Page 122 - Materials Chemistry, Second Edition
P. 122

Plume Migration in Aquifer and Soil                              105



           Example 3.22   Migration Speed of the Dissolved
                         Plume and Partition of COCs
           The toluene concentration of the groundwater in an aquifer was determined
           to be 500 ppb. Assuming no free product phase present, estimate the parti-
           tion of toluene in the two phases, i.e., dissolved in liquid and adsorbed onto
           the aquifer materials.
             From the RI work, the following parameters were determined:

              •  Retardation factor = 4.0
              •  Porosity = 0.35
              •  Dry bulk density of the aquifer matrix = 1.6 g/cm .
                                                            3

              Strategy:
              To determine the partition between the liquid phase and solid phase,
                we need to know the partition coefficient. The partition coefficient
                can be found from the retardation factor.
              Solution:

               (a)  Use Equation (3.33) to determine the partition coefficient, K :
                                                                        p
                              b ρ  K p    (1.6)  K p
                       R =+       ==   1+
                          1
                                    4
                               φ           0.35
                       So K  = 0.656 L/kg
                           p
                   Use Equation (2.24) to find the toluene concentration on the aqui-
                       fer solid, S
                   S = K C = (0.656)(0.5) = 0.328 mg/kg
                        p
               (b)  Basis: 1 L of aquifer

                  Mass dissolved in liquid = (V)(ϕ)(C)
                       	  		     	      = (1 L)(0.35)(0.5 mg/L) = 0.175 mg
                  Mass adsorbed on the aquifer solids = (V)(ρ )(S)
                                                        b
                       	  		                 	    = (1 L)(1.6 kg/L)(0.328 mg/kg)
                                                  = 0.525 mg
                                   % mass in liquid = (0.175) ÷ [(0.175) + (0.525)]
                                                  = 25% mass in liquid
              Discussion:
              This example illustrates that the majority of toluene was attached to the
                aquifer solids; only 25% was in the dissolved phase. This partially
                explains why the cleanup takes a long time for groundwater reme-
                diation using the pump-and-treat method.
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