Question

A diprotic weak base (B) has pKb values of 3.526 (pKb1) and 6.171 (pKb2). Calculate the fraction of the weak base in each of its three forms (B, BH , BH22 ) at pH 9.395.

Answer 1

For a diprotic weak base (B)

pKb = -logKb

Kb1 = 2.98 x 10^-4 ; Kb2 = 6.74 x 10^-7

pOH = 14 - pH = 4.605

[OH-] = 2.48 x 10^-5 M

[A] = [OH-]^2 + Kb1[OH-] + Kb1.Kb2

     = (2.48 x 10^-5)^2 + 2.98 x 10^-4 x 2.48 x 10^-5 + 6.74 x 10^-7 x 2.48 x 10^-5

     = 6.15 x 10^-10 + 7.40 x 10^-9 + 2.01 x 10^-10

     = 8.21 x 10^-9

So,

alpha[B] = Kb1Kb2/[A] = 2.01 x 10^-10/8.21 x 10^-9 = 0.0245

alpha[BH] = [OH-]Kb1/[A] = 7.40 x 10^-9/8.21 x 10^-9 = 0.901

alpha[BH2] = [OH-]^2/[A] = 6.15 x 10^-10/8.21 x 10^-9 = 0.075