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EDTA (ethylenediaminetetraacetic acid), a widely u
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Answer #1

pH = -log[H+] = 7

[H+] = 1 x 10^-7 M

pKa = -log[Ka]

So for EDTA,

Ka1 = 6.92 x 10^-3 ; Ka2 = 7.58 x 10^-4 ; Ka3 = 5.37 x 10^-7 ; Ka4 = 1.12 x 10^-11

[A] = [H+]^4 + Ka1[H+]^3 + Ka1Ka2[H+]^2 + Ka1Ka2Ka3[H+] + Ka1Ka2Ka3Ka4

     = 1 x 10^-28 + 6.92 x 10^-24 + 5.24 x 10^-20 + 2.81 x 10^-19 + 3.15 x 10^-23

     = 3.33 x 10^-19

alpha0 = [H+]^4/[A] = 1 x 10^-28/3.33 x 10^-19 = 3.00 x 10^-10

alpha1 = Ka1[H+]^3/[A] = 6.92 x 10^-24/3.33 x 10^-19 = 2.08 x 10^-5

alpha2 = Ka1Ka2[H+]^2/[A] = 5.24 x 10^-20/3.33 x 10^-19 = 0.157

alpha3 = Ka1Ka2Ka3[H+]/[A] = 2.81 x 10^-19/3.33 x 10^-19 = 0.844

alpha4 = Ka1Ka2Ka3Ka4/[A] = 3.15 x 10^-23/3.33 x 10^-19 = 9.46 x 10^-5

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