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.
k1 =Cwater / Cether
k2 = Cwater / Cether
k3 = Cwater / Cether
k3 = 5.5
k =( k1+k2+k3 ) /3
= (5.30 +5.46 +5.5 )/3
This data follows the distribution law because value of K1 , 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.
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 K1 , 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?
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?
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 ?
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.