Question

An important parameter is called the octanol-water partition coefficient, Kow, which is related to the tendency of molecules to accumulate in fat tissue. Consider Benzene, for which the octano-water partition coefficient is 135. Suppose you do extractions from a 75.0 mL sample of water containing.250g/L of benzene. a) What is the concentration (g/L) of benzene remaining in the water after 1 extraction using 10.0ml portions of octanol? b) how much octanol would you need per extraction, to remove 99.99% of the benzene in three extractions?

Answer 1

Amount of benzene in 75 mL before extraction = 250*75/1000 = 18.75 g

The fraction of solute remaining in the aqueous phase after the extraction is given by

(q_aq) = V_aq/(DV_org + V_aq)

= 75 mL/(135*10 mL + 75 mL) = 0.053

concentration (g/L) of benzene remaining in the water after 1 extraction

= (18.75 g* 0.053) * 1000/75 mL

= 13.25 g/L

To extract 99.99% of the solute (q_aq)₃ will be 0.0001

The fraction of solute remaining in the aqueous phase after three extractions = 0.0001

(q_aq)₃ = [V_aq/(DV_org + V_aq)]³

or, 0.0001 = [75/(135 y + 75)]³

or, y = 11.4 mL