3
$\begingroup$

I am trying to experimentally determine the $\mathrm{p}K_\mathrm{a}$ of acetic acid by titrating sodium acetate with $\ce{HCl}$. The $\mathrm{pH}$ curve starts in the alkaline region (since sodium acetate is alkaline) and the equivalence point is acidic since at the equivalence point $\ce{CH3COOH}$ is present.

$\ce{CH3COO- + HCl -> CH3COOH + Cl-}$

So, from the half equivalence point : $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$ (of acetic acid) or $\mathrm{pOH} = \mathrm{p}K_\mathrm{b}$ (of sodium ethanoate)

The literature value of the $\mathrm{p}K_\mathrm{a}$ of acetic acid at $\pu{25^{\circ} C}$ is 4.75. This means that my experimental half equivalence point should be around $\mathrm{pH}$ 4.75 which will NEVER be the case since during the titration till the equiv. point the curve lies on the basic region.

What am I doing wrong? Is there any chance the Henderson/Hasselbach equation won't apply ?

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ For better site experience, you can find useful How can I format math expressions or chem. equations on Chemistry SE. ( Not to be applied to titles ) $\endgroup$
    – Poutnik
    Commented Oct 15, 2021 at 10:48
  • 1
    $\begingroup$ The equivalence point of titration of $\ce{CH3COONa}$ by $\ce{HCl}$ definitely is not in basic region, as it will end with $\ce{CH3COOH}$ and $\ce{NaCl}$. $\endgroup$
    – Poutnik
    Commented Oct 15, 2021 at 12:29
  • $\begingroup$ Hey, welcome to Chemistry SE. Why would you do it this way? Why not titrate acetic acid with a strong base directly? $\endgroup$
    – Michael Lautman
    Commented Oct 15, 2021 at 15:50
  • 1
    $\begingroup$ Yes. The half-equivalence point is the point where the solution contains the same amount of $\ce{CH3COOH}$ and of $\ce{CH3COO^-}$ ions. So it should occur theoretically at pH = $\mathrm{p}K_\mathrm{a} = 4.74$. But it will be a little bit different if the concentration is not $1 M$, because of the activity coefficients being different from $0$. $\endgroup$
    – Maurice
    Commented Oct 15, 2021 at 19:34

1 Answer 1

Reset to default
1
$\begingroup$

When you titrate a base like ammonia (pKa of conjugate acid around 9.5) with hydrochloric acid, you can see that it buffers in the basic region and the equivalence point is in the acidic region:

enter image description here

When you try the same with acetate (pKa of conjugate acid around 4.8), if buffers in the acidic region, and the equivalence point is in the acidic region:

enter image description here (Sorry, bad copy and past job. The start of the curve should be in the basic region, with an initial sharp drop to the buffer region)

Here is a better graph, courtesy of EdV:

enter image description here

As you will find, it is difficult to distinguish the buffering region from the region past the equivalence point. Other than that, it is possible to do this, and there is nothing wrong with that.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I ran the titration curve for 0.1 M acetate ion solution versus 0.1 M HCl, in 0.01 mL acid increments. The titration curve is here: chemistry.meta.stackexchange.com/a/4757/79678. If you have any use for it, feel free to use it and annotate, etc., as you please. I did not bother to run the ammonia versus HCl, but it is trivial to do so. I upvoted, by the way. I will leave the figure up for a couple of days. Ignore the M-B stuff: it was for a physics SE comment. $\endgroup$
    – Ed V
    Commented Nov 15, 2021 at 0:39
  • $\begingroup$ @EdV I added it to my answer, imagining how you went to the lab to do the actual titration - I guess it is a simulation. $\endgroup$
    – Karsten ♦
    Commented Nov 15, 2021 at 2:09
  • $\begingroup$ Yeah, pretty easy to do the calculations. Nice that the pH at the half-equivalence point, at 12.5 mL added HCl, was the expected 4.74. $\endgroup$
    – Ed V
    Commented Nov 15, 2021 at 2:14
  • $\begingroup$ Weird that the only upvote is from me. One thing I have noticed is that pH problems, including titration curves, tend not to get much love here. Oh well. $\endgroup$
    – Ed V
    Commented Nov 15, 2021 at 23:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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