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Let's say we do the titration of a weak acid (acetic acid in this case) and a strong base (sodium hydroxide) represented by the following equation: $$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$

My question is that, since $\ce{CH3COOH}$ is partially dissociated, does $\ce{NaOH}$ react only with the partially dissociated part or the complete acetic acid solution?

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    $\begingroup$ Related: chemistry.stackexchange.com/questions/60407/… $\endgroup$
    – Ivan Neretin
    Commented Oct 13, 2017 at 4:30
  • $\begingroup$ Your formula is nonsense for a titration. Your base is already dissolved, and so stays sodium acetate. OH- reacts with undissociated acetic acid, nothing else. Write up reactions with reactants in the form you actually use them and you will never make such (stupid) mistakes. Insult goes to your teacher, not you, btw. $\endgroup$
    – Karl
    Commented Oct 13, 2017 at 8:11

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Acetic acid and acetate are connected by the dissociation equilibrium $(1)$:

$$\ce{CH3COOH + H2O <<=> H3O+ + CH3COO-}\tag{1}$$

So you basically have two acidic particles that can react with hydroxide: hydronium or acetic acid. Hydronium will actually react much faster on a molecular level even though its concentration is significantly lower as it is much more mobile in solution. However, any hydronium consumed is immediately reformed by acetic acid molecules reestablishing the equilibrium.

Therefore, on a very minute level hydroxide reacts with neither acetic acid nor acetate but with hydronium which is produced in the dissociation of acetic acid. In practice, that still means that hydroxide will be added until $n(\ce{CH3COOH})+n(\ce{CH3COO-})=n(\ce{OH-})$. From this point onwards, there will be more hydroxide than acidic particles causing the $\mathrm{pH}$ value to rise significantly, the indicator to change colour and the student to realise that they reached (and passed) the equivalence point.

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