Question

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.63. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.60 M B(aq) with 0.60 M HCl(aq).

Answer 1

before addition of any HCl

B + H2O <-------------------> BH+   + OH-

0.60                                    0           0    ------------------> initial

0.60-x                                x              x --------------------> equilibrium

Kb1 = [BH+][OH-] / [B]

7.94 x 10^-3 = x^2 / 0.60-x

x^2 + 7.94 x 10^-3 x - 4.76x 10^-3 = 0

x = 0.065

x = [OH-] = 0.065 M

pOH = -log[OH-]

         = -log(0.065)

       = 1.187

pH +pOH = 14

pH = 12.81

25 ml HCl added

base millimoles = 2 x acid millimoles

Base millimoles = 50 x 0.6 = 30

acid millimoles = 2 x 12.5 x 0.6 = 15

B + H+ ------------------> BH+

30    15                           0

15     0                             15

pOH = pKb1 + log [BH+] / [B]

pOH = 2.10 + log (15/15)

pOH = 2.10

pH + pOH = 14

pH = 11.9

note : you did not give any volume of HCl .you just mentioned following points . i do not know waht are the points. so i solved 0 ml HCl and 12.5 ml of HCl . by this process you can apply for any volume of HCl