Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Histidine has three ionisable groups: the α-COOH group (), the imidazole ring of the side chain (, positive when protonated) and the α-NH3+ group (). A solution has . Find the average charge carried by the imidazole group, by the α-NH3+ group and by the carboxyl group, and then the net charge of the molecule.
The isoelectric point is the pH at which the average charge is zero. For a neutral side chain, . For other amino acids, average the two pK values on either side of the neutral form. The fraction of a group that is still protonated is for a group that is positive when protonated.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. Find the fraction of each group that is still protonated. Add the charges.
Step 1. Imidazole. .
Step 2. Ammonium. .
Step 3. Carboxyl. The pH is more than 6 units above , so the group is fully ionised: .
Step 4. Net charge. .
Answer: , , and a net charge of .
Check. The pI of histidine is . The pH of 8.0 is a little above it, so a small negative charge is right. The molecule would move slowly to the anode.
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