Question

EDTAEDTA is a hexaprotic system with the p?apKa values: p?a1=0.00pKa1=0.00, p?a2=1.50pKa2=1.50, p?a3=2.00pKa3=2.00, p?a4=2.69pKa4=2.69, p?a5=6.13pKa5=6.13, and p?a6=10.37pKa6=10.37.

The distribution of the various protonated forms of EDTA will therefore vary with pH. For equilibrium calculations involving metal complexes with EDTAEDTA, it is convenient to calculate the fraction of EDTAEDTA that is in the completely unprotonated form, Y4−Y4−. This fraction is designated ?Y4−αY4−.

Calculate ?Y4−αY4− at two pH values

pH=3.15 ?Y4−=

pH=10.30 ?Y4−=

Answer 1

Given :

We have , pH = - log [H⁺]

$\therefore$ [H⁺] = 10 ^-pH = 10 ^{- 3.15} = 7.079 $\times$ 10 ^-04

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 $\times$ 10 -07
pKa6 = 10.37	K 6 = 10 -10.37= 4.266 $\times$ 10 -11

We have relation, $\alpha$_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

First solve Numerator value .

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

K ₁ K ₂ K ₃ K ₄ K ₅ K ₆ = 2.042 $\times$ 10 ^-23

Now ,solve denominator .

{ [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

= ( 7.079 $\times$ 10 ^-04 ) ⁶ + ( 7.079 $\times$ 10 ^-04 )⁵ 1 + ( 7.079 $\times$ 10 ^-04 )⁴ 1 ( 0.03162) + ( 7.079 $\times$ 10 ^-04 )³1 ( 0.03162) (0.01) + ( 7.079 $\times$ 10 ^-04 )² 1 ( 0.03162) (0.01) (0.002042) + ( 7.079 $\times$ 10 ^-04 ) 1 ( 0.03162) (0.01) (0.002042) (7.413 $\times$ 10 ^-07) + 1 ( 0.03162) (0.01) (0.002042) (7.413 $\times$ 10 ^-07) (4.266 $\times$ 10 ^-11)

= (1.258 $\times$ 10 ^-19 ) + (1.777 $\times$ 10 ^-16 ) + ( 7.940 $\times$ 10 ^-15 ) + ( 1.122 $\times$ 10 ^-13) + ( 3.236 $\times$ 10 ^-13) + ( 3.388 $\times$​​​​​​​ 10 ^-16) + 2.042 $\times$ 10 ^-23

= 4.442 $\times$ 10 ^-13

$\therefore$$\alpha$_Y 4- = 2.042 $\times$ 10 ^-23 / 4.442 $\times$ 10 ^-13

$\alpha$_Y 4- = 4.597 $\times$ 10 ^-11  

PART 2

Calculation of  $\alpha$_Y 4- at pH = 10.30

We have, pH = - log [H⁺]

$\therefore$ [H⁺] = 10 ^-pH = 10 ^{- 10.30} = 5.012 $\times$ 10 ^-11

Numerator is same as in PART 1 ,so need to recalculate.

We can calculate denominator part.

  [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

= ( 5.012 $\times$ 10 ^-11  ) ⁶ + (5.012 $\times$ 10 ^-11  )⁵ 1 + ( 5.012 $\times$ 10 ^-11  )⁴ 1 ( 0.03162) + ( 5.012 $\times$ 10 ^-11  )³1 ( 0.03162) (0.01) + ( 5.012 $\times$ 10 ^-11  )² 1 ( 0.03162) (0.01) (0.002042) + (5.012 $\times$ 10 ^-11 ) 1 ( 0.03162) (0.01) (0.002042) (7.413 $\times$ 10 ^-07) + 1 ( 0.03162) (0.01) (0.002042) (7.413 $\times$ 10 ^-07) (4.266 $\times$ 10 ^-11)

= 1.585 $\times$ 10 ^-62 + ( 3.163  $\times$ 10 ^-52 ) + ( 1.995 $\times$ 10 ^-43 ) + (3.981 $\times$ 10 ^-35 ) + ( 1.622 $\times$ 10 ^-27) + 2.399 $\times$ 10 ^-23) + 2.042 $\times$ 10 ^-23

= 4.441 $\times$ 10 ^-23

$\therefore$$\alpha$_Y 4- = 2.042 $\times$ 10 ^-23 / 4.441 $\times$ 10 ^-23

$\alpha$_Y 4- = 0.460