Distribution Law – Numerical Part 3

Distribution Law

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Distribution Law-

Question 1)Prove that succinic acid forms a dimer in C₆H₆ from the following data-

Gram of acid  /100ml H₂O            0.2           0.4              0.6

Gram of acid  /100ml C₆H₆         0.64           2.55            5.78

Solution )

Suppose C₁ and ₂ are the concentrations of acid in water and benzene respectively.

If succinic acid forms dimer in benzene then, distribution coefficient

K = C₁ / √C₂

K₁ = 0.2 /√0.64 = 0.25

K₂ = 0.4 / 2.55 = 0.2505

K₃ = 0.6 /√5.78 = 0.249

Since all the values of K₁,K₂ and K₃ are same, thus succinic acid forms a dimer in benzene.

Question 2) An organic substance has a normal molecular weight in water but gives a higher value in benzene. From the given data find the degree of complexity of the substance in benzene?

Conc. of substance in  H₂O(gm/litre)          0.01                      0.12                          0.24

Conc. of substance in C₆H₆ (gm/litre)       1.848 x 10^-5          2.661 x 10 ^-3       1.089 x 10^-2

Solution )

Suppose ‘n’ is complexity of succinic acid in benzene.

C₁ and C₂ are the concentrations of acid in water and benzene respectively.

K = C₁ / n√C₂

Taking log,

log K = log C₁ – 1/n(log C₂)

log K = log 0.01 – 1/n(log1.848 x 10^-5)             ——— eq 1

log K = log 0.12 – 1/n(log 2.661 x 10^-3)                ———–eq 2

log K = log 0.24 – 1/n(log 1.089 x 10 ^-2)        ———-eq 3

comparing eq (1) and eq (2)

log 0.01 – 1/n(log1.848 x 10^-5) =log 0.12 – 1/n(log 2.661 x 10^-3)

log 0.01 – log 0.12 = 1/n(log1.848 x 10^-5) – 1/n(log 2.661 x 10^-3)

-2 – (-0.9208) = 1/n [0.2667 – 5x 1  – 0.4250 + 3]

-1.0792 = 1/n (- 2.1583)

n = 2.1583/1.0792= 1.999

n = 2 Ans.

The value of n = 2 , it means above data shows that acid forms dimer in benzene.

Question 3) The distribution coefficient of an organic substance between CHCl₃ and H₂O is 20 in favour of CHCl₃. Compare the masses of organic substance remaining in aqueous layer when 100 ml containing 1 gm has been shaken with ,

i) 100 ml of CHCl₃

ii) two successive 50 ml portions.

Solution )

Distribution coefficient K = C_water / C_CHCl3 = 1/20

w_n = w[KV₁ / (KV₁ + V₂)]ⁿ

i) w = 1 , V₁ = 100 ml, V₂ = 100 ml, n =1

w_n = w[KV₁ / (KV₁ + V₂)]ⁿ

w1 = 1 [ (100/20)/(100/20 ) + 100]₁

w₁ = 0.047 gm Ans.

ii) w = 1 , V₁ = 100 ml, V₂ = 50 ml, n =2

w_n = w[KV₁ / (KV₁ + V₂)]ⁿ

w₂ = 1 [ (100/20)/(100/20 ) + 50]²

w₂ = 0.0083 gm Ans.

ratio of w₁ and w₂ = 0.047 : 0.0083  Ans.