Question

Compound A is three times more soluble in diethyl ether than in water, so its partition...

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 50mL of water, answer the questions:

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

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

3) Determine FA after four extractions, using 50mL of diethyl ether in each of the extractions.

Please explain. Thanks!

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Answer #1

1) lets say x amount of sample was extracted in diethylether, so remianing in water would be 1-x

K = 3 = (x/200)/(1-x/50)

3/50 - 3x/50 = x/200

0.06 = 0.065x

x = 0.92

so 0.92 is the mole fraction extracted in diethylether

amount of A in water = 1-x = 0.08

mole fraction of fraction A remaining in water = 0.08

2) repeating similar procedure as above.

lets x be the amount of A extracted after first extraction,

3 = (x/100)/(1-x/50)

3/50 - 3x/50 = x/100

0.06 = 0.07x

x = 0.86

so 0.86 is the mole fraction extracted in diethylether

amount of A in water = 1-x = 0.14

Lets say y is the amount extracted second time,

3 = (y/100)/(0.14-y/50)

0.42/50 - 3y/50 = y/100

y = 0.12

amount remaining in water = 0.14-0.12 = 0.02

Total amount in water F(A) = 0.02

3) Lets say x is amount extracted by first extraction,

3 = (x/50)/(1-x/50)

3/50 - 3x/50 = x/50

x = 0.75

Amount remaining in water = 1-x = 0.25

second extraction, say y amount extracted,

3 = (y/50)(0.25-y/50)

0.75/50 - 3y/50 = y/50

y = 0.187

amount remaining in water = 1-y = 0.063

Third extraction, say z amount is extracted,

3 = (z/50)(0.063-y/50)

0.189/50 - 3z/50 = z/50

z = 0.041

amount remaining in water = 1-z = 0.022

Fourth extraction say a is extracted in diethyether,

3 = (a/50)/(0.022-a/50)

0.066/50 - 3a/50 = a/50

a = 0.0165

amount remaining in water = 0.022-0.0165 = 0.0055

F(A) in water = 0.0055 after four extractions.

0.0055

amount remaining in water = 1-y = 0.063

Third extraction, say z amount is extracted,

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