EDTA is a hexaprotic system with following pKa values: pKa1=0.00, pKa2=1.50, pKa

Question

EDTA is a hexaprotic system with following pKa values: pKa1=0.00, pKa2=1.50, pKa3=2.00, pKa4=2.69, pKa5=6.13, pKa6=10.37
The distribution of protonated forms of EDTA will therefore vary woth pH.....
Calculate alphaY4- at the following two pH values: 3.10 & 10.65

Sapling Learning Ma EDTA is a hexaprotic system with the following pK, values:o0C-C2 CH2-COO pkal : 0.00 pKa2 = 1.50 pka3 = 2.00 pk-269 pkt,-6,13 pK-6-1037 00C-CH CH2-COO EDTA (or Y The distribution of protonated forms of EDTA will therefore vary with pH. For equilibrium calculations involving metal complexes with EDTA, it is convenient to calculate the fraction of EDTA that is in the completely unprotonated form, Y (see the figure). This fraction is designated ay Calculate ay at the following two pH values pH= 3.10 pH= 10.65 Number Number Previous Cheok Answer Nex ex Hint

Explanation / Answer

For EDTA,

pKa = -log[Ka]

so,

Ka1 = 0.1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 2.042 x 10^-3 ; Ka5 = 7.41 x 10^-7 ; Ka6 = 4.3 x 10^-11

Now,

at pH = 3.10

pH = -log[H+]

[H+] = 8 x 10^-4 M

[A] = [H+]^6 + Ka1[H+]^5 + Ka1.Ka2[H+]^4 + Ka1.Ka2.Ka3[H+]^3 + Ka1.Ka2.Ka3.Ka4[H+]^2 + Ka1.Ka2.Ka3.Ka4.Ka5[H+] + Ka1.Ka2.Ka3.Ka4.Ka5.Ka6

     = 2.62 x 10^-19 + 3.3 x 10^-17 + 1.3 x 10^-15 + 1.62 x 10^-14 + 4.13 x 10^-14 + 3.82 x 10^-17 + 2.06 x 10^-24

     = 1.76 x 10^-14

alpha[Y4-] = Ka1.Ka2.Ka3.Ka4.Ka5.Ka6/[A]

                 = 2.06 x 10^-24/1.76 x 10^-14

                 = 1.17 x 10^-10

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and,

at pH = 10.65

pH = -log[H+]

[H+] = 2.24 x 10^-11 M

[A] = [H+]^6 + Ka1[H+]^5 + Ka1.Ka2[H+]^4 + Ka1.Ka2.Ka3[H+]^3 + Ka1.Ka2.Ka3.Ka4[H+]^2 + Ka1.Ka2.Ka3.Ka4.Ka5[H+] + Ka1.Ka2.Ka3.Ka4.Ka5.Ka6

     = 1.26 x 10^-64 + 5.64 x 10^-55 + 8 x 10^-46 + 3.55 x 10^-37 + 3.24 x 10^-29 + 1.1 x 10^-24 + 2.06 x 10^-24

     = 3.16 x 10^-24

alpha[Y4-] = Ka1.Ka2.Ka3.Ka4.Ka5.Ka6/[A]

                 = 2.06 x 10^-24/3.16 x 10^-24

                 = 0.652

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