EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pKa3=2.00 , p K a4 = 2.69 pKa4=2.69 , p K a5 = 6.13 pKa5=6.13 , and p K a6 = 10.37 pKa6=10.37 . The distribution of the various protonated forms of EDTA will therefore vary with pH. For equilibrium calculations involving metal complexes with EDTA EDTA , it is convenient to calculate the fraction of EDTA EDTA that is in the completely unprotonated form, Y 4 − Y4− . This fraction is designated α Y 4 − αY4− . Calculate α Y 4 − αY4− at two pH values.

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temmydbrain temmydbrain · Answer 1 · 2020-05-11T05:59:53+02:00

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = [tex]5.0 X 10^{-4} M[/tex]

[tex]Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}[/tex]

[tex]alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6[/tex]

= [tex]1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}[/tex]

= [tex]2.02 X 10^{-13}[/tex]

When,

pH = -log[H+] = 10.15

[H+] = [tex]7.08 X 10^{-11} M[/tex]

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = [tex]7.4 X 10^{-7}[/tex] ; Ka6 = [tex]4.3 X 10^-11[/tex]

[tex]alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6[/tex]

= [tex]1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}[/tex]

= [tex]5.12 X 10^{-23}[/tex]