Question

Calculate the pH of a 0.382 M solution of ethylenediamine (H2NCH2CH2NH2). The pKa values for the acidic form of ethylenediamine ( H3NCH2CH2NH3 ) are 6.848 (pKa1) and 9.928 (pKa2).

pH=

Calculate the concentration of each form of ethylenediamine in this solution at equilibrium.

H2NCH2CH2NH2 =

H2NCH2CH2NH3+ =

H3+NCH2CH2NH3+ =

Answer 1

pKb1 = 14 - pKa2 = 14 - 9.928 = 4.072

Kb1 = 8.47 x 10^-5

pKb2 = 14 - 6.848 = 7.152

Kb2 = 7.5 x 10^-8

H2NCH2CH2NH2 = B

H2NCH2CH2NH2 + H2O --------------------->H2NCH2CH2NH3+ + OH-

0.382 -x x x ------------> at equilibrium

Kb1 = x^2 / 0.382-x

8.47 x 10^-5 = x^2 / 0.382-x

x^2 + 8.47 x 10^-5x -3.23 x 10^-5 = 0

x = 5.64 x 10^-3

[H2NCH2CH2NH2] = 0.382-x

[H2NCH2CH2NH2] = 0.376 M

[H2NCH2CH2NH3+] = 5.64 x 10^-3 M

pOH = -log [OH-]

pOH = -log (5.64x 10^-3)

pOH = 2.25

pH + POH= 14

pH = 11.75

[+H3NCH2CH2NH3+] = pKb2 = 7.5 x 10^-8 M

pH= 11.75

H2NCH2CH2NH2 =0.376 M

H2NCH2CH2NH3+ =5.64 x 10^-3 M

H3+NCH2CH2NH3+ = 7.5 x 10^-8 M