3) Kd for \"A\" = C (ligroin) / C (water) = 8.0 a) What weight of \"A\" can be r

Question

3) Kd for "A" = C (ligroin) / C (water) = 8.0 a) What weight of "A" can be removed from a solution that contains 6.00 grams of "A" in 50.0-mLs of water by a SINGLE EXTRACTION using 50.0- mLs of ligroin?

b) What weight of "A" can be removed in TWO SUCCESSIVE EXTRACTIONS using 25.0-mLs of ligroin in each? (Assume 6.00 g of "A" is dissolved in 50.0-mLs of water.)

c) What volume of ligroin is needed to remove 99% of "A" in a single extraction? (Assume 6.00 g of "A" is dissolved in 50.0-mLs of water.) d) Which is more effective, SIMPLE (as in part A) or MULTIPLE (as in part B) EXTRACTION?

Explanation / Answer

If we extracted n times with Vo volume organic solvent and aquos layer volume is Va and partition ratio is Kd then fraction remaining in aquos layer q wil be

qn = ( 1/ ( Kd*VR+1))n where VR = Vo/ Va

HARE ORGANIC SOLVENT IS ligroin

3A)

VR IS 50/50=1

n=1   Kd=8

SO q1 = .111 ( PUTTING ALL THE VALUE IN EQUATION )

SO WEIGHT OF A REMOVE IS (1-.111)*6 =5.3

3B)

VR = 25/50 = .5

n=2 , Kd= 5

q2 =.04

A REMOVE FROM WATER IS (1-.04)*6 = 5.76

3C)

HERE WE HAVE CALCULATE VR q =1- .99=.01, Kd = 8, n =1

VR = 12.375

VLIG = 12.375*50 =618.75 ml   THIS VOLUME IS REQUIRE TO EXTRACT 99%

MULTIPLE STEP EXTRACTION IS THE BEST.

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