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I have read the other posts on the same question posted on chemistry exchange such as this one: Shape of Weak-Strong Acid-Base Titration

and here : What does the small, steep curve at the very beginning of a weak acid strong base titration curve come from?

However, I don't see why their explanation is valid. They typically explain the little dip at the start and the following plateau as a result of creating a buffer solution. When a strong base is added to a weak acid, as more $\ce{OH-}$ is added this will react with the acid molecules in solutions and form more of the conjugate base, forming a buffer solution which I do agree with.

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

However, I do not see this is useful. Isn't having the conjugate base only useful if you want to add in more $\ce{H3O+}$ but we are not! As we only add in $\ce{OH-}$ how is having buffer capacity of $\ce{CH3COO-}$ in the other direction useful (as $\ce{CH3COO-}$ and $\ce{OH}$ don't react)? I agree this is a buffer, but how does the reverse reaction help at all? This buffer doesn't seem like it would help to reduce pH changes

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  • $\begingroup$ Just add $\ce{OH-}$ to both sides of the equation you have written. $\endgroup$ – Buck Thorn Feb 28 at 12:23
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This buffer doesn't seem like it would help to reduce pH changes

You have to be clear what you want to compare it with

  1. Adding a strong base to pure water: If you start with pure water, the concentration of hydroxide will be pretty much equal to the concentration of strong base you add. The pH will drop quickly.

  2. Adding a strong base to a strong acid: If you start with strong acid, the first few drops of strong base will be neutralized. The pH does not change much because the strong acid is still in large stoichiometric excess, and the pH is a logarithmic scale.

When a strong base is added to a weak acid, as more OH− is added this will react with the acid molecules in solutions and form more of the conjugate base, forming a buffer solution which I do agree with.

In the first approximation, the added hydroxide will be neutralized by the acetic acid, and the pH will not change. However, as you do that, the weak acid/weak base equilibrium will be disturbed, and the reaction will shift to the left, consuming some hydronium and increasing the pH. The more you change the ratio of acetic acid to acetate for a - say - milliliter of added strong base, the more the pH will change. The ratio changes fast at the beginning (where there is no acetate) and near the equivalence point (where there is almost no more acetic acid), so that is when the pH changes are largest because those are the steps where we disturb the equilibrium the most.

They typically explain the little dip at the start and the following plateau as a result of creating a buffer solution.

If you compare the slope of the dip to that of scenarios 1. and 2., you will find that it is less than in 1. (strong base added to pure water) and more than in 2. (strong base added to strong acid).

This buffer doesn't seem like it would help to reduce pH changes

The way I explained it was to say the strong base reacts with the weak acid. We could have said it reacts with the hydronium ion instead. Then the shift of the weak acid/base reaction would have been the other way. In the end, we are just rationalizing the result of two equilibria, that of water autoionization (taking a back seat here because hydronium is the dominant species and hydroxide is present at very low concentration) and that of the weak acid dissociation. The shape of the titration curve is what we get experimentally, and the two equilibria explain it fully. Any explanation on top of that is just to convince ourselves.

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  • $\begingroup$ Hm ok, so I kind of understand you. It is the excess amount of acetate ions that drives equilibrium to the left and hence makes [H+] very small. But then, why can we not say that the OH reacts with the H3O? This has been confusing me a lot with your explaination $\endgroup$ – John Hon Mar 1 at 9:29

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