At the equivalence point, why do we take volume of strong base used for titration of strong acid the same as the volume used for titration of weak acid? Shouldn't there be a difference? For example, in case of titration of CH3COOH against NaOH, even though as the small amount of H+ ions(initially) react with OH- ions and take the dissociation of CH3COOH in the forward direction, shouldn't the common ion effect of CH3COO- ion from CH3COOHNa suppress the dissociation of CH3COOH equally?


A weak acid exists in an equilibrium and once you start adding alkali $\ce H^+$ is removed disrupting the equilibrium meaning more acid molecules will dissociate - the equilibrium position shifts right. $$\ce {H_xA <=> xH^+ + A^{x-}}$$ So by the time you've added all the alkali you will have neutralised every acid molecule - even if it wasn't dissociated in the beginning. So as long as you add the same number of moles of acid and alkali no matter if they are strong or weak, at the equivalence point the number of moles of acid and alkali neutralised will be equal.

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  • $\begingroup$ What causes this to happen? Shouldn't the common ion effect of CH3COO- ion from CH3COONa suppress the dissociation of CH3COOH equally? $\endgroup$ – Arjun Dec 20 '18 at 14:00
  • $\begingroup$ When the $\ce {NaOH}$ reacts with the acid it is removing $\ce {H_3O^+}$ from the mixture which due to le chatelier principle drives the rate of the forward reaction replacing the $\ce {H_3O^+}$ which has been used up, meaning the moles of $\ce {H_xA}$decrease. this process continues until there is no $\ce {H_xA}$ left. Also if no more $\ce {H_xA}$dissocates the moles of $\ce {H^+}$ in the mixture would be zero stopping the reverse reaction meaning the $\ce A^{x-}$ acts as if it isn't there. The $\ce {H^+}$ is what is important here not $\ce A^{x-}$ $\endgroup$ – H.Linkhorn Dec 20 '18 at 14:05

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