Acid titration of sodium acetate in the presence of sodium ethoxide, is it possible?

I have an aqueous solution of ethanol of $$30 - 40\text{ V%}$$ that contains a few (unknown) $$\text{%}$$ ethyl acetate (ethyl ethanoate - $$\ce{EtOAc}$$).

I wonder if I can quantitatively determine the $$\ce{EtOAc}$$ content by back titration of a small excess of strong $$\ce{NaOH}$$.

The idea is to add a stoichiometric excess of $$\ce{NaOH}$$ to a known volume of sample solution, then reflux this mixture to achieve saponification of the $$\ce{EtAc}$$, acc.:

$$\ce{EtOAc}+\ce{NaOH} \to \ce{NaOAc}+\ce{EtOH}$$

The excess $$\ce{NaOH}$$ would form sodium ethoxide with the ethanol, acc.:

$$\ce{NaOH}+\ce{EtOH} \to \ce{NaOEt}+\ce{H2O}$$

Since as $$\ce{NaOEt}$$ is a strong base and $$\ce{NaOAc}$$ a weak one, I think it should be possible to titrate them sequentially.

Any thoughts would be appreciated.

• As ethanol is much weaker acid than water, the second reaction would occur significantly only in NaOH solution in (almost) pure ethanol. Otherwise, equilibrium is pushed leftwards. May 19 '21 at 2:23
• I though that too, later on.
– Gert
May 19 '21 at 11:31

The pH of 0.1 M NaAc is calculated to be 8.9. https://pubs.acs.org/doi/pdf/10.1021/ed079p29.1

The pH of NaOH will be much higher. The 30-40% of ethanol may have some effects, but in general, you should be able to add about twice as much NaOH as you think will react with all the EtAc, reflux and back titrate the NaOH. As Poutnik comments, there won't be significant NaOEt.

The reaction will look like this:

EtAc + 2 NaOH --> NaAc + EtOH + NaOH

If you can get some thymolphthalein, you could use that as an indicator. https://en.wikipedia.org/wiki/PH_indicator

Or, you can use pH paper strips if you are satisfied with a cruder analysis.

Possibly a more intricate back titration might work: add about twice as much NaOH as you think will react with all the EtAc in your ethanol, reflux as above, then add the same number of moles of acetic acid as you added moles of NaOH. Then the reaction will look like this:

EtAc + 2 NaOH --> NaAc + EtOH + NaOH; reflux; + 2 HAc -->

NaAc + EtOH + NaAc + H$$_2$$O + HAc

Then back titrate the excess acetic acid with phenolphthalein (common enough - you must have some). If there were no EtAc in your ethanol, the pH of the solution after adding the 2 HAc will be about 8.9; if there is any EtAc in your ethanol, you will have excess acetic acid due to the saponification of the EtAc. And you back titrate that with NaOH. So this is sort of a back to back titration, or a back and forth.

• Could you please use MathJax for the chemical equations? You have been on this site for long enough. The equations are really hard to understand as they are now. May 19 '21 at 9:35
• In addition to the rightful suggestion by @ShoubhikRMaiti regarding mhchem, note that tables are essentially textual information and should be ideally presented as such, not as images (as long as it's technically viable, which is clearly your case). Also, note that "Ac" is a symbol for actinium or acetyl, whereas it looks like you are talking about acetate with standard symbol "OAc". May 19 '21 at 10:45
• Thanks for your answer. Good news really on the non-formation of $\ce{NaOEt}$. I'll be titrating against a $\mathrm{pH}$ meter, I think.
– Gert
May 19 '21 at 11:39
• Have you any suggestions for 'optimal' reflux conditions, to ensure complete saponification?
– Gert
May 19 '21 at 16:19
• @Gert: You mean how long? Check the pH. Bring it to a boil, then cool it right down and check the pH at room temp. If there's a significant change, try it again to see if there is another change (maybe boil it for a few minutes). If no further change, you're done! Or, check it first with an artificial solution made up with reagent EtOH, EtAc and water, as a rehearsal. Maybe it doesn't even need to be boiled! May 19 '21 at 19:14