Page 29 - Materials Chemistry, Second Edition

12 Practical Design Calculations for Groundwater and Soil Remediation

(a) 5,000 m of water containing 5 ppm of toluene
3
(b) 5,000 m of soil (total bulk density = 1,800 kg/m ) having 5 ppm of
3
3
toluene
(c) An empty warehouse (indoor space = 5,000 m ) with 5 ppmV toluene
3
in air (T = 25°C; see Equation 2.1)
Solution:
(a) Mass of COC in liquid = (liquid volume)(dissolved concentration)
= [(5,000 m )(1,000 L/m )](5 mg/L)
3
3
= (5 × 10 )(5) = 2.5 × 10 mg
7
6
(b) Mass of COC in soil = [(soil volume)(total bulk density)](COC
concentration in soil)
= [(5,000 m )(1,800 kg/m )](5 mg/kg)
3
3
= (9.0 × 10 )(5) = 4.5 × 10 mg
6
7
(c) Mass of COC in air:
MW of toluene [C H (CH )] = (12)(7) + (1)(8) = 92 g/mole
3
6
5
At T = 25°C and P = 1 atm,
5 ppmV of toluene = (5 ppmV)[(MW of toluene/24.5)
((mg/m )/ppmV)]
3
= (5)(92/24.5) = 18.76 mg/m 3
Mass of COC in air = (air volume)(vapor concentration)
= [5,000 m ](18.76 mg/m ) = 9.38 × 10 mg
4
3
3
∴ The soil contains the largest amount of toluene.
Discussion:
1. Using SI units appears to be easier in these types of calculations.
However, engineers, at least in the United States, need to master
unit conversions in their job assignments because US customary
units are still commonly used in the workplace.
2. With the same volume of 5,000 m and the same concentration of
3
5 ppm, the amounts in these three media are quite different.
3. Please be aware that the equations for ppmV to mass concentra-
tion conversion are different between this example and Example
2.1 because the temperatures are different (25°C vs. 20°C).

Example 2.3: Mass and Concentration Relationship
If an adult drinks water containing 10 ppb benzene and inhales air contain-
ing 10 ppbV benzene a day, which system (ingestion or inhalation) is exposed