Question

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.37. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.55 M B(aq) with 0.55 M HCl(aq).

(a) before addition of any HCl

(b) after addition of 25.0 mL of HCl

(c) after addition of 50.0 mL of HCl

(d) after addition of 75.0 mL of HCl

(e) after addition of 100.0 mL of HCl

Answer 1

dibasic base = pKb1 = 2.10

pKb2 = 7.37.

millimoles of base = 50 x 0.55 = 27.5

(a) before addition of any HCl

pOH = 1/2 [pKb1- logC]

pOH = 1/2 [pKb1 - log 0.55]

          = 1/2 (2.10 - log 0.5)

          = 1.18

pH + pOH = 14

pH = 14 - pOH

      = 12.82

pH = 12.82

(b) after addition of 25.0 mL of HCl

millimoles of HCl = 25 x 0.55 = 13.75

it is half equivalence point. so

pOH = pKb1

pOH = 2.10

pH = 11.9

(c) after addition of 50.0 mL of HCl

it is equivalence point . at equivalence point

pOH = (pKb1 + pKb2 )/ 2

         = 2.10 + 7.37 / 2

         = 4.74

pH = 9.26
(d) after addition of 75.0 mL of HCl

moles of HCl = 75 x 0.55 = 41.25

it is second half equivalence point . so

pOH = pKb2

pOH = 7.37

pH = 6.63

(e) after addition of 100.0 mL of HCl

millimoles of HCl = 100 x 0.55= 55

it is second equivalence point.here it is only BH2+ remains.so its concentration

BH2+ millimoles = 55

BH2+ concentration = 55 / total volume

                                 = 55 / (100 + 50)

                                 = 0.37M

BH2+ + H2O ------------------> BH+   + H3O+

0.37                                             0             0

0.37 - x                                         x              x

Ka2 = x^2 / (0.37 -x)

2.34 x 10^-7 = x^2 / (0.37 -x)

x = 2.94 x 10^-4

[H3O+] = x = 2.94 x 10^-4 M

pH = -log[H3O+] = -log (2.94 x 10^-4)

       = 3.53

pH = 3.53