Assuming that the total volume of contaminated soil is 35.3 cubic foot

Question

Assuming that the total volume of contaminated soil is 35.3 cubic foot. Henry’s constant = 0.12, soil moisture content = 0.3 m3/m3, the sorption coefficient (K) = 2 cm3/g, the soil air content (the volume of soil air per volume of soil) = 0.1, the pollutant concentration in water phase = 0.25 µg/mL, and the soil bulk density = 1.2 g/cm3. Calculate the fraction of an organic contaminant present in soil water, air, and sorbed on soil particulates.

Answer 1

Step 1: Convert Given Values to Consistent Units

1. Total volume of contaminated soil Vsoil=35.3 ft3=35.3×0.0283168 m3=0.999 m3Vsoil​=35.3ft3=35.3×0.0283168m3=0.999m3
2. Soil moisture content θw=0.3 m3/m3θw​=0.3m3/m3
3. Soil air content θa=0.1 m3/m3θa​=0.1m3/m3
4. Sorption coefficient Kd=2 cm3/g=2 mL/gKd​=2cm3/g=2mL/g
5. Soil bulk density ρb=1.2 g/cm3=1.2×1000 g/L=1200 kg/m3ρb​=1.2g/cm3=1.2×1000g/L=1200kg/m3
6. Pollutant concentration in water phase Cw=0.25 μg/mL=0.25 mg/LCw​=0.25μg/mL=0.25mg/L

Step 2: Calculate the Volumes of Water and Air

1. Volume of water in soil Vw=θw×Vsoil=0.3×0.999 m3=0.2997 m3Vw​=θw​×Vsoil​=0.3×0.999m3=0.2997m3
2. Volume of air in soil Va=θa×Vsoil=0.1×0.999 m3=0.0999 m3Va​=θa​×Vsoil​=0.1×0.999m3=0.0999m3

Step 3: Calculate the Mass of Soil Particles

1. Volume of soil solids Vs=Vsoil−Vw−Va=0.999−0.2997−0.0999=0.5994 m3Vs​=Vsoil​−Vw​−Va​=0.999−0.2997−0.0999=0.5994m3
2. Mass of soil particles Ms=Vs×ρb=0.5994 m3×1200 kg/m3=719.28 kgMs​=Vs​×ρb​=0.5994m3×1200kg/m3=719.28kg

Step 4: Partitioning Coefficients

1. The contaminant concentration in air CaCa​ can be calculated using Henry's law constant HH:Ca=CwH=0.25 mg/L0.12=2.083 mg/m3Ca​=HCw​​=0.120.25mg/L​=2.083mg/m3
2. The contaminant concentration sorbed on soil particles CsCs​ can be calculated using the sorption coefficient KdKd​:Cs=Kd×Cw=2 mL/g×0.25 mg/L=0.5 mg/g=500 mg/kgCs​=Kd​×Cw​=2mL/g×0.25mg/L=0.5mg/g=500mg/kg

Step 5: Total Mass of Contaminant in Each Phase

1. Mass of contaminant in water MwMw​: Mw=Cw×Vw=0.25 mg/L×0.2997 m3×1000 L/m3=74.925 mgMw​=Cw​×Vw​=0.25mg/L×0.2997m3×1000L/m3=74.925mg
2. Mass of contaminant in air MaMa​: Ma=Ca×Va=2.083 mg/m3×0.0999 m3=0.208 mgMa​=Ca​×Va​=2.083mg/m3×0.0999m3=0.208mg
3. Mass of contaminant sorbed on soil particles MsorbedMs​orbed:Msorbed=Cs×Ms=500 mg/kg×719.28 kg=359640 mgMs​orbed=Cs​×Ms​=500mg/kg×719.28kg=359640mg

Step 6: Total Mass of Contaminant

Mtotal=Mw+Ma+Msorbed=74.925+0.208+359640=359715.133 mgMtotal​=Mw​+Ma​+Ms​orbed=74.925+0.208+359640=359715.133mg

Step 7: Fractions in Each Phase

1. Fraction in water: fw=MwMtotal=74.925359715.133≈0.000208 (or 0.0208%)fw​=Mtotal​Mw​​=359715.13374.925​≈0.000208(or0.0208%)
2. Fraction in air: fa=MaMtotal=0.208359715.133≈5.78×10−7 (or 5.78×10−5%)fa​=Mtotal​Ma​​=359715.1330.208​≈5.78×10−7(or5.78×10−5%)
3. Fraction sorbed on soil particles: fsorbed=MsorbedMtotal=359640359715.133≈0.99979 (or 99.979%)fs​orbed=Mtotal​Ms​orbed​=359715.133359640​≈0.99979(or99.979%)