Question

You have an analyte with a K(h,w) of 5890 in an aqueous sample with a volume of 100.0 mL. You extract it with 20.0 mL hexane and determine its concentration in the hexane layer to be 75 micromolar. Solve for its concentration in the original aqueous phase.

Answer 1

Here distribution coefficient Kd = 5890.

Kd =concentration in organic phase/concentration in aqueous phase.

concentration in organic phase = 75 microMolar.

Hence, 5890 = 75 microM /Caq

So concentration in aqueous phase = (75/5890) microMolar

= 0.013 microMolar.

Thus now concentration is 0.013*10^-6 M.

We have volume of aqueous phase = 100 ml.

No.of moles in 100 ml of aqueous phase after extraction = (0.013*10^-6*100)/1000 = 0.013 *10^-7 moles in 100 ml

Volume of hexane = 20 ml

No.of moles in 20 ml of hexane = 75 microMolar =(75*10^-6*20)/1000 = 1.5*10^-6 moles in 20 ml

Thus total no.of moles = 1.5*10^-6 + 0.013*10^-7

= 1.5013 * 10^-6 moles

Hence 1.5013 *10^-6 moles present in 100 ml water.

So molarity = 1.5013 *10^-6 moles/0.1 L = 1.5013 *10^-5 M.

Thus original concentration = 1.5013 *10^-5 M = 15.013 *10^-6 M

= 15.013 microMolar.