Question 3 of 5: Four-Phase Soil-Vapour Partitioning and NAPL-Mixture Equilibrium

Question 3: Four-Phase Soil-Vapour Partitioning and NAPL-Mixture Equilibrium (equal value)

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.

Given. (a) Soil-gas TCE concentration $C_g=100\ \text{mg/L}$, $K_d=2\ \text{L/kg}$, $n=0.4$, dry bulk density $\rho_b=2\ \text{kg/L}$, $S_w=0.3$, $S_g=0.7$, dimensionless Henry's constant $H=0.42$. (b) $C_{w,tol}=125\ \text{mg/L}$, $f_{oc}=0.015$, $\rho_b=1.85\ \text{g/cm}^3$, $S_{tol}=500\ \text{mg/L}$, $S_{benz}=1700\ \text{mg/L}$, $H_{tol}=0.124$, $H_{benz}=0.114$, $K_{oc,tol}=180\ \text{L/kg}$, $K_{oc,benz}=72\ \text{L/kg}$.

Find. (a) Total TCE mass per kg dry soil across all three phases (gas, water, sorbed). (b) NAPL mole fractions, water-phase benzene concentration, gas-phase and soil-phase concentrations of both compounds.

Approach. (a) Use Henry's law to get the water-phase concentration from the given gas-phase value, $K_d$ for the sorbed concentration, then convert every phase concentration to a mass per kg dry soil using the volumetric phase fractions implied by porosity/bulk density/saturation. (b) Use Raoult's law (mole fraction × pure-compound solubility = effective water solubility) to back out the NAPL composition from toluene's water concentration, then apply Henry's law and $K_{oc}f_{oc}$ to the resulting water concentrations.

1. Part (a) — water and sorbed concentrations. Henry's law (dimensionless, $H=C_g/C_w$) gives the pore-water concentration: $$C_w=\frac{C_g}{H}=\frac{100}{0.42}=238.1\ \text{mg/L}.$$ Sorbed concentration from the linear isotherm: $$C_s=K_dC_w=(2)(238.1)=\boxed{476.2\ \text{mg/kg soil}}.$$
2. Volumetric phase fractions per kg dry soil. A dry bulk density $\rho_b=2\ \text{kg/L}$ means 1 kg of dry soil occupies $1/\rho_b=0.5\ \text{L}$ of bulk (in-place) soil volume. Pore volume in that 0.5 L is $n\times0.5=0.4(0.5)=0.2\ \text{L}$, split between gas and water by saturation: $V_{gas}=S_g(0.2)=0.7(0.2)=0.14\ \text{L}$, $V_{water}=S_w(0.2)=0.3(0.2)=0.06\ \text{L}$.
3. Total TCE mass per kg dry soil. Summing the mass held in each phase for that 1 kg of soil: $$m_{gas}=C_gV_{gas}=(100)(0.14)=14.0\ \text{mg},\qquad m_{water}=C_wV_{water}=(238.1)(0.06)=14.3\ \text{mg},$$ $$m_{sorbed}=C_s=476.2\ \text{mg (already a per-kg-soil basis)},$$ $$m_{total}=14.0+14.3+476.2=\boxed{504.5\ \text{mg TCE per kg dry soil}}.$$ Over 94% of the total mass sits on the solids — typical for a compound with even a modest $K_d$, since the sorbed phase has no volume limit the way gas/water pore space does.
4. Part (b)(i) — NAPL composition and water-phase benzene. Raoult's law for an ideal binary NAPL: the effective aqueous solubility of each component equals its NAPL mole fraction times its pure-compound solubility, $C_{w,i}=x_iS_i$. Toluene's mole fraction follows directly from its given water concentration: $$x_{tol}=\frac{C_{w,tol}}{S_{tol}}=\frac{125}{500}=\boxed{0.25}.$$ Since the NAPL is a binary toluene-benzene mixture, $x_{benz}=1-x_{tol}=\boxed{0.75}$, and $$C_{w,benz}=x_{benz}S_{benz}=(0.75)(1700)=\boxed{1275\ \text{mg/L}}.$$
5. Part (b)(ii) — gas-phase concentrations. Henry's law applied to each water-phase concentration: $$C_{g,tol}=H_{tol}C_{w,tol}=(0.124)(125)=\boxed{15.5\ \text{mg/L}},\qquad C_{g,benz}=H_{benz}C_{w,benz}=(0.114)(1275)=\boxed{145.4\ \text{mg/L}}.$$
6. Part (b)(iii) — soil-phase (sorbed) concentrations. Organic-carbon-normalized partitioning: $K_d=K_{oc}f_{oc}$, then $C_s=K_dC_w$: $$K_{d,tol}=(180)(0.015)=2.70\ \text{L/kg},\quad C_{s,tol}=(2.70)(125)=\boxed{337.5\ \text{mg/kg}},$$ $$K_{d,benz}=(72)(0.015)=1.08\ \text{L/kg},\quad C_{s,benz}=(1.08)(1275)=\boxed{1377\ \text{mg/kg}}.$$