Page 70 - Materials Chemistry, Second Edition
P. 70

Site Assessment and Remedial Investigation                        53



             To estimate the solid concentration in equilibrium with the liquid concen-
           tration (or vice versa), we have to determine the value of the partition coef-
           ficient first. The following procedure can be used to determine the partition
           coefficient for a soil–water system:

              Step 1: Find K  or S  of the COC (Table 2.5).
                          ow
                                w
              Step 2:  Determine K  using correlations given in Table 2.6 or Equation
                                oc
                    (2.28).
              Step 3: Determine f  of the soil.
                               oc
              Step 4: Determine K  using Equation (2.26).
                                p
           Example 2.31:   Solid–Liquid Equilibrium Concentrations

           The aquifer underneath a site is impacted by tetrachloroethylene (PCE). A
           groundwater sample contains 200 ppb of PCE. Estimate the PCE concentra-
           tion adsorbed onto the aquifer material, which contains 1% of organic car-
           bon. Assume the adsorption isotherm follows a linear model.



              Solution:
               (a)  From Table 2.5, for PCE
                       log  K  = 2.6 → K  = 398
                           ow
                                      ow
               (b)  From Table 2.6, for PCE (a chlorinated hydrocarbon)
                         log K  = 1.00(log K ) − 0.21
                                         ow
                             oc
                       	  		= 2.6 − 0.21 = 2.39
                       K  = 245 mL/g = 245 L/kg
                        oc
                   Or, from Equation (2.28)
                   K  = 0.63 K  = 0.63(398) = 251 mL/g = 251 L/kg
                             ow
                     oc
               (c)  Use Equation (2.26) to find K :
                                             p
                           K  = f  K oc
                               oc
                           p
                       	  	= (1%)(251) = 2.51 mL/g = 2.51 L/kg
               (d)  Use Equation (2.25) to find S:
                           S = K C
                               p
                       	  	= (2.51 L/kg)(0.2 mg/L) = 0.50 mg/kg

              Discussion:

                1.  Equation (2.28) (K  = 0.63 K ), which looks very simple, yields
                                            ow
                                   oc
                   an estimate of K  (251 kg/L) that is comparable to the value (245
                                  oc
                   L/kg) from using the correlation equation in Table 2.6.
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