"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