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.
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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