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 = CW / CE
CW= [ (5 – x ) / 100] gm. / 100ml
Given ,CE = [ x / 50] gm /50ml
K = CW / CE
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’ CE’ = x gm.
Amount of acid left in water ‘ CW’ = (7 – x ) gm.
K = CW / CE
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 = CW / CE
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
CWater = (5 / x ) gm / litre
Amount of iodine in left in ether layer = (10 – 5 ) = 5 gm
CCCl4 = (5 / y ) gm / litre
According to Distribution Law,
K = CWater / CCCl4
K = 85
85 = [ (5/y )] / [5 / x]
85 = x / y
Ratio between volumes of H2O and CCl4 = 85 : 1 Ans.
