Question

100 grams of pesticide lindane (C6H6Cl6; MW=291; logKow=3.78) was dumped into a lake containing 400 m^3 of water, 400 kg of sediment with 5% foc, and 40 kg of fish. Assuming equilibrium partitioning, estimate the concentrations and mass distribution of lindane in the water, fish and sediment in the lake. Ignore any volatilization to air. Assume BCF for lindane = 1400 L/kg.

Answer 1

Lindane :

we take all wt of Lindane , to calculate, Cw

moles of lindane = 100g / 291 g/mol = 0.344 moles

Volume of water = 400 m^3 = 400000 L

Cw = concentration in water = 0.344 mole/ 400000 L = 8.6*10-7 mole/L = 8.6*10-7 M

1. these organic compound distributes between water and soil/sediment, and fish body :

KOW indicates the partitioning between lipid phases in the environment and water

KOW = CO/CW (CO = concentration in Organic/lipid phase ) (KOW = Dimensionless)

(KOW = Antilog (3.78) = 6025.6 )

KOW = CO/CW = 6025.6

CO = 6025.6 *8.6*10-7 M = 5.2*10-3 M

So, concentration in Lake sediment : 5 % organic fraction.

concentration in Lake sediment : 0.05 * 5.2*10-3 M = 2.6*10-4 M

Concentration in fish ~ CO = 5.2*10-3 M