Compound A is three times more soluble in diethyl ether than in water, so its pa
ID: 899006 • Letter: C
Question
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!
Explanation / Answer
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,
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.