Question

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

Answer 1

We have relation, _Y 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 _Y 4- we need to calculate [H⁺] from given pH values and K values from given pKa values.

We have relation , pH = - log [H⁺]

[H⁺] = 10 ^-pH

[H⁺] = 10 ^{- 3.20} = 6.310 10 ^-04 M

We have relation, pKa = - log Ka

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 10 -07
pKa6 = 10.37	K 6 = 10 -10.37= 4.266 10 -11

K ₁ K ₂ K ₃ K ₄ K ₅ K ₆ = 1 ( 0.03162) (0.01) (0.002042) (7.413 10 ^-07) (4.266 10 ^-11)

K ₁ K ₂ K ₃ K ₄ K ₅ K ₆ = 2.042 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 10 ^-04 ) ⁶ + (6.310 10 ^-04 )⁵ 1 + ( 6.310 10 ^-04 )⁴ 1 ( 0.03162) + ( 6.310 10 ^-04 )³1 ( 0.03162) (0.01) + ( 6.310 10 ^-04)² 1 ( 0.03162) (0.01) (0.002042) + ( 6.310 10 ^-04 ) 1 ( 0.03162) (0.01) (0.002042) (7.413 10 ^-07) + 1 ( 0.03162) (0.01) (0.002042) (7.413 10 ^-07) (4.266 10 ^-11)

= (6.312 10 ^-20 ) + (1.000 10 ^-16 ) + ( 5.012 10 ^-15 ) + ( 7.944 10 ^-14) + ( 2.571 10 ^-13) + ( 3.020 ​​​​​​​ 10 ^-16) + 2.042 10 ^-23

=3.420 10 ^-13

Place calculated values in the formula of _Y 4- .

$\therefore$_Y 4- = 2.042 10 ^-23 / 3.420 10 ^-13

_Y 4- = 5.970 10 ^-11

ANSWER : _Y 4- at pH = 3.20 is 5.970 10 ^-11 .

Calculation of _Y 4- at pH = 10.20

We have relation , pH = - log [H⁺]

[H⁺] = 10 ^-pH

[H⁺] = 10 ^-10.20 = 6.310 10 ^-11 M

K ₁ K ₂ K ₃ K ₄ K ₅ K ₆ = 2.042 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 10 ^-11  ) ⁶ + (6.310 10 ^-11  )⁵ 1 + ( 6.310 10 ^-11  )⁴ 1 ( 0.03162) + ( 6.310   10 ^-11  )³1 ( 0.03162) (0.01) + ( 6.310 10 ^-11  )² 1 ( 0.03162) (0.01) (0.002042) + (6.310 10 ^-11 ) 1 ( 0.03162) (0.01) (0.002042) (7.413 10 ^-07) + 1 ( 0.03162) (0.01) (0.002042) (7.413 10 ^-07) (4.266 10 ^-11)

= (6.312 10 ^-62 ) + ( 1.000 10 ^-51 ) + ( 5.013 10 ^-43 ) + (7.944 10 ^-35 ) + ( 2.571 10 ^-27) + ( 3.020 10 ^-23) + 2.042 10 ^-23

= 5.062 10 ^-23  

Place calculated values in the formula of _Y 4-

_Y 4- = 2.042 10 ^-23 / 5.062 10 ^-23

_Y 4- = 0.4040