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).

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1 Answer 1


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

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