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 32.0 mL of a 0.21 M aqueous Cu solution buffered to pH 4 is equilibrated with 19.0 mL of chloroform

Answer 1

Solution :-

Partition coefficient = 6.6

Concentration = 0.21 M

Water = 32 ml

Lets first calculate the mass of Cu in the 32.0 ml solution

Moles of Cu = 0.21 mol per L * 0.032 L = 0.00672 mol

Mass of Cu = 0.00672 mol * 63.546 g per mol = 0.427 g

First extraction using the 19.0 ml of chloroform

Kd = (x/ org solvent)/(x/water)

Kd = ( x/org solvent)/(0.427 g – x)/32 ml

6.6 = (x / 19 ml) / ((0.427-x)/90 ml)

Solving for x we get

X= 0.2486 g

Moles of Cu = 0.2486 g / 63.546 g per mol = 0.00391 mol

Concentration of the Cu in the chloroform = 0.00391 mol / 0.019 L = 0.206 M

So in the chlrofom the concentration of the Cu is 0.206 M