Question

1. Assume the water concentrations for hexachlorobenzene, ethanol, and benzyl alcohol are 5 µg/L, 5 mg/L, and 0.05 g/L, respectively. What are the expected concentrations for them in fish? Please show your calculation process when answering. Beware of log transformation and have the values in the same unit of mg/kg fish. Keep 3 significant digits for your results.

Answer 1

Ans) We know,

Log BCF = 0.76 Log K_ow - 0.23

where, BCF = bio concentration factor

K_ow = octanol water coefficient for chemical

1) Hexachlorobenzene

Log K_ow = 5.50

=> Log BCF = (0.76 x 5.50) - 0.23

= 3.95

BCF = 8912.51

Also, BCF = Concentration in fish/ Concentration in water

=> C_fish = 8912.51 x C_water

= 8912.51 x 5 x 10^-6 g/L

= 4.45 x 10^-2 g/L or 44.50 mg/L

  Density of fish is approximate to that of water = 1 g/cm³

Therefore , C_fish =44.50 mg/kg

2) Ethanol

Log Kow = -0.31

Log BCF = (- 0.76 x 0.31) - 0.23

= 0.0056

=> BCF = 1.013

Concentration in fish = 1.013 x 5

= 5.065 mg/L or 5.065 mg/kg

3) Benzyl Alchol

Log Kow = 1.10

  Log BCF = (0.76 x 1.10) - 0.23

= 0.606

=> BCF = 4.036

C_fish = 4.036 x 0.05 g/L

= 201.82 mg/L or 201.82 mg/kg