Four-Phase Partitioning of Benzene and Toluene (No NAPL): 18-Geol-B1 Contaminant Hydrogeology, December 2017, Q3

Question 3: Four-Phase Partitioning of Benzene and Toluene (No NAPL) (20 marks: a-10, b-5, c-5)

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.

Question 3 — given data
Quantity	Benzene	Toluene
Total soil concentration $C_T$ (mg/kg)	1,500	2,200
Henry's constant $H$ (atm·m³/mol)	0.0055	0.0066
$\log K_{ow}$	2.13	2.77
Common to both: $n=0.4$, $\rho_b=1.8\ \text{g/cm}^3$, $f_{oc}=0.03$, $T=20^\circ\text{C}$, $S_w=0.3$, $S_g=0.7$, no NAPL

Find. Pore-water, pore-gas, and sorbed-soil concentrations of benzene and toluene.

Approach. With no NAPL phase, the four-phase mass balance (Feenstra, Mackay & Cherry, 1991) partitions the measured total soil concentration among sorbed, dissolved, and vapour mass using $K_d$ (from $K_{ow}$) and the dimensionless Henry's constant; solving for the pore-water concentration is the pivot from which the other two phases follow directly.

Sorption coefficients ($K_{oc}$, $K_d$). Karickhoff (1981): $\log K_{oc}=\log K_{ow}-0.21$. Benzene: $\log K_{oc}=1.92\Rightarrow K_{oc}=83.2\ \text{mL/g}$, $K_d=K_{oc}f_{oc}=83.2(0.03)=2.50\ \text{mL/g}$. Toluene: $\log K_{oc}=2.56\Rightarrow K_{oc}=363\ \text{mL/g}$, $K_d=363(0.03)=10.89\ \text{mL/g}$.
Dimensionless Henry's constant. $$H'=\frac{H}{RT},\quad R=0.082\ \text{atm-L/mol-K}=8.2\times10^{-5}\ \text{atm-m}^3/\text{mol-K},\ T=293.15\ \text{K}.$$ Benzene: $H'=0.0055/(8.2\times10^{-5}\times293.15)=0.229$. Toluene: $H'=0.0066/(8.2\times10^{-5}\times293.15)=0.275$.
Volumetric water and air content. $$\theta_w=nS_w=0.4(0.3)=0.12,\qquad \theta_g=nS_g=0.4(0.7)=0.28.$$
Part (a) — pore-water concentration. Four-phase balance: $$C_T=C_w\left[K_d+\frac{\theta_w+H'\theta_g}{\rho_b}\right].$$ Benzene: bracket $=2.50+\dfrac{0.12+0.229(0.28)}{1.8}=2.598\ \text{L/kg}\Rightarrow C_w=1500/2.598=\boxed{577\ \text{mg/L}}.$ Toluene: bracket $=10.89+\dfrac{0.12+0.275(0.28)}{1.8}=11.00\ \text{L/kg}\Rightarrow C_w=2200/11.00=\boxed{200\ \text{mg/L}}.$ Both are below the respective pure-compound aqueous solubilities (benzene ≈1780 mg/L, toluene ≈526 mg/L at 20°C), consistent with the stated "no NAPL" condition.
Part (b) — pore-gas concentration. $$C_g=H'C_w.$$ Benzene: $C_g=0.229(577)=\boxed{132\ \text{mg/L}}$. Toluene: $C_g=0.275(200)=\boxed{54.9\ \text{mg/L}}$.
Part (c) — sorbed (soil-phase) concentration. $$C_s=K_dC_w.$$ Benzene: $C_s=2.50(577)=\boxed{1441\ \text{mg/kg}}$. Toluene: $C_s=10.89(200)=\boxed{2178\ \text{mg/kg}}$.

As a mass-balance check, converting all three phases back to a per-kg-soil basis and summing reproduces the stated total for each compound to within rounding ($1441+38.5+20.5\approx1500$ mg/kg for benzene; $2178+13.3+8.5\approx2200$ mg/kg for toluene) — strong confirmation that the pore-water pivot value and the two partition coefficients are self-consistent.

Question 3 — Final Results
Item	Benzene	Toluene
$K_d$ (mL/g)	2.50	10.89
(a) Pore-water concentration	577 mg/L	200 mg/L
(b) Pore-gas concentration	132 mg/L	54.9 mg/L
(c) Sorbed (soil-phase) concentration	1441 mg/kg	2178 mg/kg

Concept & Theory

Topic. Four-phase (water/gas/sorbed, no NAPL) equilibrium partitioning of dissolved organic contaminants.

Key relations. $C_T=C_w\left[K_d+\dfrac{\theta_w+H'\theta_g}{\rho_b}\right]$ (Feenstra, Mackay & Cherry 1991); $K_d=K_{oc}f_{oc}$; $\log K_{oc}=\log K_{ow}-0.21$; $H'=H/RT$.

Why this works. At local equilibrium, a fixed total mass of contaminant per kg of bulk soil distributes among the sorbed, dissolved, and vapour phases in ratios set entirely by $K_d$ and $H'$; writing every phase's contribution in terms of the single unknown $C_w$ collapses the three-phase mass balance to one linear equation, making $C_w$ the natural pivot for reconstructing $C_g$ and $C_s$.

Common pitfall. Forgetting to convert the vapour-phase and water-phase contributions to a per-kg-of-soil basis (dividing by $\rho_b$) before summing with $K_d$ — $K_d$ is already in L water/kg soil, but $\theta_w$ and $\theta_g$ are volume fractions of the BULK soil, not mass fractions, so they must be divided by $\rho_b$ to appear in the same units as $K_d$.

Source. Fetter, Contaminant Hydrogeology Ch.6 (four-phase partitioning, Feenstra-Mackay-Cherry model, K_ow-K_oc correlations, Henry's law).

Question 3 of 5: Four-Phase Partitioning of Benzene and Toluene (No NAPL)

Question 3: Four-Phase Partitioning of Benzene and Toluene (No NAPL) (20 marks: a-10, b-5, c-5)