Distribution Law – Numerical Part 2

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 H₂O and CCl₄. If distribution coefficient is 85 in favour of CCl₄.Find the ratio between volumes of  H₂O and CCl₄ such that 5 gm of iodine will be present in aqueous layer.

Solution )

Suppose, x and y are the volumes of  H₂O and CCl₄ 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  H₂O and CCl₄ = 85 : 1 Ans.