2
$\begingroup$

I understand that a buffer solution is being made, but I don't understand why the titration curve is different from a strong acid strong base one (apart from the steeper pH change). I understand the half-equivalence point and the rest of the titration curve. If Le Châtelier's applies here, please explain with it, as it makes more intuitive sense for me rather than the Henderson-Hasselbalch equation.

I included this animation if it makes it easier to explain, thanks!

http://www.chembio.uoguelph.ca/educmat/chm19104/chemtoons/chemtoons9.htm

|improve this question
$\endgroup$
2
$\begingroup$

This is part of estsblishing the buffer associated with the weak acid. To establish a buffer you need relatively large amounts of acid and conjugate base, and a weak acid can't produce the necessary conjugate base by itself. You need to push it along with a little strong base to get into the buffering condition.

Say your solution is 0.1 molar acetic acid, with $K_\mathrm{a}=1.8×10^{-5}$. Before titration the pH is given by the Henderson-Hasselbach equation. The logarithm of the above $K_\mathrm{a}$ value is about -4.74, therefore:

$\mathrm{pH}=4.74-\log_{10}\left(\frac{[\ce{HC2H3O2}]}{[\ce{C2H3O2-}]}\right)$

Now the amount of $\ce{C2H3O2-}$ in the solution is just what you get from dissociation of the acid, but acetic acid is only a weak acid and the actual amount of dissociation is small, in this case only about 1.3% of the acid you dissolved. Merely 0.01 molar strong base reacting with the acid is enough to make a lot more acetate ion than just the acid dissociation alone, so the ratio in the H-H equation makes a big change and the $\mathrm{pH}$ follows suit. Only when you have large amounts of both acid and conjugate base does the $\mathrm{pH}$ become buffered against slight additions of base or acid.

|improve this answer
$\endgroup$
  • $\begingroup$ So if the Ka value is smaller, less of the weak acid is dissociated, and the pH change at the beginning is bigger because of a larger logarithmic change in the concentration of acetate ion? (Assuming the molarity of such a solution is also 0.1 molar). Or is this canceled out by the higher initial pH? $\endgroup$ – Jesper2k Mar 30 '18 at 15:00
  • $\begingroup$ You get both effects. As long as you're above $10^{-11}$ in 0.1 M aqueous solution (to keep water autodissociation at bay), a factor of 10 decrease in Ka raises the initial pH by one half unit and raises the increment into the buffering range by another half unit. $\endgroup$ – Oscar Lanzi Mar 30 '18 at 17:44
  • $\begingroup$ If the low amount (high OH- concentration) of NaOH is enough to significantly raise pH, why does the increase slow down then? I understand that a buffer is being established, but how does it differ from the very initial solution of titrant being added? Please use Le Châteliers principle if it applies here, I find it much more intuitive than the HH equation (which I don't really understand). $\endgroup$ – Jesper2k Apr 1 '18 at 19:46
  • $\begingroup$ I can't fit this well to Le Chatelier's Principle. The principle is qualitative and we are dealing with a quantitative effect. $\endgroup$ – Oscar Lanzi Apr 1 '18 at 20:55
0
$\begingroup$

The steepness occurs when the neutralization point is reached. And it occurs early on in the case of weak acid-strong base titration because the acid is weak, and less base is required to neutralize it.

Another point to be noted is that weak acids do not dissociate to large extents as in the case of strong acids. In a solution of a strong acid, almost all of the $\ce{H+}$ ions are already liberated and very few remain connected with the anion (say $\ce{A-}$) as $\ce{HA}$. On the other side, in the case of weak acids, only a few $\ce{H+}$ ions are released and most of them remain as $\ce{HB}$ (let $\ce{B-}$ be the conjugate base of the weak acid)

So lets see what happens as you add the base to these solutions:

Case 1: Strong acid

Once the base is added into the solution, the $\ce{OH-}$ ions combine with $\ce{H+}$ ions to form water. The $\ce{OH-}$ ions continuously deplete the $\ce{H+}$ pool in the solution, increasing $\ce{pH}$.

case 2: Weak acid

Here, it's similar to the previous case, but there's a catch: There still remains some undissociated acid molecules in the solution. Once the $\ce{OH-}$ combine with $\ce{H+}$ and remove some of these ions from the solution, the equilibrium of dissociation of weak acid is shifted to the right. The undissociated acid molecules now dissociate and replenish some of the $\ce{H+}$ ions lost due to base addition. This makes the change in $\ce{pH}$ quite less as compared to the strong acid case.

|improve this answer
$\endgroup$
  • 1
    $\begingroup$ Thanks for the quick answer :) but i meant the steepness at the very beginning, when just a few milliliters of titrant is added (before the half-equivalence point) $\endgroup$ – Jesper2k Mar 30 '18 at 14:22
  • $\begingroup$ @Jesper2k I don't understand what you mean. I see the steepness at the equivalence point here: s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/… $\endgroup$ – Pritt Balagopal Mar 30 '18 at 14:27
  • $\begingroup$ The small steepness at the "Initial" point in the image $\endgroup$ – Jesper2k Mar 30 '18 at 14:35
  • 1
    $\begingroup$ Ahh I see, looks like Oscar Lanzi has pointed out that in his answer already :) $\endgroup$ – Pritt Balagopal Mar 30 '18 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.