Question

Ethylenediamine, H2NCH2CH2NH2, can interact with water in two steps, forming OH– in each step.

Part a. Write equations for each of the ionization steps for this polyprotic base. Kb1 = 8.5×10−5; Kb2 = 2.7×10−8.

Part b. Calculate the concentrations of [H3NCH2CH2NH3]2+ and OH– in a 0.15 M solution of the amine.

Answer 1

step-1

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

initial 0.15                       0             0

change   -x                        +x            +x

equil   0.15-x                      x            x

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

(8.5*10^-2) = x^2/(0.15-x)

x = 0.07815

[OH-] = X = 0.07815 M

[H2NCH2CH2NH3+] = X = 0.07815 M

STEP-2

          H2NCH2CH2NH3^+ + H2O ----> +^H3NCH2CH2NH3+ + OH-

initial   0.07815 M                                     0.07815

change       -x                            +x              +x

equil     0.07815-x                         x             0.07815

kb = [+^H3NCH2CH2NH3+][OH-]/[H2NCH2CH2NH3^+]

(2.7*10^-8) = x*0.07815/(0.07815-x)

x = [+^H3NCH2CH2NH3+] = 2.7*10^-8 M

[OH-] = 0.07815 M