Second year university biochemistry - pH and Buffers

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SUMMARY

The discussion centers on the effects of dilution on the pH and buffering capacity of a 0.1 M citrate buffer with an initial pH of 4.0. After diluting the buffer 20-fold and 50-fold, the observed pH values increased to 4.10 and 4.14, respectively, contradicting the common belief that dilution does not affect pH. The participants concluded that the changes in pH are likely due to ionic strength effects rather than mere fluctuations in measurement. The Henderson-Hasselbalch equation was referenced to explain the relationship between pH, pKa, and the concentrations of the acid and its conjugate base.

PREREQUISITES
  • Understanding of the Henderson-Hasselbalch equation
  • Knowledge of buffer systems and their components
  • Familiarity with concepts of ionic strength in solutions
  • Basic principles of pH measurement and calibration
NEXT STEPS
  • Research the effects of ionic strength on pH in buffer solutions
  • Study the Henderson-Hasselbalch equation in detail
  • Explore the concept of buffering capacity and its definitions
  • Learn about the properties and applications of citrate buffers in biochemistry
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Biochemistry students, laboratory technicians, and educators seeking to deepen their understanding of buffer systems and pH behavior in diluted solutions.

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Homework Statement


This question is for a lab write up. We prepared a 0.1 M citrate buffer with pH 4.0, we then diluted our buffer with deionized water 20 fold and then 50 fold. The subsequent changes in pH are as follows,
20 fold pH = 4.10
50 fold pH = 4.14
This confuses me because I have always been told that any water added will NOT have any affect. Is it just a fluctuation of the pH meter or am I missing something? The lab write up asks us if there is a change and says to use the Henderson Hasselbalch equation to explain our answer, no calculations necessary. It then asks if the original buffer would have the same "buffering capacity" as the 20 fold diluted one. I feel like they would but now I am confused do to the change in pH after dilution. I am really bad with chemistry so any help would be much appreciated!

Homework Equations


Henderson Hasselbalch Equation

pH = pKa + log10 ( [A-] / [HA] )

The Attempt at a Solution



Pure water is neutral with a pH of 7. The buffering capacity SHOULD be the same in my opinion, but what about the change in pH?
 
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See detailed discussion of this problem at Henderson-Hasselbalch equation and buffer capacity pages.

But you may start with just a simple thought experiment. When you dilute buffer so that its concentration is 10-20M, will it be still able to keep solution pH constant, or will the solution pH be just that of pure water, 7? And if so, is it possible that dilution doesn't matter?
 
Thank you, those pages were a big help. I struggle with this stuff no matter how much I practice. Ill be glad when I am finished with this class.
 
Note: buffer capacity has many definitions, while the one I use seems to me to be the best one, others prefer less strict definitions.
 
epenguin said:
I think it is what Borek means with his example. But I doubt such a problem would be given to someone struggling to understand buffers. Nor is any biochemist likely to use such a dilute buffer on purpose if the pH matters!

I was just pointing to the obvious flaw in the "pH of buffer doesn't depend on its concentration" statement. This is one of these things that many people take for granted, even if they have enough knowledge to understand that it must be wrong. They were just taught that buffer pH doesn't depend on the concentration and they repeat the mantra.

But you are most likely correct that in this case pH changes are due to the ionic strength change.
 
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The pH of the 0.1M, and the 20X diluted i.e. 5mM buffer are quoted.

It is not quite clear to me whether the '50X diluted' is the original 0.1M diluted 50X to 2mM, or whether the 20X was further diluted 50X to 10^-4M.

In the second case the concentration of buffer ions has become comparable with the proton concentration. When that happens you can no longer make approximations you otherwise make, e.g. that [A^-] + 2[A^{2-}] + 3[A^{3-}] = Na^+. You get an ugly quadratic equation for [H^+]. I have not time to deal with that. I think it is what Borek means with his example. But I doubt such a problem would be given to someone struggling to understand buffers. Nor is any biochemist likely to use such a dilute buffer on purpose if the pH matters!

So it is more likely it is 0.1M, 5mM and 2mM. Not totally overwhelmingly greater than [H^+] but quite a lot greater - without calculating I'm guessing the effect mentioned is not the explanation.

I'd guess more likely is an ionic strength effect. the pK's of citric acid are 3.15, 4.77 and 6.4 so at your pH's the solution is mostly A^- with some A^{2-} which makes the ionic strength quite strong - (2-) buffer ions are more charged than many you use. The more concentrated, the more these charges in solution, favouring proton association, i.e. lowering the pK. (at least it's going the right direction :smile:). The effects in that case should be quite stronger at say pH5 and still more at pH 6.5
 
Looks like you have deleted first version of your post, now I am quoting you... before you have posted.
 
Borek said:
Looks like you have deleted first version of your post, now I am quoting you... before you have posted.

Right - I needed to correct a blooper and when you have tex stuff everything is very slow :frown: at this site and it wouldn't accept corrections so I had to delete and re-post.

I think I agree with you that the mantra may even make it more difficult because people may wonder how can that really be, I think I did. Just needs to add, look we are usually talking about the buffer being much more concentrated than the H+ or the OH- , are we saying the same thing?
 
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