Question

1. Compound A is three times more soluble in diethyl ether than in water, so its partition coefficient is K = 3, for partitioning of compound A between diethyl ether and water. For a sample of compound A dissolved in 50 mL of water, answer the following questions:

A.) Determine the fraction of A that remains in the water, FsubA, after one extraction with 200 mL of diethyl ether.

B.) Determine FsubA after two extractions, using 100 mL of diethyl ether in each of the extractions.

C.) Determine FsubA after four extractionsm using 50 mL of deithyl ether in each of the extractions.

Answer 1

1. Compound A is three times more soluble in diethyl ether than in water, so its partition coefficient is K = 3, for partitioning of compound A between diethyl ether and water. For a sample of compound A dissolved in 50 mL of water, answer the following questions:

A.) Determine the fraction of A that remains in the water, FsubA, after one extraction with 200 mL of diethyl ether.

Solution :- Kd= [x/ether]/[(1-x)/water)]

                   3= [x/200 ml]/[(1-x)/50ml]

Solving for x we get x= 0.923

Therefore fraction remain is the water = 1-0.923 = 0.077.

B.) Determine FsubA after two extractions, using 100 mL of diethyl ether in each of the extractions.

Solution :- fraction remain after first extraction is

Kd= [x/ether]/[(1-x)/water)]

                   3= [x/100 ml]/[(1-x)/50ml]

Solving for x we get x= 0.857

Therefore fraction remain is the water = 1-0.857 = 0.143

Calculating after second extraction

Kd= [x/ether]/[(1-x)/water)]

                   3= [x/100 ml]/[(0.143-x)/50ml]

Solving for x we get x= 0.123

Therefore fraction remain is the water = 0.143-0.123 = 0.020

So total sample remain in the water after two extraction = 1 – (0.587+0.123) =0.02

C.) Determine FsubA after four extractionsm using 50 mL of deithyl ether in each of the extractions.

Solution

First extraction

Kd= [x/ether]/[(1-x)/water)]

                   3= [x/50 ml]/[(1-x)/50ml]

Solving for x we get x= 0.75

Therefore fraction remain in the water = 1-75 = 0.25

Extraction 2

Kd= [x/ether]/[(1-x)/water)]

                   3= [x/50 ml]/[(0.25-x)/50ml]

Solving for x we get x= 0.1875

Therefore fraction remain in the water = 0.25 -0.1875 = 0.0625

Extraction 3

Kd= [x/ether]/[(1-x)/water)]

                   3= [x/50 ml]/[(0.0625-x)/50ml]

Solving for x we get x=0.04688

Therefore fraction remain in the water = 0.0625 – 0.04688 = 0.0156

Fourth extraction

Kd= [x/ether]/[(1-x)/water)]

                   3= [x/50 ml]/[(0.0156-x)/50ml]

Solving for x we get x= 0.0117

Therefore fraction remain in the water = 0.0156 -0.0117 = 0.0039

So the amount of the compound remain in the water after 4 extraction = 0.0039