Question

A diprotic weak base (B) has pKb values of 3.737 (pKb1) and 6.540 (pKb2). Calculate the fraction of the weak base in each of its three forms (B, BH , BH22 ) at pH 8.952.

Answer 1

if pH = 8.952

pOH = 14-8.952 = 5.048

now...

pOH = pKb1 + log(BH/B)

5.048 = 3.737 + log(BH/B)

(BH/B) = 10^(5.048 - 3.737) = 20.4644

(BH/B) = 20.4644

now,

pOH = pKb2 + log(BH2/BH)

5.048 = 6.540 + log(BH2/BH)

(BH2/BH) = 10^(5.048 -6.540) = 0.03221

(BH2/BH) = 0.03221

now compare:

(BH/B) = 20.4644; (BH2/BH) = 0.03221

assume BH = 1

(BH/B) = 20.4644

(1/B) = 20.4644

B =1/(20.4644) = 0.04887

now,

(BH2/BH) = 0.03221

BH2/1 = 0.03221

BH2 = 0.03221

total

Btotal = B + BH + BH2 = 0.04887 + 1 + 0.03221 = 1.08108

fractions:

B --> 0.04887 /1.08108 = 0.04520

BH --> 1/1.08108 = 0.925

BH2 --> 0.03221 /1.08108 = 0.02978