Question

Calculate the pH of a 0.368 M solution of ethylenediamine (H2NCH2CH2NH2). The pKa values for the acidic form of ethylenediamine ( H3NCH2CH2NH3 ) are 6.848 (pKa1) and 9.928 (pKa2). Calculate the concentration of each form of ethylenediamine in this solution at equilibrium.

Answer 1

pKa1 = 6.848

- log Ka1 = 6.848

​ Ka1 = 10^-6.848 = 1.42 x 10^-7

pKa2 = 9.928

- log Ka1 = 9.928

​ Ka1 = 10^-9.928 = 1.18 x 10^-10

Now,

H₂NCH₂CH₂NH₂ + H₂O    H₂NCH₂CH₂NH₃⁺ + OH^-

Ka1 = [H₂NCH₂CH₂NH₃⁺] [OH^-] / [H₂NCH₂CH₂NH₂]

1.42 x 10^-7 = (x) (x) / (0.368 - x)

1.42 x 10^-7 = (x) (x) / (0.368) ; Since Ka1 is very small, x term in the denominator can be neglected.

x² = 5.23 x 10^-8  

x = 2.29 x 10^-4

So, [H₂NCH₂CH₂NH₃⁺] = [OH^-] = 2.29 x 10^-4

pOH = - log [OH^-] = - log (2.29 x 10^-4) = 3.64

pH = 14 - pOH = 14 - 3.64 = 10.36

Again,

H₂NCH₂CH₂NH₃⁺ + H₂O    H₃NCH₂CH₂NH₃⁺²   + OH^-

[H₂NCH₂CH₂NH₃⁺] = [OH^-] = 2.29 x 10^-4

Ka2 = [H₃NCH₂CH₂NH₃⁺²] [OH^-] / H₂NCH₂CH₂NH₃⁺]

1.18 x 10^-10 = (x) (2.29 x 10^-4 + x) / { (2.29 x 10^-4) - x }

1.42 x 10^-7 = (x) (2.29 x 10^-4 + x) / (2.29 x 10^-4)

(Since Ka2 is very small, x term in the denominator can be neglected.)

x² + (2.29 x 10^-4)x = 3.25 x 10^-11  

x² + (2.29 x 10^-4)x - 3.25 x 10^-11 = 0

Solving the quadratic equation,

x = 1.42 x 10^-7 and x = -0.00023

Negative value of x is neglected and hence

x = 1.42 x 10^-7

So, [H₃NCH₂CH₂NH₃⁺²] = [OH^-] = x = 1.42 x 10^-7

pOH = - log  [OH^-] = - log (1.42 x 10^-7) = 6.85

pH = 14 - pOH = 14 - 6.85 = 7.15