Today I stumbled across a website which defined the titration of weak acid and strong base as the following:

"The titration of a weak acid with a strong base involves the direct transfer of protons from the weak acid to the hydoxide ion. The reaction of the weak acid, acetic acid, with a strong base, NaOH, can be seen below. In the reaction the acid and base react in a one to one ratio." (https://chem.libretexts.org/Ancillary_Materials/Demos_Techniques_and_Experiments/General_Lab_Techniques/Titration/Titration_of_a_Weak_Acid_with_a_Strong_Base)


But, I clearly remember my professor for analytical chemistry saying that the neutralization of a weak acid and a strong base occurs due to the shift of equilibrium:

CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺

Where the strong base (NaOH) would continue to remove hydronium ions from the equilibrium, so that more and more of conjugated base, acetate, would form.

So the first part of my question is; what definition is correct?

Furthermore, indeed we usually formulate following equation to approximately quantify the change in concentrations of acid and base:

CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na+

This way, we find the moles of acid/base and divide by the total volume to get the pH using Hendersson-Hasselbalch equation. But is it correct this way? Are there more accurate methods? The weak acid should dissociate first before being consumed, is that correct?

Anyway, this is not a question about how to calculate the titration curve of this example, rather it is about understanding the nature of the conversion. Thanks in advance!


1 Answer 1


Your second equation is an equilibrium. Both acetic acid and acetate ions are always present in a solution, although their proportion may vary. When $\ce{CH3COOH}$ is dissolved into water, a fraction of the molecules are transformed by reaction with water, producing some $\ce{H3O+}$ But if these ions are destroyed by adding $\ce{OH-}$ ions, more molecules $\ce{CH3COOH}$ have to react with water. Acetic acid is transformed into acetate ions to compensate for the consumption of the first $\ce{H3O+}$ ions. So more and more molecules of acetic acid are transformed into ions. The final result is similar to the first equation : more and more $\ce{CH3COOH}$ is consumed by adding $\ce{NaOH}$ and more and more acetate ion and water is produced from this process.

  • $\begingroup$ I would 100% agree. But that would be hard to quantify using Hendersson-Hasselbalch wouldn't it? The same websites just uses ICE and quantitative neutralisation like equation 2. $\endgroup$
    – user115162
    Oct 16, 2021 at 16:19
  • $\begingroup$ Hendersson-Hasselbach's equation is valid for a solution that does not change in time. It does not apply in the present problem. $\endgroup$
    – Maurice
    Oct 16, 2021 at 18:32
  • $\begingroup$ How so? Why doesn't it. I learned it this way in university and don't see why $\endgroup$
    – user115162
    Oct 16, 2021 at 20:47
  • $\begingroup$ H.-H. equation is just rewritten equation for the acid dissociation constant $K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}}$, which is obviously true if and only if the reaction is in equilibrium. $\endgroup$
    – Poutnik
    Oct 18, 2021 at 12:12
  • $\begingroup$ @Maurice Sir when we reach the point where there is very very less acetic acid in the container , then how will the equilibrium be there ? In next few moments of dropping more naoh $\endgroup$ Apr 25 at 21:42

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