Question

A solution of .05 M glutamic acid is at its isoelectric pH. Calculate:
i) the approximate concentration of its major isoelectric form
ii) the approximate concentration of the completely protonated form
iii) How may forms are present in solution at even miniscule concentrations?

Answer 1

Different forms are not isomers of each other(d-andL forms)but they are conjugate acid/base.Glutamic aid has three acidic protons the alpha amino group carboxylic acid and the side cahin carboxylic acid.Each of these canexist in acid (protonated form) and base (deprotonated from) , they are theoretically 6 different formsDifferent forms are not isomers of each other(d-andL forms)but they are conjugate acid/base.Glutamic aid has three acidic protons the alpha amino group carboxylic acid and the side cahin carboxylic acid.Each of these canexist in acid (protonated form) and base (deprotonated from) , they are theoretically 6 different forms.

P^ka1 is corboxylic group=2.19 P^ka2   is ammonium ion 9.67 P^ka3 side chain group 4.25 isoelectic point =3.21

find pI using 2.19 and 4.25
pI = (pKa1+pKa2)/2 = 3.22

P^H = -log[H+]

3.22 =- log[H+]

[H+] = 10^-3.22 = 6.02*10^-4M

0.05M -6.02*10^-4M =0.0493M

using hender son equation
P^H = P^Ka+log[A-]/[HA]
3.22 = 2.19+log[A-]/[HA]

log[A-]/[HA] = 1.03

[A-]/[HA] = 10.71

[HA} =0.0853