FT_SS05-2006

FT_SS05-2006 - Fate & Transport Solution Set #5 CE/ESR...

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CE/ESR 479/579 Solution Set #5 1. (1-14) C fat = 0.1 ppm C oct /C w = C fat / C w = 10 1.96 C w = C fat / K ow = (0.1 ppm) / 10 1.96 = 0.001 ppm (mg/L) 2. (1-15) V w = 250 mL V oct = 200 mL V a = 50 mL M w = 5 mg C w = 5 mg/0.25 L = 20 mg/L Sol’y of o-xylene = 175 mg/L >> 20 mg/L so all of the xylene is dissolved: no free product AIR-WATER Ca = K H C w M a = K H C w V a = (0.22)(20 mg/L)(0.05 L) = 0.22 mg OCTANOL-WATER C oct = K ow C w M oct = K ow C w V = 10 3.12 (20 mg/L)(0.20 L) = 5,300 mg So, virtually all of the xylene is in the octanol phase. 3. (2-4) C s = f oc K oc C w (Units for C s are mg/kg: mg contaminant per kg suspended solids) f oc = 1.0 Mass of contaminant on solids = M s = (1.0)( K oc C w )(mass of susp. solids) C filt = C w C unfilt = (mass dissolved + mass on solids) / volume of solution = C w + (mass on particles) / volume of solution = C w + (K oc C w M ss )/V = C w (1 + K oc TSS) = C w + (4000 L/kg)(20 mg/L)(10-6 kg/mg) = 1.08 C w So, even though Koc is quite large at 4000 L/kg, the TSS is modest at 20 mg/L so the
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FT_SS05-2006 - Fate & Transport Solution Set #5 CE/ESR...

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