Question

The partition coefficient, K, for compound A between toluene and water is 8.15. What is the concentration of A remaining in the aqueous phase after 30.0 mL of 0.260 M A are extracted with the quantities of toluene indicated below?

One 25.0 mL portions

Two 12.5 mL portions

Five 5mL portions

Ten 2.5 mL portions

Answer 1

K = Mole of A in toulene / Mole of A in water = 8.15 (the solvents are in equal volumes****noted)

30ml water is with 0.260M of A treted with 25ml of Toulene

Toulene of 30 ml can extract A of (? Mole) = 0.26 X 8.15 = 2.119 M

So, 25 ml of toulene can extarct A of (? Mole) = 2.19 M X (25 mL / 30 mL) = 1.765 M (A is nothing left in water)

So, 12.5 ml of toulene can extarct A of (? Mole) = 2.19 M X (12.5 mL / 30 mL) = 0.882 M (A is nothing left in water)

So, 5 ml of toulene can extarct A of (? Mole) = 2.19 M X (5 mL / 30 mL) = 0.176 M (left amount of A = 0.26 - 0.176 = 0.084M)

So, 2.5 ml of toulene can extarct A of (? Mole) = 2.19 M X (2.5 mL / 30 mL) = 0.088 M (left amount of A = 0.26 - 0.088 = 0.172M)