EDTA is a hexaprotic system with the following pK_a values: pK_a1 = 0.00 pK_a2 =

Question

EDTA is a hexaprotic system with the following pK_a values: pK_a1 = 0.00 pK_a2 = 1.50 pK_a3 = 2.00 pK_a4 = 2.69 pK_a5 = 6.13 pK_a6 = 10.37 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^4- (see the figure). This fraction is designated alphaY^4-. Calculate alpha Y^4- at the following two pH values: pH = 3.30 alpha_Y^-4 = Number pH = 10.15 alpha_Y^4- = Number

Explanation / Answer

When,

pH = -log[H+] = 3.30

[H+] = 5.0 x 10^-4 M

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 x 10^-7 ; Ka6 = 4.3 x 10^-11

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

= 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

= 2.02 x 10^-13

When,

pH = -log[H+] = 10.15

[H+] = 7.08 x 10^-11 M

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 x 10^-7 ; Ka6 = 4.3 x 10^-11

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

= 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

= 5.12 x 10^-23

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