The Titration Paradox: How Can pH = 7 at Two Different Volumes of Titrant?

In summary, the problem is that you rely on an equation which does not account for the ions from dissociation of water.
  • #1
Bipolarity
776
2
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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  • #2
How many decimal digits in your "7"?
 
  • #3
Borek said:
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 7.0000000000000000000000000000000000000000000000000000000000000000000000

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

Thanks by the way.

BiP
 
  • #4
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.
 
  • #5
Bipolarity said:
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.
 

1. What is a titration paradox?

A titration paradox is a phenomenon in chemistry where the volume of a titrant added during a titration does not match the calculated volume required to reach the endpoint. This can lead to unexpected results and can be caused by various factors such as human error, chemical reactions occurring during the titration, or incorrect assumptions about the reaction taking place.

2. How does a titration paradox affect experimental results?

A titration paradox can significantly impact experimental results as it can lead to inaccurate calculations and conclusions. It can also make it difficult to determine the true endpoint of the titration, making it challenging to determine the exact concentration of a solution.

3. What are some common causes of a titration paradox?

The most common causes of a titration paradox include human error, such as incorrect measurements or misreading the volume on the burette, and chemical reactions occurring during the titration. Other factors such as impurities in the titrant or sample, or incorrect assumptions about the reaction, can also contribute to a titration paradox.

4. How can a titration paradox be avoided?

To avoid a titration paradox, it is essential to carefully plan and execute the titration experiment. This includes using accurate and precise measurements, ensuring proper mixing of solutions, and making sure the reaction conditions are consistent. It is also crucial to repeat the titration multiple times and compare results to identify any inconsistencies.

5. Can a titration paradox be corrected?

In some cases, a titration paradox can be corrected by identifying and addressing the cause of the discrepancy. This may involve repeating the titration with more accurate measurements, adjusting the reaction conditions, or using a different method or technique. However, in some cases, a titration paradox cannot be corrected, and the results may need to be interpreted with caution or repeated using a different approach.

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