Page 82 - Materials Chemistry, Second Edition
P. 82
Site Assessment and Remedial Investigation 65
Use Equation (2.37) to estimate the mass concentration of ben-
zene in the soil:
M t = φ ( w ) + ρ ( b )K p +φ () G
V H H a
0.12 (1.6)(1.28)
=
= 0.22 + 0.22 + 0.28 (0.324) 3.28 mg/L
To convert this mass concentration in soil into mg/kg, we should
divide the value by the total bulk density of the soil:
Soil concentration (X) = 3.28 mg/L ÷ 1.8 kg/L = 1.82 mg/kg
(for benzene)
(b) For pyrene, from Example 2.37, G = 0.000084 mg/L.
Use Equation (2.37) to estimate the mass concentration of pyrene in soil:
M t = φ ( w ) + ρ ( b )K p +φ () G
V H H a
= 0.12 + (1.6)(717) + 0.28 (0.000084) 482 mg/L
=
0.0002
0.0002
To convert this mass concentration in soil into mg/kg, we should
divide the value by the total bulk density of the soil:
Soil concentration (X) = 482 mg/L ÷ 1.8 kg/L = 268 mg/kg
(for pyrene)
Discussion:
1. In this example, a soil sample containing 1.82 mg/kg benzene
yields a soil vapor concentration of 100 ppmV. The soil concentra-
tion of pyrene, 268 mg/kg, is 150 times larger than that of ben-
zene, but its vapor concentration is 10,000 times smaller.
2. For a given COC concentration in soil, its soil vapor concentra-
tion will be higher if K value is smaller and the Henry’s con-
p
stant of the COC is larger. (In other words, the soil contains less
organics and the COC is less hydrophobic and more volatile.)
For sandy soil, the soil vapor concentration may be high, but the
mass adsorbed onto the sand grains may be relatively low. This
explains why PID or OVA readings on impacted sandy soil sam-
ples may be high; however, the laboratory results on the sandy
soil samples might turn out to be very low.
3. The soil concentration of pyrene in this example, 268 mg/kg,
means “268 mg pyrene per gram of dry soil plus soil moisture,”
while that in the previous example, 301 mg/kg, means “301 mg
