Question

A diprotic weak base (B) has pKb values of 4.921 (pKb1) and 8.612 (pKb2). Calculate the fraction of the weak base in each of its three forms (B, BH , BH22 ) at pH 6.466.

Answer 1

pKa1 + pKa2 = 14

pKb1 = 4.921 ---------------> pKa2 = 9.079

pKb2 = 8.621--------------------> pKa1 = 5.388

Ka1 = 4.09 x 10^-6

Ka2 = 8.34 x 10^-10

PH = 6.466

[H+] = 3.42 x 10^-7 M

[H+]^2 = 1.17 x10^-13

Ka1 Ka2 = 3.41 x 10^-15

Ka1 [H+] = 1.40 x 10^-12

[H+]^2 + Ka1 Ka2 + Ka1 [H+]   = 1.52 x 10^-12  

B = Ka1 Ka2 / [H+]^2 + Ka1 Ka2 + Ka1 [H+] = 2.24 x 10^-3

BH+   = Ka1 [H+] / [H+]^2 + Ka1 Ka2 + Ka1 [H+]    = 0.921

BH2+ = [H+]^2 / [H+]^2 + Ka1 Ka2 + Ka1 [H+] = 0.0770