Question

Calculate the pH of a 0.320 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.

1. [ H₂NCH₂CH₂NH₂ ] =

2. [ H₂NCH₂CH₂NH₃^{+ ] =}

3. [ H₃⁺NCH₂CH₂NH₃⁺ ] =

please be detailed thanks.

Answer 1

H2NCH2CH2NH2+H2O---à H2NCH2CH2CH2NH3+ +OH-

A+H2O --à A+ +OH-

Let A= H2NCH2CH2NH2 and A+ = H2NCH2CH2CH2NH3+

Pka1= 6.848, Ka1= 1.42*10^-7

Ka1= [A+] [OH-]/ [A]

At equilibrium [A+] =x and [OH-] =x and A =0.32-x

1.42*10^-7 =x²/(0.32-x)

This on solving using excel gives x= 1.418*10^-7

[OH-]= 1.418*10^-7

pOH= -log(1.418*10^-7) =6.848

pH= 14-6.848=7.152

[A+] =1.418*10-7M

[A] = 0.32- 1,418*10-7=0.3197

A+ + H2O -à A++ +OH-

Where A++ = [H3+NCH2CH2NH3+]

Ka2= [A+] [OH-]/ [A+]

1.18*10^-10=x²/(1.418*10-7-x)

Solving for x gives x= 4.04*10-9

[OH-]= 4.04*10-9

pOH= 8.39

pH= 14-8.39= 5.61 A++= 4.04*10^-9