Would you add in $\ce{NaOH}$, deprotonating the acid, and then wash the mixture in water, thereby removing the $\ce{Na^+}$ and $\ce{CH3O^-}$ ions in an aqueous layer? But then wouldn't the ethanol be dissolved in the water as well, as it is polar? In general, how do you separate two polar substances from each other using extraction?

If you don't use extraction, how else would you separate the two, aside from distillation?

All answers are appreciated.


2 Answers 2


Well, distillation is almost certainly the best method. However, since you've explicitly ruled that out, there are only a few alternatives I can think of:

  1. Ethanol can be separated from water with reasonable efficiency (at least on the microscale) by a process of "salting out." The ethanol/water mixture is saturated with $\ce{NaCl}$. The solvation of the salt ions is a highly thermodynamically favorable process, and as that occurs the molecules of water become increasingly less able and available to form hydrogen bonds with ethanol. I've never tried this with acetic acid specifically, but I can't think of any reason why it wouldn't work.
  2. I would imagine with the right choice of eluent, separation by column chromatography using silica or alumina should be feasible.
  3. Fractional freezing may be possible in theory, although I have no idea as to its efficacy for this specific case.

I'm also assuming here that you're looking strictly for methods of physical separation, not chemical (so, strictly as a general example: adding something like calcium oxide to consume water in the generation of insoluble calcium hydroxide would be out of the question).


Take the mixture of AcOH and ethanol and make it super saturated with bicarbonate, extract with ethyl acetate to remove ethanol and then acidify aqueous layer acidic and extract with 5-10% DCM in MeOH to remove acetic acid.


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.