1. Compound A is three times more soluble in diethyl ether than in water, so its
ID: 895942 • Letter: 1
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.
Explanation / Answer
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
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