Page 214 - Materials Chemistry, Second Edition

Vadose Zone Soil Remediation 197

Solution:
Basis: 1 m of soil
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(a) Determine the mass of the TPH present:

Mass of the soil matrix = (1 m )(1,800 kg/m ) = 1,800 kg
3
3
Mass of the TPH = (5,000 mg/kg)(1,800 kg)
= 9,000,000 mg = 9,000 g

(b) Use the 3.08 ratio to determine the oxygen requirements for com-
plete oxidation:

Oxygen requirement = (3.08)(9,000) = 27,720 g

(c) Determine the amount of oxygen in the soil moisture (assuming
that the moisture is saturated with oxygen and the saturated dis-
solved oxygen concentration in water at 20°C is approximately
9 mg/L):

The volume of the soil moisture = (V)(ϕ)(S )
w
= (1 m )(40%)(30%)
3
= 0.12 m = 120 L
3
The amount of oxygen in soil moisture = (V )(DO)
l
= (120 L)(9 mg/L)
= 1,080 mg = 1.08 g

(d) Determine the amount of oxygen in air (assuming that the oxy-
gen concentration in the pore void is the same as that in the
ambient air, 21% by volume, or 279 mg/L from Example 5.24):

The volume of the air void, V air void = (V)(ϕ)(1 − S )
w
= (1 m )(40%)(1 – 30%)
3
= 0.28 m = 280 L
3
The amount of oxygen in air void = (V air void )(G oxygen )
= (280 L)(279 mg/L)
= 78,120 mg = 78.1 g

(e) The total available oxygen in the soil moisture and the air void
= 1.08 + 78.1 = 79.2 g/m soil << 27,720 g/m soil
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