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Titration Paradox!

by Bipolarity
Tags: paradox, titration
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Bipolarity
#1
Feb27-12, 09:24 PM
P: 783
I'm a little boggled by a strangle titration problem that seems to contradict what I know about titration. I hope someone can resolve this seemingly strange phenomenon.

Suppose you have 0.1M of 50mL HCl.
You begin to add 0.1M of NaOH titrant.
Assume that Na and Cl ions do not hydrolyze.

Find the pH of the final solution after
a) 49.9999 mL
b) 50.0001 mL
of NaOH have been added.

Please note that the volumes described above are exactly precise. All figures are significant.

This is not a textbook problem. I made up this problem and came up with an answer of pH = 7 for both cases! How is this possible? The only equivalence point should occur when the volumes of acid/base are exactly identical, i.e. 50mL of titrant is added. But according to my calculations, pH = 7 also when these volumes of titrant are added. I don't think it's due to calculation errors. I have checked multiple times, but could still be wrong. It just seems bizarre.

I'm willing to show my work, but first I request someone can do this and confirm. If you get a different answer, then please just say so and I'll recheck my work.

Thanks!

BiP
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Borek
#2
Feb28-12, 01:56 AM
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P: 23,396
How many decimal digits in your "7"?
Bipolarity
#3
Feb28-12, 02:11 AM
P: 783
Quote Quote by Borek View Post
How many decimal digits in your "7"?
When my calculator evaluated a pH of 7, I think it was about 9-10 decimal places, so something like 7.000000000

I am pretty sure the result was exactly 7. Not completely sure, but I think if I ran it on Maple with 50 s.f. it would come out to be 7.000000000000000000000000000000000000000000000000000000000000000000000 0

I will recheck and let you know. Have you done the problem?

Thanks by the way.

BiP

Borek
#4
Feb28-12, 02:37 AM
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P: 23,396
Titration Paradox!

No, I have not done the problem, but I know what to expect. Your answer (zillions of zeros) is wrong. Show how you got it.
asym
#5
Feb29-12, 03:42 PM
P: 21
Quote Quote by Bipolarity View Post
I'm a little boggled by a strangle titration problem that seems to contradict what I know about titration. I hope someone can resolve this seemingly strange phenomenon.

Suppose you have 0.1M of 50mL HCl.
You begin to add 0.1M of NaOH titrant.
Assume that Na and Cl ions do not hydrolyze.

Find the pH of the final solution after
a) 49.9999 mL
b) 50.0001 mL
of NaOH have been added.

Please note that the volumes described above are exactly precise. All figures are significant.

This is not a textbook problem. I made up this problem and came up with an answer of pH = 7 for both cases! How is this possible? The only equivalence point should occur when the volumes of acid/base are exactly identical, i.e. 50mL of titrant is added. But according to my calculations, pH = 7 also when these volumes of titrant are added. I don't think it's due to calculation errors. I have checked multiple times, but could still be wrong. It just seems bizarre.

I'm willing to show my work, but first I request someone can do this and confirm. If you get a different answer, then please just say so and I'll recheck my work.

Thanks!

BiP
You probably did a numerical error in the total volume, but that's not important, still you get closer to 7 than you actually should. There is more fundamental fallacy behind the calculation. You rely too much on the simplified equation pH = -log c (or pOH = -log c in the basic case). This equation does not account for the ions from dissociation of water which become relevant in very dilute solutions.

Let me simplify your problem:
What is the pH of 10-7M HCl? Is it seven? No! Or pH of 10-8M HCl, that would be even basic! And pure water, being 0M HCl would have infinite pH!

Solution?
You have to do so called charge balance (I'll do it for the simple case of HCl solution):
[H+] = [OH-] + [Cl-],
then substitute from water ionic product, and assume HCl is fully dissociated:
[H+] = Kw/[H+] + c

So you have quadratic equation for [H+]. Solve it, discard physically irrelevant root, and take -log of the remaining. You will get correct pH even for very dilute solutions.


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