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Calculate the pH of a 0.393 M solution of ethylenediamine (H2NCH2CH2NH2). The pKa values for the...

Calculate the pH of a 0.393 M solution of ethylenediamine (H2NCH2CH2NH2). The pKa values for the acidic form of ethylenediamine ( H3NCH2CH2NH3 ) are 6.848 (pKa1) and 9.928 (pKa2). Calculate the concentration of each form of ethylenediamine in this solution at equilibrium.

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Answer #1

pKb1 = 14 - pKa2 = 14 - 9.928 = 4.072

Kb1 = 8.47 x 10^-5

pKb2 = 14 - 6.848 = 7.152

Kb2 = 7.5 x 10^-8

H2NCH2CH2NH2 = B

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

0.393 -x x x ------------> at equilibrium

Kb1 = x^2 / 0.393-x

8.47 x 10^-5 = x^2 / 0.393-x

x = 5.77 x 10^-3

[H2NCH2CH2NH2] = 0.393-x

[H2NCH2CH2NH2] = 0.387 M

[H2NCH2CH2NH3+] = 5.77 x 10^-3 M

pOH = -log [OH-]

pOH = -log (5.77 x 10^-3)

pOH = 2.24

pH + POH= 14

pH = 11.76

[+H3NCH2CH2NH3+] = pKb2 = 7.5 x 10^-8 M

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