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Why is it that $\ce{CO2}$ is considerably more soluble in water than $\ce{O2}$ is?

$\ce{CO2}$ is nonpolar but dissolves in water which like $\ce{CO2}$ being nonpolar, doesn't make any sense.

Is it the bond polarity and that the bond polarity is more important in solubility?

If not than why is $\ce{CO2}$ more soluble in water than $\ce{O2}$.

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Just because a molecule is non-polar, it does not mean that there are no dipole moments in the molecule. Carbon dioxide has 2 dipole moments in opposite directions and of equal strength (the dipole moment of the molecule is 0; therefore it is non-polar). –  LDC3 Jul 26 at 16:01
@LDC3 I would say it slightly differently. CO2 has no dipole moment, but that doesn't make it nonpolar. The combined opposed dipole moments give the whole molecule a "quadrupole moment" meaning that if there is a 4-pole electric field with positive at north and south and negative at east and west, the CO2 molecule will tend to turn to a north-south orientation. Larger molecules may have hexapole, octupole or higher moments (with progressively less effect.) CO2's polar bonds enable it to absorb infrared light much more easily than N2 or O2, which is what makes it an important greenhouse gas. –  steveverrill Jul 26 at 22:51
@steveverrill In collage I was taught that dipole refers to the polarization between 2 bonded atoms, not the molecule. From this definition, $CO_2$ has 2 dipoles (in opposite directions). The net effect results in $CO_2$ being non-polar. A non-polar molecule without any dipoles is oxygen, nitrogen, hydrogen, sulfur ($S_8$), ... –  LDC3 Jul 26 at 23:02
@LDC3 dipole just means a separation of charges into 2 ends, a positive end and a negative end. Bonds between dissimilar atoms have a dipole. A molecule can also have a dipole. H2O has both. A molecule's "dipole moment" is its quantitative tendency to align itself with an electric field, and tables exist of these. As CO2 has no dipole moment, many textbooks classify it as "non polar" but it does belong to the next group: molecules with quadrupole moments. When discussing solubility, equating dipole moment with "polarity" is not always helpful. For example hexamine is massively soluble in water –  steveverrill Jul 27 at 0:00
@steveverrill Since a dipole is actually a vector, then I guess that when you sum the vectors, you could still call it a dipole, although I would call the molecule polarized. I never mentioned anything about solubility in my posts (I'm not familiar enough with that topic). What I was pointing out in my first comment was that not all non-polar molecules are absent of dipole moments. –  LDC3 Jul 27 at 0:12

3 Answers 3

up vote 6 down vote accepted

Taken from my answer to your original question

There are a couple of reasons why $\ce{CO2}$ is more soluble in water than $\ce{O2}$. Because the two $\ce{C=O}$ bonds in $\ce{CO2}$ are polarized (whereas in $\ce{O2}$ the bond is not polarized) it makes it easier for the polar water molecule to solvate it and to form hydrogen bonds. Both of these factors will stabilize a $\ce{CO2}$ molecule more than an $\ce{O2}$ molecule in water; stabilization translates into greater solubility. Another factor enhancing the solubility of $\ce{CO2}$ in water is the fact that $\ce{CO2}$ reacts with water to set up an equilibrium with carbonic acid. $$\ce{CO2(aq) + H2O <=> H2CO3(aq)}$$ This reaction will also enhance $\ce{CO2}$'s solubility in water compared to oxygen which does not react with water.

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This is a good answer because it covers both possibilities - the enhanced solubility due to polarized bonds as well as the reaction of $\ce{CO2}$ with water. I think it is important to note though, that the reaction with water is primarily responsible for the increased solubility. For proof, take a look at the solubility of $\ce{CO}$ vs. $\ce{CO2}$ - $\ce{CO}$ is polar, but is 60 times less soluble than $\ce{CO2}$. –  thomij Jul 26 at 17:29
Good point thomij. Note, though, that in $\ce{CO2}$ it is the oxygen that is negatively polarized, while in $\ce{CO}$ the oxygen is positively polarized. I suspect this has a significant effect on hydrogen bonding and, consequently, solvation. As is so often the case, too many variables, too few experiments. –  ron Jul 26 at 17:41
@Jori "Both of these factors" I mean like this...Because the two $\ce{C=O}$ bonds in $\ce{CO2}$ are polarized (whereas in $\ce{O2}$ the bond is not polarized) it makes it 1) easier for the polar water molecule to solvate it and 2) to form hydrogen bonds. Private communication is fine, give me your e-mail and I'll contact you, or we could use SE Chem chat. –  ron Jul 26 at 22:54
sent you an e-mail –  ron Jul 27 at 13:00
why would the oxygen in carbon monoxide have a partially positive charge? That only seems possible for oxygen flourides. –  caters Jul 28 at 2:14

That is because $\ce{CO2}$ will react with water in equilibrium to form $\ce{H2CO3}$, which is an acid and thus will dissociate to form ions that can easily fit in the polar solvent (i.e. water). $\ce{O2}$ on the other hand does not react with water because it is not a very good electrophile in comparison to $\ce{CO2}$. The $\pi$-bond is weak in $\ce{CO2}$ (i.e. high in energy), partly because the $\ce{C=O}$ bond is polarized and because of delocalisation in remaining $\pi$-bond of $\ce{CO2}$. $\ce{O2}$ does not have that luxury.

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Though the whole molecule has a net dipole moment of zero due to its geometry, the C=O bond has a dipole moment. Compared to the O=O bond which has no dipole moment.

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