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

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?

• Very short version: a weak acid does a very good job of protecting its acidic protons against removal by water molecules. But hydroxide ions are vastly better at stealing those protons. Look around this site: there are numerous relevant detailed answers or discussions.
– Ed V
Feb 18 '20 at 22:50
• A little more. Write the dissociation equilibrium for acetic acid and note that its dissociation constant is much less than one. Next write the auto-ionization equilibrium of water, with hydrogen and hydroxide ions on the left and water on the right. So the k for that is the reciprocal of water’s auto-ionization constant. Finally, add the two equilibria, cancel common species and then you see that the equilibrium constant is acetic acid’s k divided by water’s autoionization constant. Thus it is 100 trillion times larger and much greater than zero.
– Ed V
Feb 18 '20 at 23:04
• Equivalence point and neutral point are not the same thing. I think you know the difference but are mixing up the terms.
– Karl
Feb 18 '20 at 23:24
• See the neutralization equilibrium of acetic acid and hydroxide ion, as per my second comment: chemistry.stackexchange.com/a/32781/79678 . But note that the k value is $10^{-4.75}$/$10^{-14}$ = $10^{9.25}$, not what the answer mistakenly gave. Still, a value above a billion, so the weak acid almost entirely gets converted to conjugate base.
– Ed V
Feb 19 '20 at 0:08

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.)