Distribution Law

source :Thoughtco.com

## Distribution Law-

### Question 1)Prove that succinic acid forms a dimer in C_{6}H_{6} from the following data-

### Gram of acid /100ml H_{2}O 0.2 0.4 0.6

### Gram of acid /100ml C_{6}H_{6} 0.64 2.55 5.78

### Solution )

*Suppose C _{1} and _{2} are the concentrations of acid in water and benzene respectively.*

*If succinic acid forms dimer in benzene then, distribution coefficient*

*K = C _{1} / √C_{2}*

*K*_{1} = 0.2 /**√0.64 = 0.25**

_{1}= 0.2 /

**√0.64 = 0.25**

*K*_{2} = 0.4 / 2**.55 = 0.2505**

_{2}= 0.4 / 2

**.55 = 0.2505**

*K*_{3} = 0.6 /**√5.78 = 0.249**

_{3}= 0.6 /

**√5.78 = 0.249**

**Since all the values of K _{1},K_{2} and K_{3} 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 H_{2}O(gm/litre) 0.01 0.12 0.24

#### Conc. of substance in C_{6}H_{6} (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.*

*C _{1} and C_{2} are the concentrations of acid in water and benzene respectively.*

*K = C _{1} / n√C_{2}*

*Taking log,*

*log K = log C _{1} – 1/n(log C_{2})*

*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 CHCl_{3} and H_{2}O is 20 in favour of CHCl_{3}. Compare the masses of organic substance remaining in aqueous layer when 100 ml containing 1 gm has been shaken with ,

### i) 100 ml of CHCl_{3}

### ii) two successive 50 ml portions.

### Solution )

*Distribution coefficient K = C _{water} / C_{CHCl3} = 1/20*

*w _{n} = w[KV_{1} / (KV_{1} + V_{2})]^{n}*

*i) w = 1 , V _{1} = 100 ml, V_{2} = 100 ml, n =1*

*w _{n} = w[KV_{1} / (KV_{1} + V_{2})]^{n}*

*w1 = 1 [ (100/20)/(100/20 ) + 100] _{1}*

*w*_{1} = 0.047 gm Ans.

_{1}= 0.047 gm Ans.

*ii) w = 1 , V _{1} = 100 ml, V_{2} = 50 ml, n =2*

*w _{n} = w[KV_{1} / (KV_{1} + V_{2})]^{n}*

*w _{2} = 1 [ (100/20)/(100/20 ) + 50]^{2}*

*w*_{2 }= 0.0083 gm Ans.

_{2 }= 0.0083 gm Ans.

*ratio of w*_{1} and w_{2} = 0.047 : 0.0083 Ans.

_{1}and w

_{2}= 0.047 : 0.0083 Ans.