Question

Calculate the pH of a 0.214 M solution of ethylenediamine (H2NCH2CH2NH2). The pK, values for the acidic form of ethylenediamine (H3NCH-CH2NHs') are 6.848 (pKa) and 9.928 (pka). Number Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. Number H,NCH,CH,NH, Number H,NCH,CH,NH, l- MI Number

Answer 1

solution:

pKa1 = 6.848 , pKa2 = 9.928

pKb1 = 14 - pKa2 = 14 - 9.928 = 4.072 ( p^Ka₁ + p^Ka₂ = 14)

similarly,

pKb2 = 14 - 6.848 = 7.152

Kb1 = 8.47 x 10^-5

Kb2 = 7.05 x 10^-8

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

      0.214 0                   0

0.214 - x                                                                         x                     x

Kb1 = x² / 0.214 - x

8.47 x 10^-5 = x² / 0.214 - x

x = 4.25 x 10^-3

[OH-] = 4.25 x 10^-3

pOH = -log (4.25 x 10^-3)

pOH = 2.37

pH = 11.628

[H2NCH2CH2NH3+] = 4.25 x 10^-3 M

[H2NCH2CH2NH2] = 0.214 - 0.00425 = 0.20975 M