Ethylene diamine, H₂NCH₂CH₂NH₂, can interact with water in two steps, forming OH⁻ in each step (Appendix I). If you have a 0.15 M aqueous solution of the amine, calculate the concentration of [H₃NCH₂CH₂NH₃]²⁺ and OH⁻.

Interpretation:

Concentration of [H3NCH2CH2NH3]2+ and OH- has to be calculated form 0.15 M aqueous solution of the amine.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka: it is a acid constant for equilibrium reactions.

HA + H2O⇌H3O++ A-Ka= [H3O+][A-][HA]

Kb: It is a base constant for equilibrium reaction.

BOH + H2O⇌B++ OH-Ka= [B+][OH-][BOH]

The ionization reactions of Ethylenediamine are as follows.

First ionization:

H2NCH2CH2NH2(aq) + H2O(l)⇌NH2CH2CH2NH3+(aq) + OH-(aq)Equilibrium expression:Kb1= [NH2CH2CH2NH3+][OH-]H2NCH2CH2NH2]Kb1= 8.5×10-5Second ionization:NH2CH2CH2NH3+(aq) + H2O(l)⇌[NH3CH2CH2NH3]2+(aq) + OH-(aq)Equilibrium expression:Kb2= [NH3CH2CH2NH32+][OH-]H2NCH2CH2NH3+]Kb2= 3.57×10-3

Compare the Kb1 and Kb2,Kb1  is greater than Kb2 . Therefore most of the OH- is produced from the Kb1 

Let’s calculate the concentration of OH- from Kb1 .

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

        NH2CH2CH2NH2(aq) + H2O (l)⇌ NH2CH2CH2NH3+(aq) + OH−(aq)I                      0.15                    --                       --                                 --C                       -x                      --                     +x                                 +xE                 (0.15-x)                  --                       x                                   x

Kb1= [NH2CH2CH2NH3+][OH-]H2NCH2CH2NH2]Kb1= 8.5×10-5

8.5×10-5 = (x)(x)0.15- x8.5×10-5 = (x)20.15- x(0.15-x) approximately equals to 0.158.5 ×10-5 = (x)20.15          x2=  (0.15)(8.5×10-5)          x  =  (0