How would a low concentrated acetic acid react with highly concentrated KOH solution of equal volume?

Question

If there are two solutions of 25mL volume and one contains 0.5M acetic acid and the other 1M KOH, then will every molecule of acetic acid convert into potassium acetate?

I know acetic acid is a weak acid, so it will not dissociate completely at a time. But when concentration of hydroxide ion increases due to the base it will undergo backward reaction with hydronium ion to form water. So amount of hydroxide ion will be more than hydronium ions.

But I am confused about whether the non-dissociated acetic acid molecules dissociate after addition of the base? If they do,why would they dissociate? If the existing hydroxide and hydronium ions can form water molecules to maintain the consistency of auto ionization constant, will there be any necessity for acetic acids to dissociate to supply more hydronium ions?

Another thing which is confusing me is that I used to know that acetic acid does not dissociate completely even if infinite amount of base is added to the solution. But as I have mentioned a stem at the beginning, based on that question I have seen many solution to such mathematical problems where they considered that all the acetic acid molecules will dissociate to form potassium acetate molecules. So will all the acetic acid molecules dissociate and form salts? Again salts remain ionized on aqueous solution so there remains a probability to undergo backward reaction to form acetic acid molecules.

I am really confised what happens in the reaction medium in such conditions.

Thank you.

Answer 1

Dissociated and non dissociated acetic acid react with $\ce{KOH}$ in the same way. This result does not depend on the concentrations of the initial acid and of the initial hydroxide. It will produce potassium acetate.

Now it is also right that the reaction is not total, as potassium acetate is slightly hydrolized. But this effect is rather weak. A solution $1$ molar in potassium acetate has a pH $9.37$. This corresponds to a residual $\ce{OH-}$ concentration equal to $\ce{2.35·10^{-5}}$ mole par liter.

So out of a solution containing $1$ mole potassium acetate per liter, only $23.5$ micromole are decomposed into acid and hydroxide. This ins not zero of course, but it is rather small. $23.5$ parts per million are not transformed into potassium acetate. The purity of the obtained potassium acetate is $100$% - $0.00235$% = $99.99765$ %

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As many comments have been written about my very first sentence, I think advisable to develop it in some detail. An acetic acid aqueous solution is made of a vast majority of unchanged $\ce{CH3COOH}$ molecules, plus a tiny percentage of ions due to the reaction $$\ce{CH3COOH <=> CH3COO- + H+ \tag{1}}$$ which could also be written $$\ce{CH3COOH + H2O <=> CH3COO- + H3O+ \tag{2} }$$

This reaction is an equilibrium, with an equilibrium constant which is usually derived from equation $(1)$ : $K\pu{_a = \frac{[H^+][CH_3COO-]}{[CH_3COOH]} = 1.8 10^{-5} M}$. In a $1$ molar $\ce{CH3COOH}$ solution, the concentration $\ce{[H^+]}$ is $\sqrt{K\pu{_a}} = 4.24· 10^{-3}$ M. Whatever its numerical value, it means that this concentration $\ce{[H+]}$ is rather small, with respect to $1$ M.

However, if these rare $\ce{H+}$ ions are suddenly destroyed by adding some drops of a $\ce{KOH}$ solution, the concentration $\ce{[H+]}$ should fall to zero. It does not, as immediately Nature reacts, and enough $\ce{CH3COOH}$ molecules get dissociated according to ($1$) or ($2$) to compensate for the sudden lost of $\ce{H+}$. As a consequence, the concentration of acetic acid decreases and the concentration of acetate ions increases. Apparently this reaction ($1$) or ($2$) is extremely fast, as it is impossible to follow its kinetics by spectrometric methods. Why is this reaction so fast ? It is hard to explain, because it is a reaction from or between neutral substances. And usually reactions from or between neutral substances are not immediate, and their kinetics can be followed by spectrometric methods.

On the contrary, every chemist knows that reactions due to ions (like neutralizations) are so fast that their kinetics cannot be determined. This is why some chemists have emitted the idea that the added ions $\ce{OH-}$ should react directly with the most abundant substance in the solution, namely the molecule $\ce{CH3COOH}$ in a possible reaction $\ce{OH- + CH3COOH -> CH3COO- + H2O}$. The $\ce{H+}$ ions would not wait for the equilibrium to be re-adjusted.

But I would not fight for this idea, that I have read once in a manual. It is not the point of the present discussion. I would not abject if the reader prefers thinking that the added $\ce{OH-}$ react with the ions $\ce{H+}$ (or $\ce{H3O+}$) produced by the quick adjustment of reactions ($1$) or ($2$).