Why does a strong base completely neutralize a weak acid? [closed]








**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
 
 






---








**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/128059/edit).




Closed 3 years ago.








[Improve this question](/posts/128059/edit) 



For the longest time, weak acid-strong base titration curves completely baffled me. I understood that the equivalence point of a strong acid-strong base titration is reached when equal numbers of moles of base titrant are added as there are acid analyte, but I didn't realize why the same is true for a weak acid-strong base titration; I incorrectly assumed that a strong base *will not* completely neutralize a weak acid when a stoichiometrically equal amount of base is added to acid - I thought that since a weak acid only partially deprotonates in solution, it is also only partially neutralized by strong base. Now that I realize that is an inherently incorrect assumption, my question is why does a strong base fully neutralize a weak acid if only a small percentage of the acid is deprotonated in solution? Is it an effect of Le Chatelier's principle?




The simple (high-school chemistry level) answer is that you are correct and it is indeed Le Chatelier's Principle. Consider ethanoic acid, $\ce{CH3COOH}$, which dissociates according to the equation $\ce{CH3COOH + H2O-> CH3COOH + H3O+}$. (If you haven't previously encountered hydronium ions, $\ce{H3O+}$, it's essentially just a somewhat more accurate way to represent $\ce{H+}$ ions in solution.)


As a strong base like $\ce{NaOH}$ is added, it will react with the $\ce{H3O+}$ ions in solution, so equilibrium for the dissociation of the weak acid will move to the right, forming more $\ce{H3O+}$ ions, which react with the strong base. This continues until the weak acid has entirely dissociated.
(Ed V's comment sets out a more quantitative explanation; depending on what level you're currently studying chemistry at, it may be more appropriate.)


 2