if a soln. of 1.0g of epinephrine in 50 mL of water was extracted with 100 mL of
ID: 812717 • Letter: I
Question
if a soln. of 1.0g of epinephrine in 50 mL of water was extracted with 100 mL of diethyl ether, how much epinephrine would be extracted by the diethyl ether ? KD=2.0 g/mL diethyl ether/g/mL waterb. if a soln. of 1.0 g of epinephrine in 50 mL of water was extracted with two 50 mL portions of diethyl ether. How much epinephrine would be extracted by diethyl ether? if a soln. of 1.0g of epinephrine in 50 mL of water was extracted with 100 mL of diethyl ether, how much epinephrine would be extracted by the diethyl ether ? KD=2.0 g/mL diethyl ether/g/mL water
b. if a soln. of 1.0 g of epinephrine in 50 mL of water was extracted with two 50 mL portions of diethyl ether. How much epinephrine would be extracted by diethyl ether?
b. if a soln. of 1.0 g of epinephrine in 50 mL of water was extracted with two 50 mL portions of diethyl ether. How much epinephrine would be extracted by diethyl ether?
Explanation / Answer
Kd is the ratio of solubilities of the solute in the two solvents; in this case, the epinephrine is twice as soluble in the ethyl ether as in the water on a volume basis. But note that you start with twice as much ethyl ether, so you get even more extracted in the ether:
KD=(g solute extracted in ether/mL diethyl ether)/(g solute extracted in water/mL water) = 2.0
g solute extracted in ether/g solute extracted in water = 2.0 * (mL diethyl ether/mL water) = 2.0 * 100 mL/50 mL = 4.0
g solute extracted in ether = 4.0 * g solute extracted in water
So if x grams dissolved in water, 4x grams dissolved ion ether
4x + x = 1.0 g;
x = 0.20 g dissolved in water
4x = 0.80 g dissolved in ethyl ether
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Let's repeat this, doing two portions of 50 mL ethyl ether:
First extraction:
KD=(g solute extracted in ether/mL diethyl ether)/g solute extracted in water/mL water) = 2.0
g solute extracted in ether/g solute extracted in water = 2.0 * (mL diethyl ether/mL water) = 2.0 * 50 mL/50 mL = 2.0
g solute extracted in ether = 2.0 * g solute extracted in water
So if x grams dissolved in water, 2x grams dissolved ion ether
2x + x = 1.0 g;
x = 0.33333 g dissolved in water
2x = 0.66667 g dissolved in ethyl ether
Second extraction (using the 0.33333 grams dissolved in water):
KD=(g solute extracted in ether/mL diethyl ether)/g solute extracted in water/mL water) = 2.0
g solute extracted in ether/g solute extracted in water = 2.0 * (mL diethyl ether/mL water) = 2.0 * 50 mL/50 mL = 2.0
g solute extracted in ether = 2.0 * g solute extracted in water
So if x grams dissolved in water, 2x grams dissolved ion ether
2x + x = 0.33333 g;
x = 0.11111 g dissolved in water
2x = 0.22222 g dissolved in ethyl ether
So, the total dissolved in bothe ether extractions is 0.66667 + 0.22222 = 0.88889 g = 0.89 g dissolved in ethyl ether (better than 10% improvement!)
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