See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Concept: The isoelectric point (pI) of an amino acid is the pH at which the molecule carries no net electrical charge. For a simple amino acid with one carboxylic acid group and one amino group, the pI is calculated as the average of the two relevant pKa values flanking the zwitterionic form. Step 1: Identify the given constants. - pKa of the carboxylic acid group (–CO2H) = 2.2 - pKb of the ammonium group (–NH3+) = 4.4 Step 2: Convert pKb to pKa for the ammonium group. pKa(–NH3+) = 14 – pKb = 14 – 4.4 = 9.6 Step 3: Calculate the isoelectric point. For a simple (non-ionizable side chain) amino acid, pI = (pKa1 + pKa2) / 2 where pKa1 is the pKa of the carboxyl group and pKa2 is the pKa of the ammonium group. pI = (2.2 + 9.6) / 2 = 11.8 / 2 = 5.9 Step 4: Evaluate other options. - (a) 3.3: This would be the average of two carboxyl-type pKa values, not applicable here. - (c) 9.6: This is just pKa of the ammonium group alone, not the pI. - (d) 11.8: This is the sum of the two pKa values, not the average. Therefore, the correct answer is B.