Question

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.38. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.70 M B(aq) with 0.70 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

pKb1 = 2.10 and pKb2 = 7.38

Kb1 = 7.94*10^-3 and Kb2 = 3.2*10^-8

(a). First dissociation of B is:

                  B    +        H2O                  BH+       +              OH-
Initial       0.70                                            0                            0
Change     -x                                            +x                          +x
Final     0.70 - x                                         x                            x

Kb1 = 7.94*10^-3 = x*x / (0.70 - x)

x = 0.070

[OH-] = [BH+] = 0.070 M

Second dissociation is given as:

                 BH+          +           H2O                BH2^2+          +       OH-
Initial        0.070                                                      0                          0.070
Change      -x                                                         +x                           +x
Final       0.070 - x                                                   x                        0.070 +x

Kb2 = 3.2*10^-8 = x*(0.070 + x) / (0.070 - x)

x = 3.2*10^-8

Hence, [OH-] = 0.070 M

pOH = - log (0.070) = 1.15

pH = 14 - 1.15

= 12.85

(b). after addition of 25.0 mL of HCL.

At first equivalence point:

moles of HCl = Moles of B

0.70 * V = 0.70 * 50

V = 50 mL

Volume of HCl at first equivalence point = 50 mL

So at 25 mL i.e., at half way to equivalence point:

pH = pKa1

= 14 - pKb1

= 14 - 2.1

= 11.9

(c). after addition of 50.0 mL of HCl

At first equivalence point:

pH = (pKa1 + pKa2) / 2

= (11.9 + 6.62) / 2

= 9.26

(d). after addition of 75.0 mL of HCl

Volume of HCl at second equivalence point = 2 * 50 = 100 mL

So at half way to second equivalence point:

pH = pKa2

= 6.62