At the beginning of a Weak Acid/Strong Base titration curve there is a sharp decrease in $[\ce{H+}]$ and a sharp increase in pH before the ratio between the weak acid and it's conjugate base becomes low enough that we have a buffer solution. enter image description here enter image description here

What's happening at that point?

My interpretation is that, given the balanced equation:

$$\ce{ CH3COOH~(aq) + H2O(~l) <=> CH3COO- ~(aq) + H3O+~(aq) }$$

The addition of $\ce{OH-}$ would shift equilibrium to the left by Le Chatelier's principle making the newly generated $\ce{CH3COO-}$ react with $\ce{H3O+}$ to make more $\ce{CH3COOH}$, thereby increasing pH. After that, the $[\ce{H3O+}]$ is so low that the addition of the anion is not enough to keep shifting the equilibrium position to the left, so the remaining weak acid resists the changes in pH with its dissociation.

Does this make any sense?


Unfortunately this is not correct.

$$\ce{CH3COOH~(aq) + H2O(~l) <=> CH3COO- ~(aq) + H3O+~(aq)}$$

If you have a pure acetic acid solution the equation you state governs the pH of the solution.

If you are now adding a strong base like $\ce{{}^{-}OH}$, you are removing the hydronium ion from the right hand side of the equation. According to Le Châtelier, the equilibrium adapts and shifts more to the right, i.e. more acetic acid dissociates. As you remove $\ce{H3O+}$ the pH increases.
Another way of looking at this is, that you remove acetic acid from the left hand side (similar to what you stated). $$\ce{H3CCOOH + {}^{-}OH <=> H3CCOO- + H2O}$$ According to Le Châtelier, the equilibrium (in the first equation) has to shift to the left, to regenerate more acetic acid, removing once again $\ce{H3O+}$ and the pH increases.

In any case you choose to look at, you increase the concentration of acetate in the solution and another equilibrium comes into play. $$\ce{H3CCOO- + H2O <=> H3CCOOH + {}^{-}OH}$$ This equilibrium is essentially responsible for raising the pH.

  • $\begingroup$ Isn't the strong base react both with the acid and hydronium ions? $\endgroup$
    – ado sar
    May 22 '20 at 18:58
  • $\begingroup$ @adosar Yes, but that won't make a difference in terms of equilibrium. $\endgroup$ May 22 '20 at 20:07
  • $\begingroup$ So a reaction of a strong base with the acid isn't always $100\%$? $\endgroup$
    – ado sar
    May 23 '20 at 12:35
  • $\begingroup$ @adosar I don't understand what you are asking. It is an equilibrium. I don't understand what you mean with 100%. Please ask a new question for clarification. $\endgroup$ May 23 '20 at 17:34

These systems are more complex than they might seem at first. I more or less agree with Martin's answer above but there are some subtleties at play. There are two acids present which can react with added hydroxide ions: the CH3COOH molecules and the hydronium ions. Initially, as OH- is added, it reacts with and removes the very small amount of hydronium ions since this is by far the faster reaction (H3O+ + OH- proton transfer is probably the fastest reaction that occurs in any aqueous solution). After that, it starts deprotonating CH3COOH molecules and builds up the CH3COO- concentration. As this acetate ion concentration increases it drives the equilibrium to the left, thereby suppressing any further dissociation of CH3COOH to produce hydronium ions. Similarly, as the CH3COO- concentration builds up, some of it reacts with water to produce OH- ions, and this is in general what causes the pH to rise.


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