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 = Cwater / CCHCl3 = 1/20
wn = w[KV1 / (KV1 + V2)]n
i) w = 1 , V1 = 100 ml, V2 = 100 ml, n =1
wn = 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
wn = w[KV1 / (KV1 + V2)]n
w2 = 1 [ (100/20)/(100/20 ) + 50]2
