Why is the water solubility of CO worse than that of CO2?

Question

I tried my best to ask SciFinder, Google, and Google Scholar the wrong questions ...

I found many paper on the fitting of the solubility to some equations to be able to calculate the amount of $\ce{CO}$ in water, but I could hardly find any explanation on the low solubility. Only one paper [1] dealt with it.

Why is $\ce{CO}$ way less soluble in water than $\ce{CO2}$, despite, e.g., the first having a dipole moment and the latter doesn't? What am I missing?

[1] H. Sato, N. Matubayasi, M. Nakahara, F. Hirata, "Which carbon oxide is more soluble? Ab initio study on carbon monoxide and dioxide in aqueous solution," Chemical Physics Letters 2000, 323(3-4), 257–262 (https://doi.org/10.1016/S0009-2614(00)00508-X).

related https://chemistry.stackexchange.com/questions/35603/why-is-co-practically-nonpolar https://chemistry.stackexchange.com/questions/35603/why-is-co-practically-nonpolar — Mithoron, Jun 11 '20 at 23:32
Because if it's serious then your comparing it with wrong compound - comparison between CO and N2 would be interesting, but CO2 reacts with water, which changes everything. BTW @IvanNeretin don't even start that again... — Mithoron, Jun 12 '20 at 00:01
I was going to comment on the polarity discussion here, but there's clearly some cross-talk I'm missing... :/ — Zhe, Jun 12 '20 at 00:28
Polar does not mean anything. It is among the most ambiguous words used in chemistry. — AChem, Jun 12 '20 at 05:32
@M.Farooq IMHO, "stable" is much worse. It is not used here, though. — Ivan Neretin, Jun 12 '20 at 07:20
@M.Farooq Do not confuse lack of meaning with multiple meanings :-) It's also a relative location on planet earth or an adjective, or the name of various commercial brands. — Buck Thorn, Jun 12 '20 at 12:07
Please note that OP is correct as referring to the dipole moments and not to a more ambiguous (for some) polarity. The linked article find its justification in dealing with ab initio methods rather than common sense. When we speak about CO2 solubility I do not think we intend "free" CO2. Perhaps the article does but it is not what common intuition means. That is the answer is probably that by @Oscar Lanzi integrated by Mithoron comment. — Alchimista, Jun 12 '20 at 13:41
@Alchimista, I was referring to comments. OP is using correct wordings. — AChem, Jun 12 '20 at 13:43
But it could be - discuss reaction with water and relevant equilibria, and it would be done. — Mithoron, Jun 13 '20 at 20:55
Most CO2 dissolved in water is actually CO2, not reacted to form carbonic acid. So a reaction-based explanation misses. For an alternative hypothesis see the answers. — Oscar Lanzi, Jun 14 '20 at 00:38

score 3 · Answer 1 · edited Jun 12 '20 at 10:08

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The fact that $\ce{CO2}$ does not have a dipole moment does not really mean it's nonpolar. What $\ce{CO2}$ has is a quadrupole, two opposing dipoles that occupy different portions of space (opposite sides of the carbon atom).

Ordinarily we expect an electrostatic interaction with a quadrupole to be relatively weak, but in $\ce{CO2}$ the component dipoles are strong and at close range the water molecules can "home in" on one end of the molecule and interact selectively with the corresponding component dipole. Thereby the electrostatic interaction between water and the $\ce{CO2}$ quadrupole is stronger than the interaction between water and the weak $\ce{CO}$ dipole.

edited Jun 12 '20 at 10:08

Aniruddha Deb

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answered Jun 12 '20 at 10:01

Oscar Lanzi

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Related: https://chemistry.stackexchange.com/questions/54115/quadrupole-moment-of-a-molecule – Karsten Jun 12 '20 at 11:32
Yes carbon dioxide has a weak microwave rotational spectrum as well. – AChem Jun 12 '20 at 14:48

Why is the water solubility of CO worse than that of CO2?

1 Answers1