Page 38 - Materials Chemistry, Second Edition

Site Assessment and Remedial Investigation 21

Volume of TPH in soil = (mass of TPH) ÷ (density of TPH)
= (130 kg) ÷ (0.8 kg/L) = 162.5 L = 41.9 gal
Discussion:
1. The total bulk density of soil was assumed to be 1,800 kg/m (i.e.,
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1.8 g/cm ), and the density of TPH was assumed to be 0.8 kg/L
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(i.e., 0.8 g/cm ).
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2. The TPH concentration for one of the three samples is below the
detection limit of 100 ppm. Four approaches are commonly taken
to deal with values below the detection limit: (1) use the detection
limit as the value, (2) use half of the detection limit, (3) use zero,
and (4) select a value based on a statistical approach (especially
when multiple samples are taken and a few of them are below
the detection limit). In this solution, a conservative approach was
taken by using the detection limit as the concentration.
3. We may often find that some values in tables of technical articles are
shown as ND. It would be better to show the corresponding detec-
tion limits of these samples, e.g., ND (<100 ppm), as in this example.
Example 2.11: Mass and Concentration Relationship of Excavated Soil

A leaky 1,000-gallon underground storage tank was removed. The excava-
tion resulted in a tank pit of 12′ × 12′ × 15′ (L×W×H), and the excavated soil
was stockpiled on site. Five samples were taken from the pile and analyzed
for TPH using EPA method 8015.
Based on the laboratory results, an engineer at CSUF Consulting Company
estimated that there were approximately 50 gallons of gasoline present in
the stockpile. One of the five TPH values in the report was illegible, and the
others were ND (<100), 1,000, 2,000, and 3,000 ppm, respectively. What is the
missing value?
Solution:
Let x be the missing TPH value.
Average TPH concentration = (x + 100 + 1,000 + 2,000 + 3,000)/
5 (in mg/kg)
Mass of the impacted soil = [(12)(12)(15) − (1,000/7.48) ft ](112 lb/ft )
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= (2,026)(112) = 227,000 lb = 103,000 kg
Mass of TPH in soil = (volume of gasoline) × (density of gasoline)
= [(50 gal)(ft /7.48 gal)] × [(50 lb/ft )(kg/2.2 lb)]
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= (6.68)(22.73) = 151.9 kg
151.9 kg = (average TPH concentration) × (mass of the impacted soil)
= [(x + 6,100)/5 mg/kg] × [(103,000 kg)(kg/10 mg)]
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x = the unknown TPH concentration = 1,264 mg/kg