Distribution Law source :Thoughtco.com

## Distribution Law-

### 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 = C1C2

### 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.

### 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 / nC2

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

### 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