Question

At a pH of 4, the Cu-neocuproine complex formed in the liquid-liquid extraction experiment has a partition coefficient of about 6.6 when equilibrated between the acid aqueous solution and chloroform. Calculate the concentration of Cu in the chloroform phase if 33.0 mL of a 0.23 M aqueous Cu solution buffered to pH 4 is equilibrated with 5.00 mL of chloroform

Answer 1

Initial moles of Cu in water = 33/1000 x 0.23 = 7.59*10^-3 mol

Let a be the moles of Cu in chloroform at equilibrium

Moles of Cu in water at equilibrium = (7.59*10^-3 - a)

Partition coefficient = concentration of Cu in chloroform/concentration of Cu in water

= (moles of Cu in chloroform/volume of chloroform)/(moles of Cu in water/volume of water)

= (a/5)/((7.59*10^-3 - a)/33) = 6.6

66a = 0.25047

a = 3.795*10^-3

Concentration of Cu in chloroform = 3.795*10^-3/5*10^-3

= 0.759 ≈ 0.76 M