Computing the Concentration of a Pollutant in the Different Environmental Compartments The concentration of benzo[a]pyrene in water in southern Ontario has been reported to be 2.82 × 10^4 ng/m^3. Compute the likely concentration of this chemical in ng/m^3 in the air, soil, sediment, and fish. Data:

Question

Computing the Concentration of a Pollutant in the Different Environmental Compartments

The concentration of benzo[a]pyrene in water in southern Ontario has been reported to be [latex]2.82 imes 10^{4} ng / m ^{3}[/latex]. Compute the likely concentration of this chemical in [latex]ng / m ^{3}[/latex] in the air, soil, sediment, and fish.

Data:For benzo[a]pyrene [latex]\log _{10} K_{ OW , BP }=6.04[/latex], and from Illustration 12.5-1we have [latex]K _{ AW , BP }=[/latex] [latex]5.884 imes 10^{-5}[/latex]. Also, the density of soil is approximately 1500 [latex]kg / m ^{3}[/latex] and that of sediment is about 1420 [latex]kg / m ^{3}[/latex].

Accepted Answer

Starting from

[latex]C_{ BP }^{ A }=5.884 imes 10^{-5} C_{ BP }^{ W }=5.884 imes 10^{-5} imes 2.82 imes 10^{4}=1.66 \frac{ ng }{ m ^{3}}[/latex]

and given that[latex]K_{ OW , BP }=10^{6.04}=1.096 imes 10^{6}, ext { we have }[/latex]

[latex]K_{ SW , BP }=0.4 imes 0.02 imes 1.096 imes 10^{6}=8768 \frac{\frac{ g _{ BP }}{10^{6} g _{ ext {soil }}}}{\frac{ g _{ BP }}{ m _{ ext {water }}^{3}}}[/latex]

so that

[latex]C_{ BP }^{ S }=8768 imes C_{ BP }^{ W }=8768 \frac{\frac{ g _{ BP }}{10^{6} g _{ ext {soil }}}}{\frac{ g _{ BP }}{ m _{ water }^{3}}} imes 2.82 imes 10^{4} \frac{ ng _{ BP }}{ m _{ ext {water }}^{3}}=2.47 imes 10^{8} \frac{ ng _{ BP }}{10^{6} g _{ ext {soil }}}[/latex]

[latex]\begin{array}{l}=2.47 imes 10^{8} \frac{ ng _{ BP }}{10^{6} g _{ ext {soil }}} imes \frac{1 mg }{10^{6} ng } imes \frac{10^{3} g }{1 kg }=0.247 \frac{ mg _{ BP }}{ kg _{ ext {soil }}} \=0.247 ppm ext { (by weight) }\end{array}[/latex]

and

[latex]C_{ BP }^{ S }=0.247 \frac{ mg _{ BP }}{ kg _{ ext {soil }}} imes 1500 \frac{ kg _{ ext {soil }}}{ m _{ ext {soil }}^{3}} imes 10^{6} \frac{ ng }{ mg }=3.71 imes 10^{8} \frac{ ng _{ BP }}{ m _{ ext {soil }}^{3}}[/latex]

Also,

[latex]K_{ DW , BP }=0.4 imes 0.05 imes 1.096 imes 10^{6}=21920 \frac{\frac{ g _{ BP }}{10^{6} g _{ ext {sediment }}}}{\frac{ g _{ BP }}{ m _{ ext {water }}^{3}}}[/latex]

and

[latex]C_{ BP }^{ D }=21920 imes C_{ BP }^{ W }=21920 \frac{\frac{ g _{ BP }}{10^{6} g _{ ext {sediment }}}}{\frac{ g _{ BP }}{ m _{ ext {water }}^{3}}} imes 2.82 imes 10^{4} \frac{ ng _{ BP }}{ m _{ water }^{3}}[/latex]

[latex]\begin{array}{l}=6.18 imes 10^{8} \frac{ ng _{ BP }}{10^{6} g _{ ext {sediment }}} \=6.18 imes 10^{8} \frac{ ng _{ BP }}{10^{6} g _{ ext {sediment }}} imes \frac{1 mg }{10^{6} ng } imes \frac{10^{3} g }{1 kg } \=0.618 \frac{ mg _{ BP }}{ kg _{ ext {sediment }}}=0.618 ppm ext { (by weight) }\end{array}[/latex]

so that

[latex]C_{ BP }^{ D }=0.618 \frac{ mg _{ BP }}{ kg _{ ext {sediment }}} imes 1420 \frac{ kg _{ ext {sediment }}}{ m _{ ext {sediment }}^{3}} imes 10^{6} \frac{ ng }{ mg }=8.78 imes 10^{8} \frac{ ng _{ BP }}{ m _{ ext {sediment }}^{3}}[/latex]

Finally, in the biota we have

[latex]\begin{aligned}C_{ BP }^{ B } &=0.05 imes K_{ OW , BP } imes C_{ BP }^{ W }=0.05 imes 1.096 imes 10^{6} imes 2.82 imes 10^{4}=1.55 imes 10^{9} \frac{ ng }{ m ^{3}} \&=1.55 \frac{ mg _{ BP }}{ kg _{ ext {biota }}}=1.55 ppm ext { (by weight) }\end{aligned}[/latex]

Since the density of biota is approximately [latex]1 g / cc ext { or } 1000 kg / m ^{3}[/latex] this concentration is also [latex]1.55 imes 10^{9} ng / m ^{3}[/latex] (Note that in these calculations we have used the fact that 1 [latex]m^{3}[/latex] of water is equal to [latex]10^{3}[/latex] kg.)

The values computed above together with reported values of Mackay and Paterson[latex]^{13}[/latex] are listed below.

	Concentration [latex]\left(n g / m^{3} ight)[/latex]
Environmental Compartment	Calculated	Reported
Water		[latex]2.82 imes 10^{4}[/latex]
Air	1.66	1.3 to 7.1
Soil	[latex]3.71 imes 10^{8}[/latex]	[latex]1.1 imes 10^{8}[/latex]
Sediment	[latex]8.78 imes 10^{8}[/latex]	[latex]0.8 imes 10^{8} ext { to } 3 imes 10^{8}[/latex]
Fish	[latex]1.55 imes 10^{9}[/latex]	[latex]1.4 imes 10^{8}[/latex]

Comment

The agreement between the results of the calculation here and reported data, while not perfect, is certainly reasonable given the simplicity of the thermodynamic model used and the complexity of environmental processes. In particular, the calculated results show that the large, orders-ofmagnitude difference in the concentration of a long-lived pollutant in the various environmental compartments can be explained using simple phase equilibrium concepts.

[latex]^{13}[/latex]D. Mackay and S. Paterson, Environ. Sci. Technol., 25, 427–436 (1991).

Question 12.5.3: Computing the Concentration of a Pollutant in the Different ...

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