Question

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.32. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.80 M B(aq) with 0.80 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

a)

initially, when there is no HCl:

B + H2O <-> BH+ + OH-

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

Kb1 = 10^-pKb1 = 10^-2.1 = 0.007943 = 7.94*10^-3

assume, in equilibrium, due to stoichiometry

[BH+] = [OH-] = x

[B] = M-x

7.94*10^-3 = x*x/(0.8-x)

solve for x

x = 0.075828

[OH-] = 0.075828

no need to consider 2nd ionization, since it is too low (Kb value is much lwoer)

so..

pOH = -log(0.075828) =   1.12

pH = 14-1.12 = 12.88

b)

after 25 mL of acid

this is a buffer...

BH2+ + HCl = BH+ + H2O + Cl-

so.. there is BH+ and BH2+

pOH = pKb2 + log(BH+/B)

BH+ formed = M1V1 = 0.8*25 = 20 mmol of BH+ conjguate

BH left = M2V2 - M1V1 = 0.8*50 - 0.8*25 = 20 B

so..

pOH = 2.1 + log(20/20)

pOH = 2.1

pH = 14-2.1= 11.90

pH = 11.90 after 25 mL (half equivalence point of the first protonated base)

for..

V = 50 mL

mmol of acid = MV = 0.8*50 = 40 mmol of strong acid

mmol of base = MV = 0.8*25 = 20 mmol

so this is FIRST EQUIVALENCE point

since 1 mol of acid = 1 mol of base

so...

pOH = 1/2*(pKb1 +pKb2)

pOH = 1/2*(2.1 +7.32) = 4.71

pH = 14-4.71 = 9.29

in FIRST equivalence point..

d)

after V = 75 mL..

mmol of acid = 75*0.80 = 60

mmol of base = 50*0.8 = 40

after reactions

mmol of conjugate = (60-40)= 20

mmol of base = 40-20 = 20

once again a buffer...

so

pOH = pKb1 + log(BH´+ /B2H+)

pOH = 7.44 + log(20/20)

pOH = 7.44

pH = 14-7.44 = 6.56