Page 269 - Materials Chemistry, Second Edition

252 Practical Design Calculations for Groundwater and Soil Remediation

Strategy:
The gasoline compounds in the saturated zone will be dissolved in
the groundwater or adsorbed onto the surface of the aquifer mate-
rials (assuming the free-product phase is absent). Since only the
gasoline concentration in the groundwater is known, we have to
estimate the amount of gasoline adsorbed on the aquifer materials
by using the partition equation discussed in Chapter 2. In addi-
tion, gasoline is a mixture of many compounds, and no specific
physicochemical data of gasoline are available; toluene, one of the
common constituents, will be used to represent gasoline in this
example.

Solution:
Basis: 1 m of aquifer
3
(a) From Table 2.5,
Log K = 2.73 (toluene)
ow
Use Equation (2.28) to find K :
oc
K = 0.63K = 0.63 (10 2.73 ) = (0.63)(537) = 338
oc
ow
Use Equation (2.26) to find K :
p
K = f K = (0.02)(338) = 6.8 L/kg
oc
p
oc
Use Equation (2.25) to find toluene concentration adsorbed onto
the solid:
S = K C = (6.8 L/kg)(20 mg/L) = 136 mg/kg
p
(b) Determine the total mass of gasoline present in the aquifer (per m ):
3
Mass of the aquifer materials = (1 m )(1,600 kg/m ) = 1,600 kg
3
3
Mass of gasoline adsorbed on the solid surface = (S)(M )
s
= (136 mg/kg)(1,600 kg) = 217,600 mg = 218 g
Pore volume of the aquifer = V × ϕ
= (1 m )(35%) = 0.35 m = 350 L
3
3
Mass of gasoline dissolved in the groundwater = (C)(V)
l
= (20 mg/L)(350 L) = 7,000 mg = 7.0 g
Total mass of gasoline in the aquifer = dissolved + adsorbed
= 7.0 + 218 = 225 g of gasoline/m of aquifer
3

(c) The amount of oxygen present in the groundwater = (V)(DO)
l
= (350 L)(4 mg/L) = 1,080 mg = 1.08 g