Distribution Law

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### Distribution Law-

### Question 1 ) 5 gm of succinic acid is dissolved in 100 ml water . If distribution coefficient between water and ether is 5.5. Find out how much acid can be extracted by 50 ml ether ?

### Solution )

*Suppose amount of acid extracted by ether = x gm*

*Amount of acid left in water =( 5 – x )*

*Distribution coefficient , K = C _{W} / C_{E}*

*C _{W}= [ (5 – x ) / 100] gm. / 100ml*

*Given ,C _{E} = [ x / 50] gm /50ml*

*K = C _{W} / C_{E}*

*K = 5.5*

*5.5 = [(5 – x ) / 100] / [ x / 50]*

*5.5 = 5 – x /2 x *

* x = 0.4167*

*x = 0.42 gm *

#### Amount of acid extracted by 50 ml. ether = 0.42 gm. Ans.

### Question 2 ) 7 gm of succinic acid is dissolved in water and shaken with equal volume of ether. The distribution coefficient between water and ether is 5.5 . Find the amount of acid in ether?

### Solution )

*Suppose , amount of acid in ether’ C _{E’ }= x gm. *

*Amount of acid left in water ‘ C _{W’ }= (7 – x ) gm. *

*K = C _{W} / C_{E}*

*K = 5.5*

*5.5 = (7 – x ) / x *

*5.5 = 7 – x / x *

* 5.5 x = 7 – x *

*6.5 x = 7*

*x = 1.0769 gm *

*x = 1.08 gm Ans.*

### Question 3) The ratio of solubilities of oxalic acid in water and ether is 10 . One litre of aqueous solution of oxalic acid in water having 10 gm / litre was mixed with 100 ml of ether. Calculate the concentration of oxalic acid in gm. in ether layer ?

### Solution )

*K = 10*

*Suppose, amount of oxalic acid extracted by ether layer = [y / 100 ] gm / ml*

*Amount of oxalic aid left in water layer = ( 10 – y )/ 1000 gm / ml*

*K = C _{W} / C_{E}*

*K = 10.0*

*10 = [ (10 – y )/ 1000] / [y / 100] *

*10 = 10- y / 10 y *

* 100 y = 10 – y *

*101 y = 10*

*y = 0.099 gm *

*concentration of oxalic acid in ether layer = 0.099 gm Ans.*

### Question 3 ) 10 gm of iodine is allowed to distribute between H_{2}O and CCl_{4}. If distribution coefficient is 85 in favour of CCl_{4}.Find the ratio between volumes of H_{2}O and CCl_{4} such that 5 gm of iodine will be present in aqueous layer.

### Solution )

*Suppose, x and y are the volumes of H _{2}O and CCl_{4} respectively*

*Amount of iodine taken = 10 gm*

*Amount of iodine in water layer = 5 gm*

*C _{Water} = (5 / x ) gm / litre *

*Amount of iodine in left in ether layer = (10 – 5 ) = 5 gm*

*C _{CCl4} = (5 / y ) gm / litre *

*According to Distribution Law, *

*K = C _{Water} / C_{CCl4}*

*K = 85*

*85 = [ (5/y )] / [5 / x] *

*85 = x / y *

* Ratio between volumes of H _{2}O and CCl_{4} = 85 : 1 Ans.*