Distribution Law
Distribution Law

source :Chromatography-online.org

Numericals of Distribution Law

Q.1      From the following data for concentration of succinic acid in water & ether at 250C, find the value of K of succinic acid between water & ether?

 

Conc. in Water (in gm/20ml)          0.0244            0.071              0.121

Conc. in ether (in gm/20ml)           0.0046            0.013              0.022

Prove that this data follow the distribution law.

Ans.

 k1 =Cwater / Cether

=   0.0244/.0046

k=5.30

k2 = Cwater /  Cether   

  =0.071 /0.013

k2= 5,46

k3 = Cwater /  Cether

=0.121 /0.022  

k= 5.5

k =( k1+k2+k3 ) /3

= (5.30 +5.46 +5.5 )/3

=16.26/3

K=5.42

This data follows the distribution law because value of K, K2 & K3 are nearly equal

Q.2.     In one experiment, the concentration of the acid in water & ether were found as follows-

Conc. in Water (in gm/20ml)          25.4                33.2                42.6

Conc. in ether (in gm/20ml)           4.2                   5.5                   7.1

show that the above results are in favour of distribution law.

Ans.

   K1 = Cwater /  Cether

=25.4 / 4.2 =6.05

K2 = 33,2/5.5 =6.04

K3 = 42.6 / 7.1 =6.00

The above results are in favour of distribution law because values of K, K2 & K3 are nearly equal.

Q.3      Iodine (5.0 gm) was dissolved in a mixture containing equal volumes of water & CCl4. If K of iodine between CCl4 & H2O is 82, find the amount of I2 in aqueous layer?

Ans.

Suppose,

Amount of I2 in aqueous layer = x

Amount of I2 in aqueous CCl4 layer =5- x

K =C CCl4 / C H2O

82  = ( 5-x) /x

82 x =5-x

83 x =5

x=5/83 =0.0602  Ans.

Q.4      Solubility of I2 in water is 0.8 gm/litre. If distribution coefficient of I2 between CCl4 & H2O is 82. Find the solubility of I2 in CCl4?

Ans.

  SWater               = 0.8 gm/l

SCCl4     =          ? gm/l

K=  SCCI4  / Swater

82 = SCCI4 /0.8

SCCI4  =82 X0.8 =65.6 gm. / l   Ans.

Q 5) Succinic acid is distributed between water and ether  at 18 0 C . If ether layer has 0.046 gm. of acid in 10 ml. and K for this system is 5.2. What will be the concentration of acid in 25 ml. water layer ?

Solution )

Distribution coefficient , K = CW / CE

K = 5.2

CW = x  gm / 25 ml

Given ,CE  = 0.046 gm / 10 ml

CE  = 0.046 x 25 / 10 = 0.115 gm / 25 ml

K = CW / CE

5.2 = x / 0.115

x = 5.2 x 0.115= 0.598

x ( conc. of acid in water layer )   = 0.598  gm / 25 ml Ans.