Soils, sediments, adsorption and geochemical transport

A soil constituent can occupy several phases

A soil or sediment system can contain mineral solids, organic material, pore water and gases. A constituent may be dissolved, sorbed, precipitated or associated with a separate phase. State the chemical species and amount basis used in an inventory.

A solid-phase concentration per dry mass and a dissolved concentration per water volume cannot be compared without the relevant mass and volume. Wet mass, dry mass and pore-water volume are different denominators.

Sorption redistributes an inventory and can delay a model front

For an original closed equilibrium model, prescribe total symbolic constituent 0.880 mg, water volume 0.220 L and dry soil mass 0.0600 kg. Assign reversible linear relation S=K_d C_w with K_d=5.40 L kg⁻¹, S in mg kg⁻¹ dry soil and C_w in mg L⁻¹ water. No transformation or other phase is included.

The total balance is m_total=C_w(V_w+K_d m_soil). Its effective volume is 0.544 L, so C_w=1.61764706 mg L⁻¹ and S=8.73529412 mg kg⁻¹. Water and solid hold 0.355882353 and 0.524117647 mg; their sum is 0.880 mg. Sorbed concentration per dry mass is not directly comparable to dissolved concentration per volume.

Add 0.220 L clean water, retain all constituent and solids, and allow the same conditional model to re-equilibrate. With V_w=0.440 L, C_w=1.15183246 mg L⁻¹ and S=6.21989529 mg kg⁻¹. Water and solid amounts become 0.506806283 and 0.373193717 mg. A lower water concentration accompanies a larger dissolved amount; total inventory still closes.

For a separate saturated, homogeneous one-dimensional ideal transport-front model, prescribe dry bulk density ρ_b=1.55 kg L⁻¹ of total bulk volume, water-filled porosity φ=0.310 and K_d=0.850 L kg⁻¹. Assume instantaneous local reversible linear sorption, no dispersion, decay or preferential flow. Retardation factor is R_f=1+ρ_bK_d/φ=5.25.

Prescribe Darcy flux q_D=0.0620 m day⁻¹ and path length 8.40 m. Pore-water velocity is q_D/φ=0.200 m day⁻¹; the ideal constituent front moves at 0.200/5.25=0.0380952381 m day⁻¹. Water travel is 42.0 days and front travel 220.5 days. Halving K_d to 0.425 L kg⁻¹ gives R_f=3.125 and front time 131.25 days; the water time remains 42.0 days. Retardation is a storage delay, not a destruction half-life or permanent isolation.

Sorption relations are conditional models

A linear partition relation can connect a sorbed amount per dry mass to a dissolved concentration when its conditions and coefficient units are specified. A nonlinear isotherm has different concentration dependence. Neither is universally valid over every composition and range.

Surface chemistry, competing species, pH and mineral or organic composition can affect the relation. Sorption changes distribution; it does not automatically destroy the constituent or establish permanent immobilization.

A finite capacity and an unresolved recovery need explicit boundaries

For a different symbolic equilibrium pool, assign capacity q_max=18.0 mg kg⁻¹ and coefficient b=0.400 L mg⁻¹. Let prescribed free concentration C enter q=q_max bC/(1+bC). At C=2.00 mg L⁻¹, sorbed concentration q=8.00 mg kg⁻¹, occupied capacity fraction 4/9 and apparent K_d=q/C=4.00 L kg⁻¹.

At C=8.00 mg L⁻¹, the same model gives q=13.7142857 mg kg⁻¹ and apparent K_d=1.71428571 L kg⁻¹. Extrapolating the earlier apparent K_d linearly would give 32.0 mg kg⁻¹, exceeding the assigned capacity 18.0 mg kg⁻¹. The graph describes the stipulated nonlinear relation; it is not an environmental fit, rate law or destruction mechanism.

For another original inventory audit, take initial whole-boundary parent mass 1.26 mg and earlier water-phase mass 0.980 mg. The boundary includes water, soil and vessel. Assign exact teaching final observations: water 0.310 mg, extracted solid 0.740 mg and vessel wall 0.0800 mg.

Recovered parent totals 1.13 mg, leaving 0.130 mg unresolved; recovered fraction is 0.896825397. Water-phase decrease is separately 0.670 mg. A drop in water amount does not uniquely establish degradation: transfer, products, volatile loss, incomplete extraction or measurement error could require further evidence.

If a separately justified improved wall inventory is 0.210 mg, with the other final observations unchanged, recovered parent totals all 1.26 mg and the unresolved amount becomes zero. Water still holds less than before. This revision illustrates a possible closed parent inventory across phases, rather than a rule for correcting a real observation.

A constituent can travel more slowly than the water

A simplified equilibrium transport model can account for reversible storage on a solid phase as water moves. Its retardation relation needs bulk density, water-filled porosity and a compatible partition coefficient. The assumptions exclude processes that require a fuller account.

Slow equilibration, preferential flow, dispersion, transformation and irreversible changes can alter transport. An equilibrium retardation factor is different from a reaction half-life or proof that the constituent cannot reach another location.

Local conditions and history affect geochemical behaviour

Oxygen availability, electron donors, microbial activity and mineral phases can affect redox transformations. A measured potential alone does not prove that every dissolved redox couple is equilibrated. Species-specific observations can test a more limited account.

Geochemical conditions can vary over short distances and change over time. The original soil ledgers below illustrate phase and transport reasoning without specifying an actual contaminated site or remediation procedure.

Worked examples

Try it yourself

1. Double water volume to 0.440 L, retaining all constituent and soil. Re-equilibrate the same linear model. Find concentrations and masses.

   Show the answer and reasoning

   C_w=1.15183246 mg L⁻¹; S=6.21989529 mg kg⁻¹; water/solid 0.506806283/0.373193717 mg.

   The denominator becomes 0.440+5.40×0.0600=0.764 L and the total remains 0.880 mg.

2. Halve only transport-model K_d to 0.425 L kg⁻¹. Find retardation and ideal-front travel time.

   Show the answer and reasoning

   R_f=3.125; front time 131.25 days; water time 42.0 days.

   Only the sorbed-storage term halves. The water-storage term one remains in R_f.

3. Prescribe C=8.00 mg L⁻¹ in the finite-capacity model. Find q and apparent K_d; compare the earlier linear extrapolation.

   Show the answer and reasoning

   q=13.7142857 mg kg⁻¹; apparent K_d=1.71428571 L kg⁻¹; the old linear prediction is 32.0 mg kg⁻¹.

   The nonlinear result stays below capacity 18.0 mg kg⁻¹, while the linear extrapolation exceeds it.

4. Use a separately justified wall inventory 0.210 mg, leaving the other measured phases unchanged. Find whole-boundary parent recovery.

   Show the answer and reasoning

   All 1.26 mg is recovered; unresolved amount zero; parent destruction is not demonstrated.

   The corrected phase sum closes the original parent inventory despite a smaller water-phase pool.

What to carry forward

• A phase concentration needs its own mass or volume denominator.
• Conditional reversible sorption can redistribute material without destroying it.
• Retardation and a chemical half-life are different model quantities.
• Whole-boundary recovery needs every relevant phase and unresolved pool.