Biochemistry: pH of L-glutamic acid

AI Thread Summary
The discussion centers on calculating the pH of a solution of L-glutamic acid, where the concentration of net negatively charged species is five times greater than that of the isoelectric structure. Participants clarify the concept of isoelectric point (pI) and its relevance to the problem, noting that at pI (pH 3.22), the net charge is zero. The calculation involves determining the concentration of negative charges and adjusting for the specified ratio, leading to a calculated pH of 2.52. Further exploration of the relationship between pH and the distribution of charged species is encouraged, emphasizing the importance of understanding the underlying chemical equilibrium. The conversation highlights the complexity of glutamic acid's behavior in solution and the need for careful consideration of pH in relation to charge states.
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Homework Statement



Calculate the pH of a solution of L-glutamic acid in which the ionic species with a net charge of -1 is 5 times greater than that of the isoelectric structure. (Refer to pKa Table at end of problem set).


Homework Equations



pKa for glutamic acid:

-COOH = 2.19
-NH3+ = 9.67
-R group = 4.25
pI = 3.22

The Attempt at a Solution



My understand of pI is that it is the pH of a particular molecule when the molecule has no net charge. However, I guess I am not understanding the question. What does "5 times greater than the isoelectric structure" mean?
 
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I guess what they mean is "concentration is 5 times larger". Otherwise it doesn't make sense to me.
 
It is slightly obscure. You can always write "I think the question means..." and then answer that.

The isoelectric structure is pretty clear if there is only one acidic and one basic group, as in glycine where it is -OOC-CH2-NH3+ .

It's true that HOO-CH2-NH2 is also an isoelectric structure - the one with 0 net charge. But you can work out that there will be very little of it at the isoelectric point. (And that it wouldn't affect issues like this even if there were a lot of it, which would happen if the two pKs were close to each other.)

Now glutamic acid is more complicated and you can work out that there are in a solution a total of 8 different species with charges in various places or not.

I think "the isoelectric structure" means the structure with one - and one +. That is not really one structure in this case, but I would take the words to mean what you've got at the IEP which is at pH 3.22. At that pH net - charged species = net + charged ones (and you could say the average net charge on the molecules is 0). That is the definition of pI (it is certainly not "pH of a particular molecule" - pH is not a property that a particular molecule can have). At that pH you should be able to argue that most of the 8 structures are in negligible concentrations.

Now we're being asked to calculate at what pH is the concentration of net- charge molecules 5X that of net 0 charged ones.

I would use drastic approximation in the first place, and when we see what you do with that we might think about refinements.
 
Okay. So it's been about 5 years since I had general chemistry, so please forgive me if this is wrong.

After reading your explanation (thanks so much), I decided to the determine the concentration of the negative charges (so that I could multiply it by 5) and then determine the pH. So here's what I did:

pH = -log[H+]
10-3.22 = [H+]
0.00060256 = [H+]

And then:

0.00060256 * 5 = 0.003012798

So:

pH = - log(0.003012798)
pH = 2.52

What do you think?
 
Well it is also a long time since I did this sort of calculation except here.
But you should also be installing your own alarm systems which I still have. :smile:
In this case ask yourself on which side of the IEP should the pH be if there are more net negative-charged molecules than neutral ones? In fact which side of the R-group pK should it be?

Also I think it would help to see your way if you wrote out equations for the system. I will write out one, otherwise it is hard to think about. Maybe we can reason about this system representing it

H3Y+\stackrel{}{\leftrightarrow}H2Y0\stackrel{}{\leftrightarrow}HY-\stackrel{}{\leftrightarrow}Y2-

Now helpful to you will be to write a simple equation that states the question.

You may then begin to see the problem as not so very different from ones you have done before.
 
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and in fact really very simple despite all the distracting stuff gong on.
 
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