15
$\begingroup$

The electronics youtuber bigclivedotcom has an on-and-off-again series where he carbonates various types of alcohol and comments on the taste. One thing he's noticed is that the stronger the alcohol, the more carbon dioxide it will absorb; so, he recently tried carbonating 'moonshine' (70% lab ethanol, 30% water).

This absorbed a staggering 22g of CO₂ per litre (and ended up completely undrinkable, or at least more so than it was previously).

I know that when you carbonate water you end up with a weak carbonic acid solution based on the H₂O + CO₂ ←→ H₂CO₃ equilibrium. But the references I've found says that carbon dioxide is much less soluble in ethanol than in water because ethanol is less polar. So, I'd expect that the stronger the ethanol solution, the less carbon dioxide would be absorbed. This is the complete opposite of what he actually observed.

So, what's happening here?

Reference: https://www.youtube.com/watch?v=yArcH80PiP4

$\endgroup$
4
  • 8
    $\begingroup$ The thought of drinking this mixture gave me rather intense mixed feelings. What can be more noble than experimenting on oneself? $\endgroup$ Mar 15 at 13:50
  • 2
    $\begingroup$ Well, that's one way to stay sober. $\endgroup$ Mar 15 at 15:01
  • 2
    $\begingroup$ @IvanNeretin - my personal favorite is jstor.org/stable/1660704 "Concerning the taste of heavy water" by Urey... $\endgroup$
    – Jon Custer
    Mar 15 at 16:06
  • 1
    $\begingroup$ @JonCuster I'd note the more recent people who have tasted heavy water and disagreed with Urey's conclusion. $\endgroup$
    – matt_black
    Mar 15 at 16:20

1 Answer 1

17
$\begingroup$

Carbon dioxide is in fact roughly ten times as soluble in ethanol as in water [1](https://doi.org/10.1016/j.fluid.2006.04.017) (meaning you need more dissolved under pressure to get that bubbling effect; alcoholic beverages that bubble or sparkle are mostly water). The figure below from the reference illustrates the increasing solubility of carbon dioxide under pressure as we increase the ethanol content in an ethanol-water mixture.

enter image description here

Carbon dioxide is actually non-dipolar, and while the strong quadrupole enhances its solubility in water, it still fits better with less polar solvents such as ethanol.

Reference

  1. I. Dalmolin, E. Skovroinski, A. Biasi, M.L. Corazza, C. Dariva, J. Vladimir Oliveira (2005). "Solubility of carbon dioxide in binary and ternary mixtures with ethanol and water". Fluid Phase Equilibria 245, 2, 193-200, ISSN 0378-3812, https://doi.org/10.1016/j.fluid.2006.04.017.
$\endgroup$
5
  • $\begingroup$ Do you know if diethylcarbonate formation is at all a factor with high ethanol? $\endgroup$
    – Andrew
    Mar 15 at 15:51
  • 2
    $\begingroup$ The authors do not seem to say. In water carbonic acid formation is not the main driver of carbon dioxide solubility, so I would guess no. $\endgroup$ Mar 15 at 16:10
  • $\begingroup$ Huh. Google clearly lied to me (although it's possible I saw 'carbon dioxide is only weakly soluble in ethanol' and forgot to ask whether that was more or less soluble than in water). I can't see the article content, sadly --- is there anything interesting about the chemistry caused by the carbon dioxide reacting with the ethanol, in a way analogous to the production of carbonic acid? $\endgroup$ Mar 16 at 0:29
  • 1
    $\begingroup$ @DavidGiven That seems unlikely. The ethyl analogue of carbonic acid would be ethyl hydrogen carbonate, a.k.a. "etabonic acid"; it's apparently unstable, if it exists as a bound molecule at all, although salts of the corresponding etabonate anion $\ce{C2H5OCO2-}$ are apparently stable enough to be synthesized. $\endgroup$ Mar 16 at 16:45
  • $\begingroup$ @IlmariKaronen I agree. Carbonic acid is only a minor apecies in the water solution and I would expect nothing different in alcohols. $\endgroup$ Mar 16 at 16:49

Your Answer

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

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