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14. The concentration of methadone in trout in Salt Spring Lake is 2.3 ppb on average. In the lake's water, the concentration of methadone is 55 ng/L, Assume that the fish contain 5.0% fat and that the water's density is 1.0 g/mL. Calculate methadone's octanol-water partitioning coefficient and its log value. 2 pts Kow 836 Kow-10 2.92

Answer 1

Methadone undergo chloramination to form NDMA (N-Nitroso dimethylamine) which is human carcinogen.

Bioconcentration factor (BCF) is the ratio of concentration of chemical/methadone in fish in a lake and concentration of substance/methadone in water. It's units is L/Kg

BCF = 2.3 ppb/55 ng/l

Concentration of substance/methadone in water = 55 x 10^-9 g/l = 55 x 10^-3 ppb

Concentration of /methadone in fish in a lake = 2.3 ppb =

BCF = 2.3 ppb / 55 x 10^-3 ppb = 41.8

Amount of fat in fish = 5 %

BCF = Ko/w x (percent of fat in fish/body fat/100)

41.8 = Ko/w x (5/100)

41.8 x 100/5 = Ko/w = 836

Partition coefficient of octanol-water, Ko/w = 836

Log value of Ko/w = pKo/w

10^x = 836

Apply log on both sides

log 10^x = log 836

x log 10 = 2.92

x = 2.92

So, log value of Ko/w = 10^2.92