What is the pH of a 0.01M solution of arginine that has a net charge of -.65?
5 years later and almost to the day
To find the pH of a solution of arginine, we need to consider the dissociation of the amino acid in water. Arginine is a zwitterion, which means it has both a positive and a negative charge.
The net charge of -0.65 suggests that the arginine is predominantly in its deprotonated form (negative charge). This means that the carboxyl group (-COOH) is deprotonated (charged -1) and the amino group (-NH2) is protonated (neutral). The pH of the solution can be calculated based on the pKa values of these groups.
Arginine has three pKa values: pKa1 = 2.18, pKa2 = 9.04, and pKa3 = 12.48. The pKa1 corresponds to the deprotonation of the carboxyl group, pKa2 to the deprotonation of the α-amino group, and pKa3 to the deprotonation of the guanidino group.
Since the net charge is closest to zero when the carboxyl group is negatively charged and the α-amino group is neutral, we can assume that these two groups are mostly in their respective deprotonated and protonated forms.
To calculate the pH, we can consider the equilibrium between the deprotonated and protonated forms of the carboxyl group:
pH = pKa1 + log([A-]/[HA])
Where [A-] is the concentration of the deprotonated form (carboxylate) and [HA] is the concentration of the protonated form (carboxylic acid). For a 0.01 M solution of arginine, we can assume that the concentration of the deprotonated form is equal to the concentration of arginine itself, and the concentration of the protonated form is negligible.
Therefore, the pH of the solution can be approximated as:
pH ≈ pKa1 + log(0.01/0)
Since the concentration of the protonated form ([HA]) is negligible, log(0) is undefined. Thus, the pH of the 0.01 M arginine solution with a net charge of -0.65 cannot be accurately determined based on the given information.