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?
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2$\begingroup$ 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. $\endgroup$– Ed VFeb 18, 2020 at 22:50
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$\begingroup$ 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. $\endgroup$– Ed VFeb 18, 2020 at 23:04
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$\begingroup$ Equivalence point and neutral point are not the same thing. I think you know the difference but are mixing up the terms. $\endgroup$– KarlFeb 18, 2020 at 23:24
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1$\begingroup$ 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. $\endgroup$– Ed VFeb 19, 2020 at 0:08
1 Answer
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.)
