# Does H2CO3 exist in solution?

When $\ce{CO2}$ reacts with water, it supposedly forms $\ce{H2CO3}$, which is the aqueous form of $\ce{CO2}$. But it is hard to isolate, but its $K_\mathrm{a}$ value is pretty low (on the order of $10^{-7}$) indicating it should be found in a non dissociated form? So does $\ce{H2CO3}$ exist in solution?

• The equilibrium of $\ce{CO2(aq) + H2O(l) <=> H2CO3 (aq)}$ lies extremely far to the left (you might find data on the Internet... I remember reading it in Skoog et al) so if you were to have $\ce{H2CO3}$ in solution, it won't dissociate to $\ce{H+ + HCO3-}$ but instead to $\ce{CO2 + H2O}$ (I believe, not entirely sure). – orthocresol Oct 24 '15 at 20:21
• @orthocresol so how does CO2 exist in solution? – TanMath Oct 24 '15 at 20:28
• The given low $K_\mathrm a$ value applies to $\ce{[CO2(aq)] + [H2CO3]}$. True $\ce{H2CO3}$ is a much stronger acid with $\mathrm pK_{\mathrm a}\approx3.5$, which is stronger than acetic acid and formic acid. – user7951 Oct 24 '15 at 20:32

Aqueous carbonate solutions contain four different solute species:

• $\ce{CO2(aq)}$
• $\ce{H2CO3}$
• $\ce{HCO3-}$
• $\ce{CO3^2-}$

The equilibrium for the hydration reaction

$$\ce{H2O + CO2(aq) <=> H2CO3}\tag1$$

lies to the left. The corresponding equilibrium constant $K$ at $25\ \mathrm{^\circ C}$ is about

$$K=\frac{\left[\ce{CO2(aq)}\right]}{\left[\ce{H2CO3}\right]}\approx650$$

(literature values of $K$ range from $350$ to $990$.)

The commonly used first acid dissociation constant of carbonic acid $\mathrm pK_{\mathrm a1}=6.35$ (at $25\ \mathrm{^\circ C}$) actually is a composite constant that includes both the hydration reaction $\text{(1)}$ and the protolysis of true $\ce{H2CO3}$ $\text{(2)}$.

$$\ce{H2CO3 <=> H+ + HCO3-}\tag2$$

The constants $K$, $K_{\mathrm a1}$, and the actual acid dissociation constant $K_{\ce{H2CO3}}$ of true $\ce{H2CO3}$ are interrelated according to

$$K_{\mathrm a1}=\frac{K_{\ce{H2CO3}}}{1+K}\approx\frac{K_{\ce{H2CO3}}}{K}$$

Calculated values for $\mathrm pK_{\ce{H2CO3}}$ range from $3.8$ to $3.4$. Thus, true $\ce{H2CO3}$ is stronger than acetic acid and formic acid.

The second acid dissociation constant of carbonic acid is $\mathrm pK_{\mathrm a2}=10.33$ (at $25\ \mathrm{^\circ C}$).

Therefore, the predominant species in aqueous carbonate solutions (at $25\ \mathrm{^\circ C}$) are

• $\ce{CO3^2-}$ at $10.33\lt\mathrm{pH}$
• $\ce{HCO3-}$ at $6.35\lt\mathrm{pH}\lt10.33$
• $\ce{CO2(aq)}$ and $\ce{H2CO3}$ at $\mathrm{pH}\lt6.35$

Thus, true $\ce{H2CO3}$ primarily exists in acidic solutions. However, because of the hydration equilibrium $\text{(1)}$ and the corresponding equilibrium constant $K\approx650$, most of the true $\ce{H2CO3}$ is converted to $\ce{CO2(aq)}$.

(All values taken from: Stumm, W.; Morgan, J. J. Aquatic Chemistry, Third Edition; John Wiley & Sons: New York, NY, 1996; pp 150–152.)

• No doubt that you've written the "traditional" model correctly in equations 1 and 2. The math and chemical equations work. The real question is how long does a particular molecule of H2CO2 on average last? What is the mean time between collisions for molecules in water? So if a molecule on average lasts for trillions of collisions then it is stable. Certainly if the "molecule" last for only a trillionth of the average time between collisions then it is unstable. But where is the absolute cutoff between stable and unstable? // In other words the kinetics of your equation 1 are very very fast. – MaxW Oct 25 '15 at 20:50

I think that the answer is rather orthogonal to your question. If you think of any sort of acid $\ce{H+}\ce{A-}$ then the "molecule" of HA exists so that a Ka can be tabulated. But let's tag the particular ions $\ce{H+}$ and $\ce{A-}$) of one molecule and drop it into the mixture. It won't stay as as one molecule but rather due to the transfer of protons (hydronium ions really) the two ions will become separated. Due to mathematical combinatorial possibilities it is unlikely that our two tagged ions would get together again.

From wikipedia

"The long-held belief that carbonic acid could not exist as a pure compound has reportedly been recently disproved by the preparation of the pure substance in both solid and gas form by University of Innsbruck researchers."

So the question is really about how long does a particular "molecule" of HA last in solution? For $\ce{H2}\ce{C}\ce{O2}$ the answer would be "not long." So in solution I think you really have to define "exists" in terms of the average time (T) between molecular collisions. So how many multiples of T are necessary for the molecule to "exist" in solution?

• It hasn't been isolated in an aqueous solution, and that's why they say it doesn't "exist" in there. Wikipedia tries to be skeptical of the widely held belief by linking to a news article that doesn't link to the journal article? I'm a bit skeptical of Wikipedia's skeptical skeptic skepticism. – M.A.R. Oct 25 '15 at 20:16
• Wikipedia links to this article phys.org/news/… which has journal information. // I'm certainly not an expert on this particular controversy, but I think my explanation is reasonable. The "stability" of a molecule which is in equilibrium is fuzzy since the "equilibrium" doesn't account for the kinetics. – MaxW Oct 25 '15 at 21:08