# Isoelectric point

Tags: isoelectric, point
 P: 320 How is the isoelectric point for an amino acid or peptide mathematically defined? Does it require that the solution be buffered to the pI? I raise this concern because the charge balance must be invalid here - I have seen pI defined as pI=1/2(pKa1+pKa2) for some amino acids, but then we find that at pI, [H2A+]=[A-] (the protonated form of the acid is equal in concentration to the completely deprotonated form of the acid, for some only) - so using this in the charge balance means [H+]=[OH-] unless the charge balance must be dropped for some reason. My guess is, this means that 1) the amino acid, alone in solution, can never be brought to the isoelectric point, regardless of dilution (this in fact suggests to me that the pH of the pure amino acid solution is independent of initial concentration), unless pI=7 for that acid; 2) to reach pI, we are adding other ions to the amino acid solution to change the pH, and so without knowing their identity we can't use the charge balance.
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P: 1,317
 Quote by Big-Daddy How is the isoelectric point for an amino acid or peptide mathematically defined? Does it require that the solution be buffered to the pI? I raise this concern because the charge balance must be invalid here - I have seen pI defined as pI=1/2(pKa1+pKa2) for some amino acids, but then we find that at pI, [H2A+]=[A-] (the protonated form of the acid is equal in concentration to the completely deprotonated form of the acid, for some only) - so using this in the charge balance means [H+]=[OH-] unless the charge balance must be dropped for some reason. My guess is, this means that 1) the amino acid, alone in solution, can never be brought to the isoelectric point, regardless of dilution (this in fact suggests to me that the pH of the pure amino acid solution is independent of initial concentration), unless pI=7 for that acid; 2) to reach pI, we are adding other ions to the amino acid solution to change the pH, and so without knowing their identity we can't use the charge balance.
The pI is defined as the pH at which the average charge of an amino acid is zero (i.e. it is electrically neurtral). The exact formula depends on how many ionizable groups are present in the amino acid, but for a diprotic amino acid pI=1/2(pKa1+pKa2) works. This is equivalent to saying [H2A+]=[A-].

Now, this does not imply that at the pI, [H+]=[OH-]. In fact, if you look at the Henderson-Hasselbalch equation you'll see that [H2A+]=[A-] and [H2A+]=[A-] are incompatible for most amino acids. Why is this?

Well, experimentally, if we want to get an amino acid to the point that it is electrically neutral, we would take a solution of that amino acid then adjust the pH. How do we adjust pH? By adding acid (e.g. HCl) or base (e.g. NaOH) to the solution. Adjusting the pH of the solution necessarily introduces additional counter-ions like Na+ or Cl-.
 P: 320 Ok so this is the same conclusion I thought possible earlier. But (1) do we also take it that dilution of the solution, no matter how great, will not lead to a shift in the pH significant enough to bring the solution to isoelectric point, for most amino acids (because if the pure amino acid, alone in solution, manages to reach its isoelectric point, [H+]=[OH-] is a forced condition)? (2) Is there any direct formula to apply for other amino acids or molecules, with more ionisable groups? Or do we just drop the charge balance (as with this amino acid, since we're using unknown ions to change pH) and use the equivalence of charge over all forms of the amino acid/protein, e.g. for glutamic acid (3 protons, with the form with 2 protons being the neutral one), we'd use [H3A+]=[HA-]+2[A2-], and then solve the system of equations as with any other equilibrium scenario? (3) What is the theoretical definition of the isoionic point?
P: 3,261

## Isoelectric point

If the aminoacid is neutral on the pI, it does not contribute to charge balance.
You need a buffer to get the amino acid to its isolelectric point. The buffer will take care of the charge balance.
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P: 1,897
 Quote by Big-Daddy . for glutamic acid (3 protons, with the form with 2 protons being the neutral one), we'd use [H3A+]=[HA-]+2[A2-], and then solve the system of equations as with any other equilibrium scenario?
Yes. And to work out those things is not very difficult. To work out a general equation as you ask is not very difficult, no principles not already there in your example.

In the glutamic acid case notice that when [H3A+]=[HA-], 2[A2-] is very small and can be ignored in what you are calculating, and you can just use the mean formula you mentioned. The first two pKs are 2.19 and 4.25 (the next is 9.67). The mean of the first two is 3.22 and that is what the isionic point for glu is given as in tables.

Yes there is as general formula, this is just an application of stuff you have already done a problem or two on, you could work out an impressive-looking formula with Ʃ's. It won't be a simple mean of pKs, as the last example shows.

Rather more important than general formal formulations would be to realise that the neutral form of glu for example is not one thing, it is three, and so is the -1 charged form, because of the three ionizing groups. So there are really more than three dissociation constants. Titration can be treated as if there were three, for e.g. kinetics or spectroscopy the relevant constants may not be the same - you'd need to read a section on 'macroscopic' and 'microscopic' dissociation constants. You won't need to worry much when the constants are well separated as they are for glu.
P: 320
 Quote by epenguin In the glutamic acid case notice that when [H3A+]=[HA-], 2[A2-] is very small and can be ignored in what you are calculating, and you can just use the mean formula you mentioned. The first two pKs are 2.19 and 4.25 (the next is 9.67). The mean of the first two is 3.22 and that is what the isionic point for glu is given as in tables.
Ok, good one.

 Quote by epenguin Yes there is as general formula, this is just an application of stuff you have already done a problem or two on, you could work out an impressive-looking formula with Ʃ's. It won't be a simple mean of pKs, as the last example shows.
You mean a general exact formula? I wouldn't bother. So long as I can write the equations that determine the system I'm happy.

So, in a general case, we just need to drop the charge balance (since we don't necessarily know what ions are being used to shift the pH to isoelectric point), and write a new charge equivalence, Ʃ([Amino Acid Form] * z(Amino Acid Form))=0 when summed over all amino acid forms, where z(Amino Acid Form) is the charge on that form and [Amino Acid Form] is the concentration of that form. Then I'd use this together with the mass balance and equilibrium expressions to find the isoelectric point.

Must be noted that it looks like, for any cases except the most basic one (HA hydrolyses to H2A+ and A-, which is governed exactly by pI=1/2(pKa1+pKa2)), the initial concentration of the acid in solution may still have an effect on the isoelectric point.

 Quote by epenguin Rather more important than general formal formulations would be to realise that the neutral form of glu for example is not one thing, it is three, and so is the -1 charged form, because of the three ionizing groups. So there are really more than three dissociation constants. Titration can be treated as if there were three, for e.g. kinetics or spectroscopy the relevant constants may not be the same - you'd need to read a section on 'macroscopic' and 'microscopic' dissociation constants. You won't need to worry much when the constants are well separated as they are for glu.
Good point. Do you know of any good quantitative treatments of this?
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P: 1,897
 Quote by Big-Daddy Good point. Do you know of any good quantitative treatments of this?
(Macroscopic vs. microscopic or 'intrinsic' pK's)
I would think any comprehensive book on biophysical chemistry would treat this. At least for two dissociations where the issue already presents itself. A classical case is cysteine where the pK's of the -SH and of the -NH3+ are not far separated.
If you can't find it in any other textbook it is discussed in Gutfreund "Enzymes: Physical Principles" publ Wiley 1972, pp 39-42.

But frankly there are some things it is easier to work out yourself than to wade through someone else's treatment of, T least the attempt will clarify to you what they are doing.
P: 320
 Quote by epenguin (Macroscopic vs. microscopic or 'intrinsic' pK's) I would think any comprehensive book on biophysical chemistry would treat this. At least for two dissociations where the issue already presents itself. A classical case is cysteine where the pK's of the -SH and of the -NH3+ are not far separated. If you can't find it in any other textbook it is discussed in Gutfreund "Enzymes: Physical Principles" publ Wiley 1972, pp 39-42. But frankly there are some things it is easier to work out yourself than to wade through someone else's treatment of, T least the attempt will clarify to you what they are doing.
Thanks for the recommendations.

I would think the issue presents itself wrt any amino acid. Even a simple acid HA still exists in 4 possible forms rather than 3 - H2A+, A-, undissociated HA, and HA with a carboxylate and ammonium.

What are you actually suggesting it would be good practice to work out?
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P: 1,897
 Quote by Big-Daddy Thanks for the recommendations. I would think the issue presents itself wrt any amino acid. Even a simple acid HA still exists in 4 possible forms rather than 3 - H2A+, A-, undissociated HA, and HA with a carboxylate and ammonium. What are you actually suggesting it would be good practice to work out?
Yes it does for any amino-acid. In principle. In practice since the pK's are in all cases well separated it is not such a big issue for natural amino acids, but in some other systems it could be. That is to say that e.g. pK's of gly are 2.34 and 9.60, so pKa1 is for all practical purposes the pK for the equilibrium
NH3+CH2COOH ⇔ NH3+CH2COO- + H+

with negligible NH2C... forms participating, and similarly for the second pK there are negligible -COOH forms.

Since you seem to like very general formulations you could just consider the equilibria in the quadrilateral between a di-acid HXH and -XH, XH-, and X2-. There are four K's of which three are independent. But you should find that the relation of pH and say added base in a titration can be represented by just two apparent or 'macroscopic' pK's which are functions of your three or four 'microscopic' ones. That is, titrations would be indistinguishable from if you had an equimolar mixture of two acids, one with a pK equal to one of your macroscopic constants and the other with the other.

...contd
 Sci Advisor P: 3,261 Honestly, I don't quite understand what you are discussing now. The pKa gives the equilibrium between aminoacid(ions) with different charge. Whether in a molecule of given charge one or the other group is protonated or not is a tautomeric equilibrium which is not of much interest as it equilibrates extremely rapidly and you cannot influence it, say, changing the pH.
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 Quote by DrDu Honestly, I don't quite understand what you are discussing now. The pKa gives the equilibrium between aminoacid(ions) with different charge. Whether in a molecule of given charge one or the other group is protonated or not is a tautomeric equilibrium which is not of much interest as it equilibrates extremely rapidly and you cannot influence it, say, changing the pH.
Is there an official definition of tautomerism? Most of us first meet it with keto-enol tautomerism and think of it according to this dictionary definition:
noun Chemistry .
the ability of certain organic compounds to react in isomeric structures that differ from each other in the position of a hydrogen atom and a double bond.

Other definitions I saw do not specify anything about double bonds.

Of course it is rapid equilibrium but not what I thought of as 'tautomerism'.

Then there are many perfectly observable things that depend where a proton is, not just on overall charge or protonation of the molecule and so on 'microscopic' not 'macroscopic' pK's.