Question

# The log P value for nicotine is 1.09. If 20 mL of 0.10 g/mL nicotine solution...

The log P value for nicotine is 1.09. If 20 mL of 0.10 g/mL nicotine solution in water is extracted with 20 mL of octanol, how much nicotine is removed from the water layer into the octanol layer? Of the same water solution is extracted with (2 x 10 mL) of octanol, how much nicotine is removed from the water layer? Both methods employ a total of 20 mL octanol for the extraction, but are they equally effective for removing nicotine from a water solution? Which method is more effective?

log P = 1.09

P = 12.30

let x be the amount of nicotine extracted in octanol

12.30 = (x/20)/[(0.1 - x)/20]

0.0615 - 0.615x = 0.05x

x = 0.092 mg is extracted by one 20 ml extraction with octanol

Double extraction

First extraction

12.30 = (x/20)/[(0.1 - x)/10]

0.123 - 1.23x = 0.05x

x = 0.096 mg is extracted by one 10 ml extraction with octanol

amount of nicotine remaining in water = 0.1 - 0.096 = 0.004 mg

second extraction

12.30 = (0.004/20)/[(0.004 - x)/10]

0.005 - 1.23x = 250x

x = 0.00002 mg is extracted by one 10 ml extraction with octanol

Total nicotine extracted by two 10 ml extraction = 0.09602 mg

Double extraction is more effective extraction method