Question

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.65. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.65 M B(aq) with 0.65 M HCl(aq).

Answer 1

(a) When no HCl is added

Kb1 = 7.94 x 10^-3 = x^2/0.65

x = [OH-] = 0.072 M

pOH = -log[OH-]= 1.14

pH = 14 - pOH = 12.86

(b) When 10 ml 0.65 M HCl added

[BH+] = 0.65 x 10/60 = 0.11 M

[B] = (0.65 x 50 - 0.65 x 10)/60 = 0.43 M

pH = pKa1 + log[(base)/(acid)]

pKa1 = 14 - pKb1 = 11.9

pH = 11.9 + log(0.43/0.11)

     = 12.49

(c) when 25 ml of 0.65 M HCl added

First half equivalence point

[BH+] = [B]

pH = 11.9

(d) Whe 50 ml og 0.65 M HCl added

First equivalence point

pH = 1/2(pKa1 + pKa2)

     = 1/2(11.9 + 6.35)

     = 9.125

(e) when 75 ml of 0.65 M HCl added

second half equivalence point

pH = 6.35

(f) When 100 ml of 0.65 M HCl added

[BH2+] = 0.65 x 50/150 = 0.22 M

Ka2 = 2.24 x 10^-8 = x^2/0.22

x = [H+] = 7.02 x 10^-5 M

pH = -log[H+] = 4.15

(g) When 110 ml of 0.65 M HCl added

[H+] = 0.65 x 10/160 = 0.04 M

pH = -log[H+] = 1.39