EXTRACTION EXERCISE. The Kd (coeficient of distribution) between Dietilic eter and Water is 1.0 if 50mg...

EXTRACTION EXERCISE. The Kd (coeficient of distribution) between Dietilic eter and Water is 1.0 if 50mg of A is dissolve on 1.0ml of water. How much of A is going to be extracted in a portion of 0.50ml of dietilic eter? How much of A is going to remain on the water? Remember Kd=1.0

Expert Solution

Let x mg of A be extracted in diethyl ether. Therefore, concentration of A per mL of ether is (x mg)/(0.50 mL).

Mass of A remains in water = (50.0 – x) mg; the concentration of A per mL of water is (50.0 – x) mg/(1.0 mL).

The partition co-efficient K_d is defined as

K_d = concentration of A in ether/concentration of A in water

===> 1.0 = (x mg/0.50 mL)/(50.0 – x)mg/1.0 mL

===> 1.0 = x/0.50(50.0 – x)

===> 1.0*0.50*(50.0 – x) = x

===> 25.0 – 0.5x = x

===> 1.50x = 25.0

===> x = 25.0/1.50 = 16.6667 ≈ 16.67

The mass of A extracted in ether is 16.67 mg (ans).

The mass of A remaining in the water layer is (50.0 – 16.67) mg = 33.33 mg (ans).

queen_honey_blossom answered 2 years ago

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