Page 72 - Materials Chemistry, Second Edition

Site Assessment and Remedial Investigation 55

(b) From Table 2.5, H = 14.4
Convert H to dimensionless Henry’s constant, using Table 2.4:
H = H RT = 14.4 = H (0.082)(273 + 20)
*
*
H = 0.60 (dimensionless)
*
Use Equation (2.20) to find the liquid concentration:
G = HC = 732.2 mg/L = (0.60)C
So, C = 1,220 mg/L = 1,220 ppm

(c) From Table 2.5, for TCA
log K = 2.49 → K = 309
ow
ow
From Table 2.6, for TCA (a chlorinated hydrocarbon)
log K = 1.00(log K ) − 0.21
oc
ow
= 2.49 − 0.21 = 2.28
K = 191 mL/g = 191 L/kg
oc
Or, from Equation (2.28)
K = 0.63 K = 0.63(309) = 195 mL/g = 195 L/kg
oc
ow
Use Equation (2.26) to find K :
p
K = f K oc
oc
p
= (2%)(191) = 3.82 mL/g = 3.82 L/kg
Use Equation (2.25) to find the soil concentration, S:
S = K C
p
= (3.82 L/kg)(1,220 mg/L) = 4,660 mg/kg

Discussion:
1. The calculated liquid concentration, 1,220 mg/L, is smaller than
the solubility, 4,400 mg/L, as given in Table 2.5.
2. The simple equation, Equation (2.28), again yields an estimate
of K (195 L/kg) that is comparable to the value (191 L/kg) from
oc
using the correlation equation in Table 2.6.
3. The calculated concentrations are the maximum possible values;
the actual values would be lower if the system is not in equilib-
rium and not a confined system.

Example 2.33: Solid–Liquid–Vapor Equilibrium
Concentrations (Absence of Free Product)

For a subsurface impacted by 1,1,1-trichloroethane (1,1,1-TCA), the soil vapor
concentration at a location was found to be 1,320 ppmV. The soil is silty, with