Question

Consider a solution of 10 mM aspartic acid at its isoelectric pH. (Its -COOH group has a pka of 1.99; -NH₃ group pka of 9.90; side chain pka of 3.90)

1.) What is the concentration in mM of the major isoelectric form?

2.) What is the concentration in mM of the species with all groups unprotonated?

3.) How many total forms are present in solution, including those that are present even at very low concentration?

Please show how you got answer so I can properly understand how to do this. Thumbs up given to ones that are correct.

Answer 1

For aspartic acid

isoelectric pH = 1/2(1.99 + 3.90)

                       = 2.945

at this pH the side chain -CH2COOH group remains in protonated form

1) concentration of major isoelectric form

using Hendersen-Hasselbalck equation,

pH = pKa + log(minor form/major form)

2.945 = 3.90 + log(minor form/major form)

(minor form/major form) = 0.111

aspartic acid = 10 mM

aspartic acid in solution is major unprotonated form and minor protonated form

major form + minor form = 10 mM

0.111(major form) + major form = 10 mM

major isoelectric form = 10/1.111 = 9 mM

2) concentration of minor form at this pH = 10 - 9 = 1 mM

3) Total two forms are present in solution major unprotonated and minor protonated form