Page 206 - Materials Chemistry, Second Edition

Vadose Zone Soil Remediation 189

Solution:
(a) From Table 2.5,

log(K ) = 1.53 for 1,2-DCA → K = 34
ow
ow
log(K ) = 4.88 for pyrene → K = 75,900
ow
ow
(b) Using the given relationship, K = 0.63K , we obtain
ow
oc
K = (0.63)(34) = 22 (for 1,2-DCA)
oc
K = (0.63)(75,900) = 47,800 (for pyrene)
oc
(c) Using Equation (2.26), K = f K , and f = 0.005, we obtain
oc
p
oc
oc
K = (0.005)(22) = 0.11 L/kg (for 1,2-DCA)
p
K = (0.005)(47,800) = 239 L/kg (for pyrene)
p
(d) Use Equation (5.32) to find the mass of dry soil:
M s,dry = (1,000) × (1.6/1.8) = 889 kg

Use Equation (5.30) to find the final concentration as (1,000 gal
= 3,785 L):
1
X final ≈ × X initial
1 ( M s,dryp)
+
V l
K
1
X final ≈ × 500 = 12.6 mg/L for 1,2-DCA
1 ( (889)(0.11))
3,785
+
1
X final ≈ 1 ( (889)(239)) × 500 = 491 mg/L forpyrene
3,785
+
Discussion:
1. Pyrene is very hydrophobic, and its K value is very high. This
p
example demonstrates that water washing is essentially ineffec-
tive in removing pyrene from soil. Addition of surfactants into
the washing fluid, using organic solvents, or raising the tempera-
ture of the washing fluid should be considered.
2. The calculated values are based on an assumption that the liquid
and the soil are in equilibrium. For a practical reactor design,
an equilibrium condition is seldom reached. Consequently, the
actual final concentration would be higher.