Question

Compound A has a partition coefficient (Kp) of 3.5. If you start with a solution of 150 mg of compound A in 1.0 mL water, the amount that is recovered by extraction with one 300 uL portion of ether is 76.8 mg. How much A would be recovered after extraction with TWO 150 uL portions of ether instead of one 300 uL portion of ether?

Answer 1

Kp = amount of solute in ether/amount of solute in water

let x amount of solute extracted by one extraction with 150 ul (0.15 ml) ether

3.5 = (x/0.15)/[(150 - x)/1]

525 - 3.5x = 6.67x

So mass extracted in ether, x = 51.62 mg

Second extraction with 150 ul ether,

amount remaining in water = 150 - 51.62 = 98.38 mg

3.5 = (x/0.15)/[(98.38 - x)/1]

344.33 - 3.5x = 6.67x

amount extracted in ether = 33.86 mg

Total amount extracted by two extractions = 51.62 + 33.86 = 85.48 mg