Download The pH/pI/pKa problems are straightforward if you

Note on pH/pI/pKa Problems for Amino Acids
The pH/pI/pKa problems are straightforward if you apply these rules:
pI (electrophoresis problems), like 27.2
If the pH is less than the pI, the amino acid will move toward the negative electrode.
If the pH is greater than the pI, the amino acid will move toward the positive electrode.
If the pH equals the pI, the amino acid will not move
pH/pKa problems, like 27.24
If the pH is less than the pKa of a group, the predominant form will be the conjugate acid.
If the pH is greater than the pKa of a group, the predominant form will be the conjugate
base.
If the pH is equal to the pKa of a group, the conjugate acid and conjugate base will be present
in equal amounts.
Potential for confusion in pI problems
Note that in problem 27.2, the pI for histidine is 7.64, and the experiment is conducted at pH
7.00. In this case the pH is less than the pI, so the histidine molecules move toward the negative
electrode. However, note that the side-chain pKa is 6.10, which means that most of the side-chain
groups will be unprotonated at pH 7.0, and since the amino group is protonated and the carboxyl
group is deprotonated at pH 7.0, most (about 90%) of the molecules of histidine will have no
charge.
So why does the histidine migrate to the negative electrode? At pH 7.00, a little less than 10%
of the histidine molecules will have a protonated side chain, giving them a positive charge. As this
small fraction of postitively charged amino acids moves toward the negative electrode, the acidbase equilibrium for the rest of the sample will shift to produce more of the positive form of the
histidine, which will then also move toward the negative electrode. This sequence of events will
continue until all of the histidine moves in this direction. See the figure on the next page
1
O
+
H 3N
H
N
O
+
H 3N
O−
N
H
MAJOR form of His at pH=7.0
(net neutral)
N
O−
N H
+
pKa = 6.1
Minor form of His at pH=7.0
(net + charge)
Moves to the negative
electrode and drags the
equilibrium to the right.
2