirely in a vacuum. Maybe the ions would become gaseous and float around until they hit a surface. $\endgroup$ – Nicolau Saker Neto Nov 16 '13 at 4:40
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    $\begingroup$ Here are a few sources: 1 2 3 $\endgroup$ – Nicolau Saker Neto Nov 16 '13 at 4:45
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I would love to add some quantum chemical bone to the stew we are trying to cook here.

Let us assume that it is possible to have a solution of carbonate ions, like already discussed by Nicolau in the comments. Then Rauru is also right that you will have to consider several equilibria: \begin{aligned}\ce{ (1) && 2H2O &<=> H3+O + {}^{-}OH \\ (2) && CO3^{2-} + H2O &<=> HCO3- + {}^{-}OH\\ (3) && HCO3- + H2O &<=> H2CO3 + {}^{-}OH }\end{aligned}

Now $(3)$ seems a little far fetched, so we will ignore it.

When you do now remove water from the equilibrium through (gentle) evaporation, you will have a mixture of only $\ce{CO3^{2-} + HCO3- + {}^{-}OH}$. Further assuming this is possible you will always have some collisions within this ensemble. Statistical it is therefore always possible to also remove $\ce{HCO3- + {}^{-}OH}$ via $(2)$ from the system. Also single entities of carbonate ions will repel each other as states by G M. So basically it all boils down to on question: Is carbonate stable in vacuum? What happens to carbonate if you put it into vacuum?

Now we have some quite good approximations to these particular conditions (gas phase and 0K), but I do not want to go into detail. I did a calculation on the carbonate with Gaussian09 and as a cheap shot BP86/cc-pVTZ level of theory. The only thing I want to show is the highest occupied molecular orbital.

homo of carbonate

What can clearly see is that this is no bonding orbital. There are lone pair contributions from all Oxygen atoms (due to imposed symmetry constraints). In addition to the high negative charge, electrons will not be stabilized by the field of the nuclei. Also, Orbitals only vanish at infinity, giving electrons a non-zero probability to actually leave the compound. Furthermore, there are two electrons in this orbital, so the spins of these are paired, which also costs energy.

My suspicion therefore is that the carbonate at least loses one electron. To where? To anywhere. (Also mentioned by Nicolau in the comments.) I do not have any actual proof for this, as it quite hard to find people actually doing experiments and/ or calculations on that topic. I remember that one question of a Professor of mine in the first semester was: "What happens to sulphate if you shoot it into space?" The answer was given without any further explanation and I never bothered to ask: "It looses an electron."


Note:

Another point that should always be considered when dealing with gas phase reactions is, that when ions have enough kinetic energy they might react in the strangest ways. One of them might be $$\ce{2CO3^{2-} -> 2CO2- + O2^{2-}}.$$

Also collision with any particle may cause electrons to be transferred.

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Theoretically you should have an explosion. Because every $\ce{CO_3^{2-}}$ ions exerts a force against the other. You should be able to calculate it with coulomb equation:

$$|F|=K_e\frac{q_{\ce{CO_3^{2-}}}^2}{r^2}$$

Of course you have to consider the force exerts from every particle to each others, summing every force field assuming a certain distance: a very hard calculation. By the way your assumption are quite hard to reach in great scale as say Raur Ferro. But a practical application of electrostatic forces that could be enlightening is Electrospray ionization (ESI), a technique used in mass spectrometry to produce ions. In this technique the solvent from an areosols that hold the analyte is forced to evaporate (with various expedients) from a charged droplet until the drop becomes unstable due the charges. At a certain point the droplet undergoes to what is called Coulomb fission, so the original droplet explodes creating many smaller droplets. I know this is not the same thing but this can show you how electrostatic forces can act. I think however that this question is more related to physic than chemistry.

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As a theoretical experiment, you must think on different aquieus equilibria:

$\ce{H2O <=> H+ + OH-}$

$\ce{CO3^{2-} <=> HCO3- <=> H2CO3}$

$\ce{H2CO3}$ will be formed with water's $\ce{H+}$, and in vacuum, possibly will decompose to water and $\ce{CO2}$.

At the end, may be a mixture of $\ce{OH-}$ and $\ce{CO3^{2-}}$ will remain.

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