Question

Calculate the ph of a 0.377 m solution of ethylenediamine ( h2nch2ch2nh2 ). the pa values for the acidic form of ethylenediamine ( h 3nch2ch2nh 3 ) are 6.848 ( pa1 ) and 9.928 ( pa2 ).

Answer 1

The final answer is 11.71

The concentration of a given solution (H2NCH2CH2NH2)  is 0.314 M

pa1 = 6.848

a1 = 1.42 X $10^{-7}$

Pa2 = 9.928

a2 = 1.18 X $10^{-10}$

The equation is given as;

NH2CH2CH2NH2 + H2O ---> NH2CH2CH2NH3+ + OH-

Initial 0.314 0 0

Change -x +x +x

Equilibrium 0.314 -x x x

b1 = Kw / Ka2 = 10^-14 / 1.18 X 10^-10 = 8.47 X 10^-5 = [NH2CH2CH2NH3+][OH-] / [NH2CH2CH2NH2]

8.47 X 10^-5 = x^2 / 0.314 -x

8.47 X 10^-5 X 0.314 - 8.47 X 10^-5x = x^2

x^2 - 2.66 X 10^-5 + 8.47 X 10^-5x = 0

on solving

x = 0.0051 M = [NH2CH2CH2NH3+] = [OH-]

[NH2CH2CH2NH2] = 0.314 - 0.0051 = 0.3089 M

For the second dissociation,

NH2CH2CH2NH3+ + H2O ---> +NH3CH2CH2NH3+ + OH-

Initial 0.0051 0 0.0051

Change -x +x +x

Equilibrium 0.0051 -x x 0.0051 +x

b2 = Kw / Ka1 = 10^-14 / 1.42 X 10^-7 = 7.04 X 10^-8 = [+NH3CH2CH2NH3+] [OH-] / [+NH3CH2CH2NH2]

7.04 X 10^-8 = x (0.0051+x) / (0.0051-x)

We may ignore x in the denominator as b2 is very low

7.04 X 10^-8 = x (0.0051+x) / (0.0051)

3.59 X 10^-10 = 0.0051x + x^2

x = 7.039 X 10^-8 M

[OH-] = 0.0051 + x = 0.0051 approx

[+NH3CH2CH2NH3+] = x = 7.039 X 10^-8 M

NH2CH2CH2NH3+ = 0.0051 +x = 0.0051 M approx

pOH = --log [OH-] = -log 0.0051 = 2.29

pH = 14 - pOH = 11.71

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