Question

100 mg of a non-polar drug X (MW = 510) was shaken with 100 mL of 1:1 (v/v) of octanol/water mixture for a P-value determination. The concentration of the drug in the aqueous layers was found to be 5.2 x 10-4 M. Calculate, a) The partition coefficient P of the drug. b) Log P

Answer 1

Determination of P (or log P) values involves the placing of a drug compound along with the two immiscible solvents in a separation funnel. Molecules of the solute will distribute in each phase until equilibrium is established. The ratio of the two concentrations is the partition coefficient or distribution coefficient P, i.e. P = Co/Cw.

In these case, we have the concentration of the drug, in the aqueous layer and the innitial so:

Co = 100 mg * 1 g / 1000 mg * 1 mol / 510 * 1000 mL/1L * 1/100 mL = 1.96x10^-3 M

P = 1.96x10^-3 / 5.2x10^-4

P = 3.77

log P = log 3.77 = 0.5762

Hope this helps