Question

The distribution coefficient between methylene chloride and water for solute Y is 6. An amount of 78.0 g of Y is dissolved in 120 mL of water.

a) What weight of Y would be removed from water with a single extraction with 120-mL of methylene chloride? Report to 1 decimal place.

b) What weight of Y would be removed from water (the original solution) with two successive extractions with 60-mL portions each of methylene chloride? Report to 1 decimal place.

Answer 1

The ratio of the concentration, in mole fractions, of the solute in one solvent when compared to a second solvent is the partition coefficient.

Here, in (A) equal amounts of solvent are used with a partition coefficient of 6, so 6 times as much solute in the CH2Cl2 will be there than the water.

Therefore the CH2Cl2 will have

78 gms X 0.6 = 46.8 gms in CH2Cl2

78-46.8 = 31.2 gms in water phase.

(B) 120 ml of water and 60 ml of CH2Cl2 is used

  first extraction:   Kc  = 6.0  =      X / 60 mL CH2Cl2 / (78.0 ‑ X) / 120 mL water

X = 58.5 gms extracted

second extraction:  Kc = 6.0 =   X / 60 mL CH2Cl2 / ((78.0-58.5) ‑ X) / 120 mL water

6.0=  X / 60 mL CH2Cl2 / (19.5 ‑ X) / 120 mL water

Y = 14.63 grams extracted

total extracted =  X + Y = 58.5 + 14.63 = 73.13 gms