Question

A diprotic weak base (B) has pKb values of 4.301 (pKb1) and 7.527 (pKb2). Calculate the fraction of the weak base in each of its three forms (B, BH , BH22 ) at pH 7.768.

Answer 1

pKa1 + pKa2 = 14

pKb1 = 4.301 ---------------> pKa2 = 9.699

pKb2 = 7.527 --------------------> pKa1 = 6.473

Ka1 = 3.365 x 10^-7

Ka2 = 2.0 x 10^-10

PH = 7.768

[H+] = 1.706 x 10^-8 M

[H+]^2 = 2.91 x 10^-16

Ka1 Ka2 = 6.73 x 10^-17

Ka1 [H+] = 5.74 x 10^-15

[H+]^2 + Ka1 Ka2 + Ka1 [H+]   = 6.10 x 10^-15

B = Ka1 Ka2 / [H+]^2 + Ka1 Ka2 + Ka1 [H+] = 0.0110

BH+   = Ka1 [H+] / [H+]^2 + Ka1 Ka2 + Ka1 [H+]    = 0.941

BH2+ = [H+]^2 / [H+]^2 + Ka1 Ka2 + Ka1 [H+] = 0.0477