# HOW TO PERFORM BUFFER AND ISOELECTRIC POINT CALCULATIONS by

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"HOW TO PERFORM BUFFER AND ISOELECTRIC POINT CALCULATIONS by"

```					       HOW TO PERFORM BUFFER AND ISOELECTRIC POINT CALCULATIONS
by Frank Deis

First, visualize what will happen at low and high pH. At low pH there are a lot of protons
around (H+), and things tend to be protonated. At high pH, there are very few protons around, and
functional groups are stripped of ionizeable protons. What happens between the extremes of pH is
determined by the pKa's of the groups involved. You are usually told the relevant pKa's in a problem.

It is best to think "on paper" about all of this. Try drawing the low pH, high proton form on the
far left of the page, and the high pH, low proton form on the far right. Then arrange the pKa numbers
in increasing sequence from left to right between the two forms. Now start from the left hand, fully
protonated form; as you increase the pH to the first pK, the group whose pK it is has become 50%
protonated. Draw the protonated form minus that proton. Continue with this process until you have
passed all of the pKa's and you will have drawn all possible ionization states for that compound. Now
figure the charge on each ionization form. The zero charge form is the iso-electric form, and the
isoelectric point is the average of the two pKa values immediately on both sides. All other pKa values
are irrelevant for this calculation.

On past exams, dipeptides or tripeptides have been used in problems of this sort. In this case,
you have to remember that amides are not ionizeable in aqueous solution, and that you have to identify
and ignore the pKa's of carboxyl and α-amino groups bound in peptide bonds. Another useful thing to
remember is that it is not really necessary to draw the entire compound, just the ionizeable groups.

EXAMPLE: What is the isoelectric point of Alanyl Histidine? pKa values: Ala COOH = 2.3,
α-NH 2 = 9.9, His COOH = 1.8, α-NH 2 = 9.2, R = 6.0 This problem is solved below, but please try to
solve it by yourself.

This sort of visualization of chemical change in response to shifting pH can also help in solving
buffer problems with the Henderson Hasselbalch Equation. It is useful to remember that at the pKa,
a compound tends to be a good buffer – that is, you can add acid or base without significantly
changing the pH. The iso-electric point, or pI, has the opposite property. Adding small amounts of
acid or base at the pI or any of the midpoints between two adjacent pKa values (where 100% of some
specific charge form is present) produces a large change in pH. The 100% charge forms could be
thought of as "anti-buffers." If you bought a bottle of alanyl-histidine, it would come in the zero charge
"iso-electric" form. That is one reason why buffer problems start at the pI and then add acid or base.

Adding a strong acid or base to an amino acid has a very predictable effect. A strong acid
completely protonates anything which can be protonated, whereas a strong base completely strips away
protons – the amino acid reacts quantitatively with however much base or acid is added. You don't have
to worry about it "fighting back." There is an example of a problem using the Henderson Hasselbalch
equation below.

The name of the compound, "alanyl histidine" tells us that alanine's α-NH 2 is free, and the
histidine carboxyl is also free. Alanine's carboxyl and histidine's α-NH 2 group are bound in a peptide
bond. The only ionizeable side chain is histidine's.

You fill in the intermediate forms and figure out what their charges are. Be careful – good math
can be frustrated by bad chemistry in problems like this. For example, if cys or tyr are involved, you
must know that the two forms of the side chain are zero and minus (not plus!) charged. The answer to
this example is:

pI = (6.0 + 9.9) / 2 = 7.95
BUFFER PROBLEM
To one mole of ala his at its pI are added a) 0.1 moles of HCl or b) 0.1 moles of NaOH or c)
0.5 moles of HCl or d) 1.5 moles of HCl. What pH results in each of these examples?
The Henderson Hasselbalch equation is pH = pKa + log(A/HA). The numerator, A, is a form
with one less proton than HA.

a)     pH = pK a + log (A/HA). NEVER use the pI where the pKa is called for! You must
remember that pI and pKa have opposite meanings. Always insert the pKa value toward which you are
moving. Here you are starting at the isoelectric point and adding acid so (see your charge form
drawings on the previous page) you will be moving left from the zero charge form, or toward 6.0. To
determine A and HA: you will be adding a proton to 10% of the dipeptide, and leaving the other 90%
unchanged, so A/HA should equal 9/1 or 0.9/0.1 (and NOT 10/1!!). Remember that A + HA =
original concentration of amino acid or peptide.

pH = 6.0 + log (0.9/0.1) = 6.0 + log 9 = 6.95
b)    Here you are moving to the right, deprotonating 10% of the available dipeptide, and
leaving 90% unchanged (as HA).
pH = 9.9 + log (0.1/0.9) = 9.9 – log 9 = 8.95
c)     Here you are moving farther to the left – protonating half of the peptide, and leaving the
other half unchanged.
pH = 6.0 + log (0.5/0.5) = 6.0 + log 1 = 6.0
d)       Here you move beyond 6.0. The first mole completely protonates the first ionizeable
group, taking the compound from the zero to the (+1) charge form, and then the second 1/2 mole of
acid moves it toward 1.8.
pH = 1.8 + log (0.5/0.5) = 1.8 + log 1 = 1.8

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