Three-Phase Equilibrium Partitioning of Toluene: 18-Env-B4 Site Assessment and Remediation, December 2014, Q7

Question B-2: Three-Phase Equilibrium Partitioning of Toluene (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.

Given. Basis: 1 m³ of bulk soil. Wet-basis toluene concentration 30 g/kg; dry bulk density $\rho_b=1950\ \text{kg/m}^3$; porosity $n=0.35$; volumetric water content $\theta_w=0.20\%=0.0020$ (as printed); $f_{oc}=0.03$; toluene $K_{ow}=537$, dimensionless Henry’s constant $H'=0.235$, solubility $=515$ mg/L.

Find. Mass of toluene (g) in the dissolved (water), vapour (air) and sorbed (soil) phases, assuming linear equilibrium partitioning and no separate NAPL phase.

Given data
Quantity	Value
Wet-basis concentration	30 g/kg
Dry bulk density, $\rho_b$	1950 kg/m³
Porosity, $n$	0.35
Volumetric water content, $\theta_w$	0.0020
Organic carbon fraction, $f_{oc}$	0.03
$K_{ow}$ / $H'$ / Solubility	537 / 0.235 / 515 mg/L

Approach. Work per 1 m³ of bulk soil: convert the given phase fractions to phase volumes/masses, estimate the soil–water partition coefficient $K_d$ from $K_{ow}$ via the Karickhoff correlation, then solve the linear mass balance $M_T=C_w(V_w+H'V_a+K_dM_{s,dry})$ for the equilibrium water concentration $C_w$ and back out each phase mass.

Phase volumes in 1 m³ of bulk soil. Air-filled porosity $\theta_a=n-\theta_w=0.35-0.0020=0.348$. Water volume $V_w=\theta_w\times 1000=2.0$ L; air volume $V_a=\theta_a\times 1000=348$ L; dry-solids mass $M_{s,dry}=\rho_b\times 1\ \text{m}^3=1950$ kg.
Total toluene mass in the reference volume. Wet-soil mass in 1 m³ $=M_{s,dry}+V_w\rho_{water}=1950+2.0=1952$ kg. Total toluene mass: $M_T=30\ \text{g/kg}\times 1952\ \text{kg}=\boxed{58{,}560\ \text{g}}$.
Soil–water partition coefficient. Using the Karickhoff (1981) correlation $K_{oc}=0.411\,K_{ow}=0.411\times 537=220.7$ L/kg, so $K_d=f_{oc}K_{oc}=0.03\times 220.7=\boxed{6.62\ \text{L/kg}}$.
Solve the mass balance for $C_w$. $M_T=C_w\left(V_w+H'V_a+K_dM_{s,dry}\right)=C_w\left(2.0+0.235\times 348+6.62\times 1950\right)=C_w\times 12{,}995$. So $C_w=58{,}560/12{,}995=\boxed{4.51\ \text{g/L}}$ (4506 mg/L).
Back out the phase masses. Water: $M_w=C_wV_w=4.51\times 2.0=\boxed{9.0\ \text{g}}$. Air: $M_a=H'C_wV_a=0.235\times 4.51\times 348=\boxed{368.5\ \text{g}}$. Soil: $M_s=K_dC_wM_{s,dry}=6.62\times 4.51\times 1950=\boxed{58{,}182\ \text{g}}$. Check: $9.0+368.5+58{,}182=58{,}560$ g $=M_T$. ✓

Toluene phase distribution (per 1 m³ bulk soil)
Phase	Mass	Share of total
Water (dissolved)	9.0 g	0.015%
Air (vapour)	368.5 g	0.63%
Soil (sorbed)	58,182 g	99.35%
Total	58,560 g	100%

Check: the computed equilibrium water concentration (4506 mg/L) is nearly 9× toluene’s aqueous solubility (515 mg/L) — physically, dissolved toluene cannot exceed solubility, so this result shows that a bulk concentration of 30 g/kg cannot in reality exist purely as dissolved + sorbed + vapour phases as the question instructs us to assume; free-phase (NAPL) toluene would actually be present at this loading. The distribution above is reported as the linear-equilibrium-partitioning answer the question explicitly asks for, with this physical inconsistency flagged rather than silently reconciled.

Topic: Linear equilibrium partitioning (LEP) of an organic contaminant among soil, water and air.

Key relations: $K_{oc}\approx 0.411\,K_{ow}$ (Karickhoff 1981); $K_d=f_{oc}K_{oc}$; $C_s=K_dC_w$; $C_a=H'C_w$; mass balance $M_T=C_w(V_w+H'V_a+K_dM_{s,dry})$.

Why this works: At equilibrium every phase concentration is linked to the same aqueous concentration $C_w$ through its own partition coefficient, so the total mass balance collapses to one linear equation in the single unknown $C_w$ — this is the standard first-cut “where did the contaminant go” calculation before more sophisticated fate-and-transport modelling.

Common pitfall: Forgetting that $K_d$ must be built from $f_{oc}\times K_{oc}$ (not $K_{ow}$ directly) for a mineral soil with organic carbon content this low; and not sanity-checking the resulting $C_w$ against the compound’s solubility, which is exactly the check that exposes the NAPL inconsistency here.

Source: Karickhoff (1981), Chemosphere; Schwarzenbach, Gschwend & Imboden, Environmental Organic Chemistry, Ch. 9–11.

Question 7 of 8: Three-Phase Equilibrium Partitioning of Toluene

Section A — Three of Five Questions

Question B-2: Three-Phase Equilibrium Partitioning of Toluene (20 marks)