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18-Env-B4 Site Assessment and Remediation · December 2019

Question 6 of 7: Section B — answer ONE of TWO (both answered)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2019 — 18-Env-B4: Site Assessment and Remediation (3 hours, open book). Instructions: answer FOUR of the FIVE Section A questions and ONE of the TWO Section B questions; all eight questions (A-1–A-5, B-1, B-2) are answered in full below as a complete study resource. Each question is worth 20 marks.

Reference texts: Nyer, E.K., Practical Techniques for Groundwater and Soil Remediation; Fetter, C.W., Contaminant Hydrogeology (2nd ed.); Leeson, A. & Hinchee, R.E. (1997), Soil Bioventing: Principles and Practice; CSA Z768-01, Phase I Environmental Site Assessment; British Columbia Contaminated Sites Regulation (Environmental Management Act); Ontario Regulation 153/04 (Records of Site Condition); Canadian Council of Ministers of the Environment (CCME), National Classification System for Contaminated Sites and Canada-Wide Standard for Petroleum Hydrocarbons in Soil; Karickhoff, S.W. (1981), organic-carbon partitioning correlations.

Section A — answer FOUR of FIVE (all five answered)

Section B — answer ONE of TWO (both answered)

Question B-1: TCE Three-Phase Partitioning (Water / Air / Soil) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Check: two source-data corrections were needed to make this problem solvable. (1) A soil concentration of "15,000 g/kg" is physically impossible (>100% by mass); this is read as 15,000 µg/kg wet (15 mg/kg), a realistic trace-VOC soil concentration, with the micro-symbol evidently dropped in print. (2) The supplied sorption formula's coefficient, 6.3×10−7, gives an implausibly small Kd (~5×10−6 L/kg, essentially no sorption) for an organic solvent with Kow=263; it is read as a misprinted exponent of the well-known Karickhoff-family correlation Koc = 0.63·Kow (i.e. 6.3×10−1), which gives Koc=165.7 L/kg — matching TCE's documented literature Koc range (~126–217 L/kg, ATSDR/EPA) closely. Both corrections are used below; the volumetric water content of 0.20% (not 20%) is taken literally, per the exam's own printed value.

Given.

QuantitySymbolValue
Soil concentration (wet basis)Cs15 mg/kg (see the check note)
Dry bulk densityρb1900 kg/m³
Total porosityn0.35
Volumetric water contentθw0.0020
Organic carbon fractionfoc0.03
Octanol–water coefficientKow263
Dimensionless Henry's constantH′0.38
Solubility (20°C)S1,100 mg/L

Find. The mass, in grams, of TCE partitioned into the pore-water, soil-air, and sorbed-soil phases, per unit (1 m³) of bulk soil — no total impacted volume is given, so results are reported per m³ and scale linearly with whatever total volume the investigation later defines.

Approach. Set up the standard three-phase linear-equilibrium partitioning mass balance (Karickhoff sorption + Henry's Law volatilization) on a 1 m³ bulk-soil basis, solve for the equilibrium pore-water concentration Cw, then recover each phase mass from Cw.

  1. Sorption coefficient. Koc = 0.63·Kow = 0.63×263 = 165.7 L/kg (see the check note). Then $$K_d = f_{oc}\cdot K_{oc} = 0.03\times165.7 = 4.97\ \text{L/kg}$$
  2. Phase volumes per 1 m³ of bulk soil. Air-filled porosity $$\theta_a = n-\theta_w = 0.35-0.0020 = 0.348$$ so $$V_w = 0.0020\times1000 = 2\ \text{L}, \qquad V_a = 0.348\times1000 = 348\ \text{L}, \qquad M_{s,dry}=1900\ \text{kg}$$
  3. Total TCE mass in the 1 m³ basis. The wet-basis concentration applies to the total (soil + pore water) mass: wet mass = 1900 kg dry solids + (0.0020×1000) kg water = 1902 kg, so $$M_T = 15\ \text{mg/kg}\times1902\ \text{kg} = 28{,}530\ \text{mg} = 28.53\ \text{g}$$
  4. Solve for the equilibrium pore-water concentration. Total mass splits linearly across the three phases via $$M_T = C_w\left(V_w + H'V_a + K_dM_{s,dry}\right)$$ $$28{,}530 = C_w\left(2 + 0.38\times348 + 4.97\times1900\right) = C_w\times9578.6\ \text{L}$$ $$\boxed{C_w = 2.98\ \text{mg/L}}$$ This is well below the 1,100 mg/L solubility limit, so the "no pure TCE (NAPL)" assumption is internally consistent with this concentration.
  5. Recover each phase mass. $$M_w = C_w V_w = 2.98\times2 = 5.96\times10^{-3}\ \text{g (water)}$$ $$M_a = H'C_wV_a = 0.38\times2.98\times348 = 0.394\ \text{g (soil air)}$$ $$M_s = K_dC_wM_{s,dry} = 4.97\times2.98\times1900 = 28.13\ \text{g (sorbed on soil)}$$ Check: $$0.00596+0.394+28.13 = 28.53\ \text{g} = M_T\ \checkmark$$
PhaseMass (g / m³ bulk soil)% of total
Dissolved in pore water0.005960.02%
Soil-gas (vapour)0.3941.38%
Sorbed on soil solids28.1398.60%
Total28.53100%

Nearly all of the TCE mass (~99%) resides sorbed to the soil organic carbon rather than dissolved or vapour-phase, which is the expected outcome for a compound with a moderate Koc in a soil of only 3% organic carbon and very low water content — the small water volume (0.2% of the bulk) leaves almost nothing available to dissolve, while the much larger air-filled pore space still carries over 60× more mass than the water phase despite TCE's modest Henry's constant.