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EDTA is a hexaprotic system with the pKa values: pKa1=0.00, pKa2=1.50, pKa3=2.00, pKa4=2.69, pKa5=6.13, and pKa6=10.37....

EDTA is a hexaprotic system with the pKa values: pKa1=0.00, pKa2=1.50, pKa3=2.00, pKa4=2.69, pKa5=6.13, and 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, it is convenient to calculate the fraction of EDTA that is in the completely unprotonated form, Y4−. This fraction is designated αY4−.

Calculate αY4− at two pH values; ph=3.20 and ph=10.20

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Answer #1

We have relation, phpz6Y6KM.pngY 4- =K 1 K 2 K 3 K 4 K 5 K 6 / { [H+] 6 + [H+] 5 K 1 + [H+] 4 K 1 K 2 +  [H+] 3 K 1 K 2 K 3 +  [H+] 2 K 1 K 2 K 3 K 4 + [H+]  K 1 K 2 K 3 K 4 K 5 + K 1 K 2 K 3 K 4 K 5 K 6

To calculate phpz6Y6KM.pngY 4- we need to calculate [H+] from given pH values and K values from given pKa values.

We have relation , pH = - log [H+]

phpLnG4oK.png [H+] = 10 -pH

phpLnG4oK.png [H+] = 10 - 3.20 = 6.310 phpCrn69h.png 10 -04 M

We have relation, pKa = - log Ka

phpLnG4oK.png Ka = 10 -pKa

By using this relation, we can calculate K values as shown below.

pKa Ka = 10 - pKa
pKa1 = 0.00 K 1 = 10 - 0.00 = 1
pKa2 = 1.50 K 2 = 10 -1.50 = 0.03162
pKa3 = 2.00 K 3= 10 -2.00 =0.01
pKa4 = 2.69 K 4 = 10 -2.69 =0.002042
pKa5 = 6.13 K 5 = 10 -6.13 = 7.413 php0Laqsl.png 10 -07
pKa6 = 10.37 K 6 = 10 -10.37= 4.266 phpP4FuyV.png 10 -11

K 1 K 2 K 3 K 4 K 5 K 6 = 1 ( 0.03162) (0.01) (0.002042) (7.413 phpvFECpM.png 10 -07) (4.266 phpLp9nZr.png 10 -11)

K 1 K 2 K 3 K 4 K 5 K 6 = 2.042 phpxsI3G4.png 10 -23

{ [H+] 6 + [H+] 5K 1 + [H+] 4 ​​​​​​​K 1 K 2 +  [H+] 3 K 1 K 2 K 3 +  [H+] 2 K 1 K 2 K 3 K 4 + [H+]  K 1 K 2 K 3 K 4 K 5 + K 1 K 2 K 3 K 4 K 5 K 6

= ( 6.310 phpCrn69h.png 10 -04 ) 6 + (6.310 phpCrn69h.png 10 -04 )5 1 + ( 6.310 phpCrn69h.png 10 -04 )4 1 ( 0.03162) + ( 6.310 phpCrn69h.png 10 -04 )31 ( 0.03162) (0.01) + ( 6.310 phpCrn69h.png 10 -04)2 1 ( 0.03162) (0.01) (0.002042) + ( 6.310 phpCrn69h.png 10 -04 ) 1 ( 0.03162) (0.01) (0.002042) (7.413 phpnc1S72.png 10 -07) + 1 ( 0.03162) (0.01) (0.002042) (7.413 phpY0ArnJ.png 10 -07) (4.266 phpM2avnS.png 10 -11)

= (6.312 php4XBgZV.png 10 -20 ) + (1.000 php4XBgZV.png 10 -16 ) + ( 5.012 phpnixfSg.png 10 -15 ) + ( 7.944 phpB6ADVt.png 10 -14) + ( 2.571 phpeZIEiY.png 10 -13) + ( 3.020 phpp33TuC.png​​​​​​​ 10 -16) + 2.042 phpouy615.png 10 -23

=3.420 phpeZIEiY.png 10 -13

Place calculated values in the formula of phpz6Y6KM.pngY 4- .

\thereforephpz6Y6KM.pngY 4- = 2.042 phpouy615.png 10 -23 / 3.420 phpeZIEiY.png 10 -13

phpz6Y6KM.pngY 4- = 5.970 phpeZIEiY.png 10 -11

ANSWER : phpz6Y6KM.pngY 4- at pH = 3.20 is 5.970 phpeZIEiY.png 10 -11 .

Calculation of phpz6Y6KM.pngY 4- at pH = 10.20

We have relation , pH = - log [H+]

phpLnG4oK.png [H+] = 10 -pH

phpLnG4oK.png [H+] = 10 -10.20 = 6.310 phpCrn69h.png 10 -11 M

K 1 K 2 K 3 K 4 K 5 K 6 = 2.042 phpxsI3G4.png 10 -23

[H+] 6 + [H+] 5 ​​​​​​​K 1 + [H+] 4​​​​​​​K 1 K 2 +  [H+] 3 K 1 K 2 K 3 +  [H+] 2 K 1 K 2 K 3 K 4 + [H+]  K 1 K 2 K 3 K 4 K 5 + K 1 K 2 K 3 K 4 K 5 K 6

= ( 6.310 phpWJGN2Z.png 10 -11  ) 6 + (6.310 phpvbzCgV.png 10 -11  )5 1 + ( 6.310 phpCxun4C.png 10 -11  )4 1 ( 0.03162) + ( 6.310  phpnbqNqk.png 10 -11  )31 ( 0.03162) (0.01) + ( 6.310 phpjfbsg5.png 10 -11  )2 1 ( 0.03162) (0.01) (0.002042) + (6.310 php1Ophc3.png 10 -11 ) 1 ( 0.03162) (0.01) (0.002042) (7.413 php0t71EL.png 10 -07) + 1 ( 0.03162) (0.01) (0.002042) (7.413 phpqnrPa5.png 10 -07) (4.266 php0v12ly.png 10 -11)

= (6.312 phpHNqOwJ.png 10 -62 ) + ( 1.000 phpAwex1q.png 10 -51 ) + ( 5.013 phpvZUHoP.png 10 -43 ) + (7.944 phpyJfAvC.png 10 -35 ) + ( 2.571 phpOKI9ox.png 10 -27) + ( 3.020 php1XVwWW.png 10 -23) + 2.042 php5vDEES.png 10 -23

= 5.062 phpbijd4X.png 10 -23  

Place calculated values in the formula of phpz6Y6KM.pngY 4-

php3Xw7TQ.pngphpPcaXLS.pngY 4- = 2.042 phpbFbkLb.png 10 -23 / 5.062 phpUzuTMf.png 10 -23

phpAs9ea7.pngY 4- = 0.4040

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EDTA is a hexaprotic system with the pKa values: pKa1=0.00, pKa2=1.50, pKa3=2.00, pKa4=2.69, pKa5=6.13, and pKa6=10.37....
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