Question

For the titration of 50.00 mL of
a 0.1000 M solution of compound ethylenediamine with a 0.2000
M solution of compound HCl in the following table.
For each titration, calculate the pH after the addition
of 0.00, 20.00, 25.00, 37.50,

50.00, and 60.00 mL of com-
pound HCl.

Please explain all the steps.

Answer 1

let Ethylenediamine be represented as B

B + HCl --> BH+ + Cl- ----- first neutralization step

BH+ + HCl ---> BH2^2+ + Cl- ------ second neutralization step

titration with HCl

(a) 0.00 ml HCl added

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

let x amount of base is hydrolyzed

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

8 x 10^-5 = x^2/0.100

x = [OH-] = 2.83 x 10^-3 M

pOH = -log(2.83 x 10^-3) = 2.55

pH = 14 - 2.55 = 11.45

(b) 20.0 ml HCl added

initial B = 0.10 M x 50 ml = 5 mmol

added HCl = 0.20 M x 20 ml = 4 mmol

BH+ formed = 4 mmol

B remained = 1 mmol

Using Hendersen-Hasselbalck equation,

pH = pKa1 + log(B/BH+)

      = 9.9 + log(1/4)

      = 9.3

(c) 25.0 ml HCl added

initial B = 0.10 M x 50 ml = 5 mmol

added HCl = 0.20 M x 25 ml = 5 mmol

BH+ formed = 5 mmol

First equivalence point

pH = 1/2(pKa1 + pKa2)

      = 1/2(9.9 + 6.85)

      = 8.40

(d) 37.50 ml HCl added

initial B = 0.10 M x 50 ml = 5 mmol

added HCl = 0.20 M x 37.50 ml = 7.5 mmol

BH2^2+ formed = 2.5 mmol

BH+ remained = 2.5 mmol

Second half-equivalence point

Using Hendersen-Hasselbalck equation,

pH = pKa2 + log(BH+/BH2^2+)

      = 6.85 + log(2.5/2.5)

      = 6.85

(e) 50.0 ml HCl added

initial B = 0.10 M x 50 ml = 5 mmol

added HCl = 0.20 M x 50 ml = 10 mmol

BH2^2+ formed = 5 mmol

[BH2^2+] = 5 mmol/100 ml = 0.05 M

Second equivalence point

BH2^2+ + H2O <==> BH+ + H3O+

let x amount of salt is hydrolyzed

Ka1 = [BH+][H3O+]/[BH2^2+]

1.42 x 10^-7 = x^2/0.05

x = [H3O+] = 8.40 x 10^-5 M

pH = -log(8.40 x 10^-5) = 4.07

(f) 60.0 ml HCl added

excess [HCl] = [H+] = 0.20 M x 10 ml/110 ml = 0.0182 M

pH = -log(0.0182) = 1.74