Question #9237e

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Question #9237e

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Answer 1 · 2016-08-11T13:26:28

The BCF (bioconcentration factor) is $60256$ $60256$ . Then let us calculate the rest.

Explanation:

According to Connell, D. W. (Bioaccumulation of Xenobiotic Compounds, 1990 CRC Press), BCF of DDT is $60256$ $60256$ (or $10^{4.78}$ $10^4.78$ ).

So the DDT concentration in plankton is

$BCF \cdot concentration in water = 60256 \cdot 0.2 = 12051 mg/kg$ $"BCF" * "concentration in water" = 60256 * 0.2 = "12051 mg/kg"$

Now we can calculate the concentration of DDT in fish (eating plankton). This is

$10 \cdot 12051 = 120510 mg/kg or 121 g/kg$ $10 * 12051 = "120510 mg/kg"" ""or" " 121 g/kg"$ .

In the end, the concentration of DDT in a bird will be $1210 g/kg$ $"1210 g/kg"$ . This cannot pass 1000 g per kg. It is highly recommended to check site-specific conditions (such as water hardness, pH, dissolved or suspended solids, etc.). And this concentration (computed one) cannot be observed in reality. I only assumed that a ten-fold increase in DDT in food chain occur. Maybe it is less than ten-fold.

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Question #9237e

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