Question

Compound A is released to a groundwater aquifer that has a volume of water of 109 m3. The pressure of the compound in the vadose soil gas immediately above the groundwater table and at equilibrium is 3 × 10^-8 atm. Calculate the total mass of compound A on aquifer material that is 5 × 10⁵ ppm in water

There are no other contaminants in the system.
- Density of aquifer material = 1.5 g/cm3
- Organic carbon of aquifer material = 4%
- Organic carbon partitioning coefficient = 1 (dimensionless volume based)
- Universal gas constant (R) = 0.082 atm·L/mol·°K

Answer 1

The soil organic carbon-water partitioning coefficient is the ratio of the mass of a chemical that is adsorbed in the soil per unit mass of organic carbon in the soil per the equilibrium 
chemical concentration in solution

Pressure=equilibrium Chemical concentration*R*T

T, Temperature = 298 K (room temperature assumed)

=>3*10^-8=concentration*.082*298

=>Concentration =1.22*10^-9 mol/litre

Soil concentration = 5*105 ppm = 5*105*10^-6 litre/litre = 5*105*10^-6*1000*1.5*1000=787.5 mg/l

Organic carbon = 4% of soil

=> Organic carbon concentration = 4/100*787.5=31.5 mg/l

=> Total Organic carbon= 31.5*109*1000=3433.5 g

Mass of chemical adsorbed on soil = m gram

=> Organic carbon partitioning coefficient = 1=(m/3433.5)/(1.22*10^-9 mol/litre)

(Molecular weight of A is required to solve above equation and find m, mass of chemical A adsorbed on soil)