Question

Calculate the pH of a 0.228 M solution of ethylenediamine (H2NCH2CH2NH2). The pKa values for the acidic form of ethylenediamine ( H3NCH2CH2NH3 ) are 6.848 (pKa1) and 9.928 (pKa2). AND Calculate the concentration of each form of ethylenediamine in this solution at equilibrium.

Answer 1

pKb1 = pKw -pka1= 14-6.848 = 7.512

Kb1 = 10^-pKb1 = 3.076*10^-8

pKb2 = pKw - pKa2 = 14- 9.928 =4.072

Kb2 = 10^-4.072 = 8.47*10^-5

H2NCH2CH2NH2 + H2O <==> H2NCH2CH2NH3+ + OH-

	H2NCH2CH2NH2	H2NCH2CH2NH3+	OH-
inital	0.228	0	0
change	-x	+x	+x
equilibrium	0.228-x	x	x

Kb1 = x*x/0.228-x

or, 3.076*10^-8 = x^2/0.228-x

or, 0.7013*10^-8 - 3.076*10^-8x -x^2 = 0

or, x = 8.37*10^-5

[OH-] = 8.37*10^-5 M

[H2NCH2CH2NH2] = 0.228- 8.37*10^-5 M= 0.2279

[H2NCH2CH2NH3+] = 8.37*10^-5 M

H2NCH2CH2NH3+ + H2O <==> +NH3CH2CH2NH3+ + OH-

	H2NCH2CH2NH3+	+NH3CH2CH2NH3+	OH-
inital	8.37*10^-5	0	0
change	-x	+x	+x
equilibrium	8.37*10^-5 -x	x	x

Kb2 = x^2/ 8.37*10^-5 -x

8.47*10^-5 = x^2/ 8.37*10^-5 -x

or, 70.89 *10^-10 - 8.47*10^-5x -x^2 = 0

or, x = 5.189 *10^-5 M

[OH-] = 5.189 *10^-5 M

[NH3CH2CH2NH3+] = 5.189 *10^-5 M

[NH2CH2CH2NH3+] = (8.47*10^-5)- (5.189 *10^-5 M) = 3.281 *10^-5 M

Total [OH-] = 5.189 *10^-5 + 8.47*10^-5 = 13.66*10^-5 M

pOH = -log[13.66*10^-5] = 3.86

pH = 14-pOH = 10.14

[H2NCH2CH2NH2] = 0.2279 M

[NH2CH2CH2NH3+] = 3.281 *10^-5 M

[NH3CH2CH2NH3+] = 5.189 *10^-5 M