Why is hydrated carbon dioxide - the predominant form of acid that one gets upon dissolution of carbon dioxide in solution - so unstable?

Is the below rationale valid?

  1. Carbon in carbon dioxide has two (empty? no, but still vulnerable to attack) p-orbitals and bears a strong partial positive charge.
  2. Oxygen's lone pair can attack an empty p-orbital and form a formal charge-separated complex with the carbon dioxide.
  3. This is "hydrated" carbon dioxide or "carbonic acid." This form is extremely unstable and subject to disproportionation due to an unfavorable charge and the unfavorable nature of charge separation itself. The change in entropy also favors the products of disproportionation.
  4. However, there exists a pathway to stability - that is - protonation of the oxygen with the negative formal charge by the oxygen bearing the positive formal charge.
  5. This, however, is akin to a forbidden fruit; the ephemeral three-membered ring that would have to be formed exhibits "ring strain" (if you object to this term, can you please elaborate on your objection), and as a result, disproportionation is overwhelmingly favored, especially from an entropic standpoint (reconstitution of carbon dioxide gas is highly entropically favorable).

enter image description here


2 Answers 2


The situation is described by the following equilibrium $$\ce{H2O + CO2 <=> HO-CO-OH}$$ In a closed system, using Le Chatelier’s principle you can manipulate the equilibrium in either direction (e.g. pump in carbon dioxide and increase the concentration of carbonic acid, etc.). However in an open system, as carbon dioxide escapes, the amount of carbonic acid will decrease.

When all is said and done, water and carbon dioxide are two extremely stable molecules. Since $$\Delta G = -RT\ln K$$ Their stability is what drives the equilibrium to the left and why carbonic acid appears to be (relatively) so unstable.

  • $\begingroup$ I had the same issue in class today; I don't see how we get from water and carbon dioxide to (HO)2CO directly. I think my proposed intermediate explains the instability; it is very hard to actually form (HO)2CO unless under pressure; only 0.4% of the CO2 pumped into water turns into actual carbonic acid or (HO)2CO at STP. The rest is in the charge separated form. $\endgroup$
    – Dissenter
    Sep 8, 2014 at 23:20
  • 3
    $\begingroup$ "I don't see how we get from water and carbon dioxide to (HO)2CO directly" Your scheme is correct, it's just another SN2 reaction of water and a carbonyl. $\endgroup$
    – ron
    Sep 8, 2014 at 23:24
  • $\begingroup$ SN2 but no leaving group? $\endgroup$
    – Dissenter
    Sep 8, 2014 at 23:36
  • 1
    $\begingroup$ After water reacts with one C=O, the oxygen in the remaining C=O (let's view the C=O as a two-membered ring) is the leaving group, more specifically, one of the two C=O bonds is the leaving group. The remaining C-O bond keeps the leaving group connected to the carbon. $\endgroup$
    – ron
    Sep 8, 2014 at 23:46
  • 4
    $\begingroup$ Yes. (but my comment needs to be at least 15 characters) $\endgroup$
    – ron
    Sep 9, 2014 at 0:04

This answer looks at the molecular structure characteristics behind the thermodynamics reported by Ron

We are dealing with the reaction

$\ce{CO2 +H2O <=> H2CO3}$

or more accurately

$\ce{O=C=O +H2O <=> (HO)2C=O}$

One might expect the equilibrium to favor the right side through the replacement of a pi bond with a sigma bond and the ability of the hydroxylated compounds to fit better with the hydrogen-bonded water structure. Such bonding features are common drivers of reactions between high oxidation-state nonmetaloxides and water. But the carbon dioxide molecule on the left has two particular features that disfavor such a reaction thermodynamically.

Carbon dioxide polarity: the whole truth

Many a chemistry textbook describes carbon dioxide as nonpolar, but it isn't. It is non-dipolar. Carbon dioxide has two strong opposing dipoles that cancel out as dipoles, but combine to make a powerful quadrupole. This quadrupole is well-sized to be solvated strongly with both the water dipole and the water quadrupole (look along the line from one hydrogen atom to the other, you see a quadrupole). Indeed the water quadrupole, whose dipole components are directed oppositely from the dipole components of the carbon dioxide quadrupole, fits like a hand in a glove when it comes to solvation. The resulting solvation capability is one reason carbon dioxide, despite its "nonpolarity", has fairly good solubility in water in the first place.

The revenge of 4n pi electrons

Carbon dioxide has an especially strong pi bonding stabilized by delocalization; it might be thought of as a linear-molecule counterpart to aromaticity. (In contrast with the usual form of aromaticity, the interaction favors $4n$ electrons; carbon dioxide with all atoms $sp$ hybridized gives $n=2$.) Reaction with water to break one of these pi bonds upsets this delocalization much like benzene resisting addition because of a similarly strong stabilization by delocalization.

  • $\begingroup$ Note that waterfree H2CO3 is reportedly kinetically very stable, as solid or gas, subliming near -55 deg C. A molecule of H2CO3 has reportedly halflife 180 000 years. But there is extremely strong autocatalytic effect of water. // At ambient temperatures, pure carbonic acid is a stable gas.[5] There are two main methods to produce anhydrous carbonic acid: reaction of hydrogen chloride and potassium bicarbonate at 100 K in methanol and proton irradiation of pure solid carbon dioxide.[2] $\endgroup$
    – Poutnik
    Oct 12, 2023 at 9:27
  • $\begingroup$ @Poutnik all true. But I do not discuss this in the answer because the water-free case seems outside the scope of the question. Aqueous carbonic acid seems to be stable at high pressure and temperature.. $\endgroup$ Oct 12, 2023 at 9:53
  • $\begingroup$ Sure, I have mentioned it as a side note. This On the Surprising Kinetic Stability of Carbonic Acid (H2CO3) may be worthy to read too, assuming with the access. $\endgroup$
    – Poutnik
    Oct 12, 2023 at 9:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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