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For the buffer region in this, why does it stop the pH from increasing?

At the start:

$$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$

Which means there will be a high amount of salt in the solution, but why is this stopping the pH from rising? I thought it would actually make it rise further as the disassociated salt it reacts with water to form $\ce{OH-}$?

What's reacting with the $\ce{OH-}$ in solution, that isn't there anyway? Seeing as, with a buffer solution, the acid is needed ($\ce{CH3COOH + OH- -> CH3COO- + H2O}$).

The acid concentration hasn't been increased?

Or is the presence of the dissasocated salt: $\ce{CH3COONa(aq) -> CH3COO- + Na+}$

Causing the acid to reform: $\ce{CH3COO- + H2O -> CH3COOH + OH-}$ <<< but there's more $\ce{OH-}$!!

Surely the $\ce{Na+}$ isn't going to react with the $\ce{OH-}$, as it will dissociate in the water.


2 Answers 2


The linked question's answers were too long for me, so I'll attempt to summarize quickly.

The equation that tells you everything you need to know is the Henderson-Hasselbalch equation:

$$\mathrm{pH} = \mathrm{p}K_{\mathrm a} - \log\frac{\ce{[A-]}}{\ce{[HA]}}$$

Here $\ce{HA}$ is a weak acid, and it's conjugate base is also weak. But we can imagine also a weak base with a weak conjugate acid. The magic happens when you've half neutralized a weak acid or base, creating a mixture of a weak acid and a weak base. The fraction on the right is fairly close to 1 (within an order of magnitude, let's say).

As you add more acid, the weak base can neutralize it. As you add more base, the weak acid can neutralize it. Assuming these amounts are reasonably small, the logarithm of the fraction on the right doesn't change by much, so the overall pH of the solution does not change by much. For example, changing the fraction on the right by a factor of 10 only affects the pH value by $\pm 1$.

Of course, if you add enough acid or base to neutralize (or close to neutralize) one of the buffer components, then the buffer stops working. This is also when the fraction gets very large or goes to zero and even the logarithm can't change that.


At the very start of the titration you have $\ce{CH3COOH}$ — or so you may think! Actually, you have an equilibrium of water with acetic acid as shown in equation $(1)$.

$$\ce{CH3COOH + H2O <<=> CH3COO- + H3O+}\tag{1}$$

This equilibrium is strongly shifted to the left since hydronium ($\mathrm pK_\mathrm a = 0$) is a very strong acid while acetic acid ($\mathrm pK_\mathrm a = 4.76$) is a rather weak one. What follows from here is basically equilibrium chemistry: as you add hydroxide ions, there is a great tendency for these to recombine with hydronium ions as in equation $(2)$ to form water. Thus, the equilibrium of equation $(1)$ is slowly shifted to the right as more and more hydroxide is added, formally consuming the acetic acid. Since equilibrium $(1)$ is still at equilibrium, the concentration of $\ce{H3O+}$ won’t change that much.

$$\ce{H3O+ + OH- <=>> 2 H2O}\tag{2}$$

Towards the end of the titration, the concentration of acetic acid ($\ce{CH3COOH}$) becomes very low, meaning there are less molecules to donate protons to water to uphold the equilibrium. Therefore, the $\mathrm{pH}$ value starts rising much more rapidly than previously — as you would expect from the titration of a strong acid with a strong base. The curve’s turning point is the equivalence point, right where the amount of hydroxide added matches the amount of acetic acid originally present.

  • $\begingroup$ Sir why concentration of H+ wouldnot change that much ? I didnt get the reason as such $\endgroup$ Commented Apr 25, 2022 at 21:27
  • $\begingroup$ And yes towards ends conc of ch3cooh is low so if eqb is not achevable what is done so that eqb is again reached or it will not reach at all again . So all the left ch3cooh will quickly dissociate to give that graph at equivalence point? $\endgroup$ Commented Apr 25, 2022 at 21:29

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