## Log Cw

FIGURE 4.5 Freundlich isotherm for benzene partitioning between water and soil.

The equation of the least-squares fitted line is log Cs = 0.941 log Cw - 0.468. Therefore, n = 0.941 log Kd = -0.468 Kd = 0.340 L/kg.

### The Role of Soil Organic Matter

Values for Kd are extremely site and chemical specific because the extent of sorption depends on several physical and chemical properties of both the soil and the sorbed chemical. For dissolved neutral organic molecules, such as fuel hydrocarbons, sorption to soil is controlled mostly by interactions with the organic portion of the soil. Therefore, the value of the soil-water partition coefficient, Kd, depends on the amount of organic matter in the soil. Expressing the soil-water partition coefficient in terms of soil organic carbon (Koc), rather than total soil mass (Kd), can eliminate a large part of the site variability in Kd.

The amount of organic matter in soil is usually expressed as either the weight fraction of organic carbon, foc, or the weight fraction of organic matter, fom. The amount of organic matter in typical mineral soils is generally between about 1 and 10% but is typically less than 5%. In wetlands and peat-soils, it can approach 100%. Since soil organic matter is approximately 58% carbon, foc typically ranges between 0.006 and 0.06. Some characteristic values for foc for a range of different soil types are given in Table 4.5.

There is a critical fraction of organic carbon in the soil, foc* (dependent on the dissolved organic compound), below which sorption to inorganic matter becomes dominant. Typical values for foc* are between about 10-3 and 10-4 (0.1% to 0.01%).

If we define Koc = —^, where Coc is the concentration of contaminant sorbed to organic carbon

Cw and Cw is the concentration of contaminant dissolved in water, then the soil/water partition coefficient becomes

This relation is useful as long as foc is greater by about 0.001.

### Rules of Thumb

1. It is often found that typical soil organic matter is about 58% carbon. Using this value to convert between the fraction of organic carbon and the fraction of organic matter leads to the relation foc = 0.58 fom.

2. Soil organic carbon content can vary by a factor of 100 for similar soils. Although there have been tabulations of foc values according to soil type,7 it is much better to use values measured at the site for use in Equation 4.16.

Example 4.8

Find Koc for benzene in the soil of Example 4.7. The soil contained 1.1% organic matter.

Answer: The soil's organic matter was measured to be 1.1%, or fom = 0.011. This may be converted to fraction of organic carbon (foc) by the approximate rule of thumb foc = 0.58 fom

Therefore, foc = 0.58 x 0.011 = 0.0064. Since Kd = Kocfoc, we have: Koc(benzene) = 0.340/0.0064 = 53 L/kg.

TABLE 4.5

Typical Values of Fraction of Organic Carbon in Different Soilsa

Type of Soil Typical foc (wt. fraction)

Coarse soil 0.04

Silty loam 0.05

Silty clayey loam 0.03

Clayey silty loam 0.005

Clayey loam 0.004

Sand 0.0005

### Glaciofluvial 0.0001

a These typical values were collected from many sources. They should be used only as crude estimates when site-specific measurements cannot be obtained. The range of measured values for a single soil type can span a factor of 100.

The Octanol/Water Partition Coefficient, Kow

For laboratory experiments, the liquid compound octanol, an eight carbon organic alcohol, CH3(CH2)7OH, is a good surrogate for the organic carbon fraction of soils. The partitioning behavior of organic compounds between octanol and water is similar to that between the organic carbon fraction of soil and water. The basic steps for measuring the octanol/water partition coefficient, Kow, are as follows:

1. Combine octanol and water in a bottle. Octanol forms a separate phase floating on top of the water.

2. Add the organic contaminant (e.g., carbon tetrachloride, CCl4), shake the mixture, and let the phases separate.

3. Measure the contaminant concentrations in the octanol phase and in the water phase. Then,

water

An empirical equation that relates Kow and organic carbon to Kd is

Kd = focbKowa, or log Kd = a log Kow + log foc + log b, (4.18)

where the empirically determined constants a and b depend on the organic compound. Equation 4.18 defines a relation that allows a calculation of K,c in terms of Kow. Since

water

Consider the pesticide DDT. Its Kow = 3.4 x 106, which is well into the high value range for Kow. Using the rules of thumb for Kow, one can predict that DDT has low water solubility (solubility = 0.025 mg/L), is slowly biodegraded, persistent in the environment (low mobility and biodegradation), strongly adsorbed to soil, and strongly bioaccumulated.

On the other hand, phenol has Kow = 30.2, a low value. Phenol has high water solubility (solubility = 8.3 x 104 mg/L), biodegrades rapidly, is not persistent, is weakly sorbed to soil, highly mobile, and weakly bioaccumulated.

 Rules of Thumb for Kow (High Kow = >1000) (Low Kow = <500) The higher Kow is for a compound, The lower Kow is for a compound, • The higher is the sorption to soil • The lower is the sorption to soil • The higher is the bioaccumulation • The lower is the bioaccumulation • The lower is the biodegradation rate • The higher is the biodegradation rate • The lower is the water solubility • The higher is the solubility • The lower is the mobility • The higher is the mobility

Estimating Kd Using Solubility or Kow

Literature values for Kd measurements vary considerably because they are so site specific. Considerable effort has been expended in finding more consistent approaches to soil sorption. EPA has recently published a comprehensive evaluation of the soil-water partition coefficient, Kd.12

The following empirical observations have led to methods for estimating Kd from more easily measured parameters:

Water solubility is inversely related to Kd; the lower the solubility, the greater is Kd. Since molecular polarity correlates with solubility, molecules whose structure indicates low polarity (hence, low solubility) may be expected to have a high Kd. For compounds of low solubility, such as fuel hydrocarbons, sorption is controlled primarily by interactions with the organic portion of the solid sorbent. The surface area of the solid is important. The larger the surface area, the larger will be Kd. A simple way to estimate the tendency for a compound to partition between water and organic solids in the soil is to measure Kow, the partition coefficient for the compound between water and octanol. The larger is Kow, the larger is Kd.

Because Kd is site specific, it is generally preferable to calculate it from the more easily obtained quantities Kow, Koc, foc, and solubility. Kow and solubility (S) are easily measured in a laboratory and Koc may be normalized to foc, percent organic carbon in soil, which is an easily measured site parameter. There are linear relationships between log Koc, log Kow, and S that vary according to the class of compounds tested6a (e.g., chlorinated compounds, aromatic hydrocarbons, ionizable organic acids, pesticides, etc.). These relations can be used to calculate Koc from Kow or S in the absence of measured Koc data. The EPA has reviewed the soil-water partitioning literature and selected or calculated the most reliable values in their judgment. In these calculations, the EPA used the following relations.

For nonionizable, semivolatile organic compounds:

For nonionizable, volatile organic compounds:

In addition, the EPA suggests the use of the following equation for estimating Koc from solubility (S) or bioconcentration factors (BCF):

Values for Koc, Kow, S, and BCF for many environmentally important chemicals have been collected in look-up tables for use in the EPA soil screening guidance procedures.11 Table 4.6 is adapted from these EPA tables.

Example 4.9

### Benzene spill

A benzene leak soaked into a patch of soil. To determine how much benzene was in the soil, several soil samples were taken in a grid pattern across the area of maximum contamination. When the samples were analyzed, the average benzene concentration in the soil was 2422 mg/kg (ppm). The soil contained 2.6% organic matter. From Table 4.6, log Kow (benzene) = 2.13. If rainwater percolates down through the contaminated soil, what concentration of benzene might initially leach from the soil and be found dissolved in the water? Assume the water and soil are in equilibrium with respect to benzene.

Calculation: The general approach to this type of problem is

1. Find a tabulated value of Koc for the chemical of concern, or calculate it from tabulated values for Kow or S.

2. Obtain a measurement of foc or fom in soil at the site or estimate it from the soil type.

3. Calculate Kd from the above quantities.

Benzene is a volatile, nonionizable organic compound (refer to Group 2 in Table 4.6). C

Cw log Koc = 0.7919 log Kow + 0.0784. log Koc = 0.7919(2.13) + 0.0784 = 1.7651. Koc = 58.2.

Since foc = 0.58 fom = 0.58(0.026) = 0.015, we have C

When 1 L of water is in equilibrium with 1 kg of soil, Csoil + Cw = 2422 ppm. 2422 - C

2422 mg/kg

Cw = = 1288 mg/L = 1300 mg/L to 2 significant figures.

Note that a substantial portion of the sorbed benzene partitions into the water, indicating that benzene has sufficient solubility to be mobile in the environment. This is predictable from the rules of thumb for Kow, which classify Kow(benzene) = 2.13 (from Table 4.6) as a low value giving benzene correspondingly high solubility and mobility.

### 4.7 MOBILITY OF CONTAMINANTS IN THE SUBSURFACE

The sorption of a contaminant from a liquid to a solid is a reversible reaction. Just as a contaminant has some probability of sorbing from water to a surface it comes in contact with, a sorbed contaminant also has a probability of desorbing from the surface back into the water. For strongly sorbed contaminants, the probability of sorption is much greater than the probability of desorption, but both processes continually take place. The rates of sorption and desorption depend on the strength of the bond holding the sorbed compound to the surface, and on the concentrations of the dissolved and sorbed contaminant.

Rate of sorption = ksorb Cw (4.24)

Rate of desorption = kdesorb Cs (4.25)

where Cw is the contaminant concentration in water, Cs is the contaminant concentration sorbed to soil, and the rate constants, ksorb and kdesorb, depend on the binding strength of sorption. For strongly sorbed contaminants, ksorb >> kdesorb. For weakly sorbed contaminants, ksorb << kdesorb.

The partition coefficient Kd quantifies the equilibrium condition of sorption, where sorption and desorption occur at the same rate. When both rates are equal, k C = k C

desorb w

Equation 4.26 shows that if ksorb >> kdesorb, which is the case of strong sorption, then at equilibrium Cs >> Cw and more of the contaminant is sorbed than is dissolved, as expected. If ksorb << kdesorb, the reverse is true. Thus, strongly sorbed contaminants accumulate to higher concentrations in the soil than do more weakly sorbed contaminants.

In terms of contaminant mobility, strongly sorbed contaminants remain sorbed to soil surfaces longer than do weakly sorbed contaminants and consequently move more slowly through the subsurface than does the groundwater in which they are dissolved. The movement of contaminants dissolved in groundwater through the subsurface is analogous to the movement of analytes through a chromatograph column. Because each analyte binds to the column wall (stationary phase) with a unique binding strength, different contaminants move through the column at different velocities and eventually become separated in space.

Contaminants dissolved in groundwater are similarly retarded in their downgradient movement relative to the flow of groundwater. Soil serves as the chromatographic stationary phase. The extent of retardation is related to the contaminants' value of Kd. At the front of a groundwater contaminant plume one will find the fastest moving contaminants with the lowest values of Kd (or weakest

CAS No.

Compound

Chemical Groupb

Solubility of Pure Compound (mg/L)

Henry's Law Constant (atmm3/L)

Henry's Law Constant (unitless)

log Kow log Ko,

Organic Carbon Partition Coefficient (L/kg)

Tbp Normal Boiling Point

Normal Melting Pointc

 83-32-9 Acenaphthene 1 4.24E+00 1.55E- 04 120-12-7 Anthracene 1 4.34E-02 6.50E- 05 71-43-2 Benzene 2 1.75E+03 5.55E- 03 56-55-3 Benzo(a)anthracene 1 9.40E-03 3.35E- 06 205-99-2 Benzo(£)fluoranthene 1 1.50E-03 1.11E- 04 207-08-9 Benzo(A)fluoranthene 1 8.00E-04 8.29E- 07 50-32-8 Benzo(a)pyrene 1 1.62E-03 1.13E- 06 117-81-7 Bis(2-ethylhexyl)phthalate 1 3.40E-01 1.02E- 07 56-23-5 Carbon tetrachloride 2 7.93E+02 3.04E- 02 57-74-9 Chlordane 2 5.60E-02 4.86E- 05 108-90-7 Chlorobenzene 2 4.72E+02 3.70E- 03 67-66-3 Chloroform 2 7.92E+03 3.67E- 03 218-01-9 Chrysene 1 1.60E-03 9.46E- 05 50-29-3 DDT 1 2.50E-02 8.10E- 06 53-70-3 Dibenzo(a,h)anthracene 1 2.49E-03 1.47E- 08 84-74-2 Di-n-butylphthalate 1 1.12E+01 9.38E- 10 95-50-1 1,2-Dichlorobenzene 2 1.56E+02 1.90E- 03 106-46-7 1,4-Dichlorobenzene 2 7.38E+01 2.43E- 03 75-34-3 1,1-Dichloroethane 2 5.06E+03 5.62E- 03 (1,1-DCA) 107-06-2 1,2-Dichloroethane 2 8.52E+03 9.79E- 04 (1,2-DCA) 75-35-4 1, 1 -Dichloroethlyene 2 2.25E+03 2.61E- 02 156-59-2 cis-1,2-Dichloroethlyene 2 3.50E+03 4.08E- 03 156-60-5 trans-1,2- 2 6.30E+03 9.38E- 03 Dichloroethlyene (DCE)

6.36E-03

3.92

3.85

7.08E+03

288

93

2.67E-03

4.55

4.47

2.95E+04

324

215

2.28E-01

2.13

1.77

5.89E+01

178

5.5

1.37E-04

5.70

5.60

3.98E+05

376

84

4.55E-03

6.20

6.09

1.23E+06

380

168

3.40E-05

6.20

6.09

1.23E+06

401

217

4.63E-05

6.11

6.01

1.02E+06

380

176

4.81E-06

7.30

7.18

1.51E+07

347

-55

1.25E+00

2.73

2.24

1.74E+02

177

-23

1.99E-03

6.32

5.08

1.20E+05

329

106

1.52E-01

2.86

2.34

2.19E+02

207

-45

1.50E-01

1.92

1.60

3.98E+01

168

-64

3.88E-03

5.70

5.60

3.98E+05

379

258

3.32E-04

6.53

6.42

2.63E+06

278

109

6.03E-07

6.69

6.58

3.80E+06

395

269

3.85E-08

4.61

4.53

3.39E+04

323

-35

7.79E-02

3.43

2.79

6.17E+02

234

-17

9.96E-02

3.42

2.79

6.17E+02

231

53

2.30E-01

1.79

1.50

3.16E+01

166

-97

4.01E-02

1.47

1.24

1.74E+01

181

-36

1.07E+00

2.13

1.77

5.89E+01

152

-123

1.67E-01

1.86

1.55

3.55E+01

168

-80

3.85E-01

2.07

1.72

5.25E+01

161

Compound

Chemical Groupb

Solubility of Pure Compound (mg/L)

Henry's Law Constant (atmm3/L)

Henry's Law Constant (unitless)

Organic Carbon Partition Coefficient (L/kg)

Tbp Normal Boiling Point

Normal Melting Pointc

CAS No.

Compound

78-87-5

1,2-Dichloropropane

2

2.80E-03

2.80E-03

1.15E-01

1.97

1.64

4.37E+01

187

-70

60-57-1

Dieldrin

2

1.95E-01

1.51E-05

6.19E-04

5.37

4.33

2.14E+04

323

176

121-14-2

2,4-Dinitrotoluene

1

2.70E+02

9.26E-08

3.80E-06

2.01

1.98

9.55E+01

310

71

115-29-7

Endosulfan

2

5.10E-01

1.12E-05

4.59E-04

4.10

3.33

2.14E+03

357

106

72-20-8

Endrin

2

2.50E-01

7.52E-06

3.08E-04

5.06

4.09

1.23E+04

381

200

100-41-4

Ethylbenzene

2

1.69E+02

7.88E-03

3.23E-01

3.14

2.56

3.63E+02

209

-95

206-44-0

Fluoranthene

1

2.06E-01

1.61E-05

6.60E-04

5.12

5.03

1.07E+05

347

108

86-73-7

Fluorene

1

1.98E+00

6.36E-05

2.61E-03

4.21

4.14

1.38E+04

299

115

76-44-8

Heptachlor

1

1.80E-01

1.09E-03

4.47E-02

6.26

6.15

1.41E+06

318

96

58-89-9

y-HCH (Lindane)

2

6.80E+00

1.40E-05

5.74E-04

3.73

3.03

1.07E+03

314

113

193-39-5

Indeno(1,2,3- cfl)pyrene

1

2.20E-05

1.60E-06

6.56E-05

6.65

6.54

3.47E+06

432

162

74-83-9

Methyl bromide

2

1.52E+04

6.24E-03

2.56E-01

1.19

1.02

1.05E+01

136

-94

75-09-2

Methylene chloride

2

1.30E+04

2.19E-03

8.98E-02

1.25

1.07

1.17E+01

156

-96

91-20-3

Naphthalene

1

3.10E+01

4.83E-04

1.98E-02

3.36

3.30

2.00E+03

255

80

108-95-2

Phenol

1

8.28E+04

3.97E-07

1.63E-05

1.48

1.46

2.88E+01

235

41

129-00-0

Pyrene

1

1.35E-01

1.10E-05

4.51E-04

5.11

5.02

1.05E+05

353

151

100-42-5

Styrene

1

3.10E+02

2.75E-03

1.13E-01

2.94

2.89

7.76E+02

215

-33

79-34-5

1,1,2,2-Tetrachloroethane (PCA)

2

2.97E+03

3.45E-04

1.41E-02

2.39

1.97

9.33E+01

215

-44

127-18-4

Tetrachloroethylene (PCE, PERC)

2

2.00E+02

1.84E-02

7.54E-01

2.67

2.19

1.55E+02

201

-22

108-88-3

Toluene

2

5.26E+02

6.64E-03

2.72E-01

2.75

2.26

1.82E+02

195

-95

8001-35-2

Toxaphene

1

7.40E-01

6.00E-06

2.46E-04

5.50

5.41

2.57E+05

347

65 to 90

120-82-1

1,2,4-Trichlorobenzene

2

3.00E+02

1.42E-03

5.82E-02

4.01

3.25

1.78E+03

252

17

71-55-6

1,1,1-Trichloroethane

2

1.33E+03

1.72E-02

7.05E-01

2.48

2.04

1.10E+02

175

Compound

Chemical Groupb

Solubility of Pure Compound (mg/L)

Henry's Law Constant (atmm3/L)

Henry's Law Constant (unitless)

log Kow log Ko,

Organic Carbon Partition Coefficient (L/kg)

Tbp Normal Boiling Point

Normal Melting Pointc

CAS No.

Compound log Kow log Ko,

 79-00-5 1,1,2-Trichloroethane (1,1,2-TCA) 2 4420 0.000913 0.0374 2.05 1.7 50.1 197 -37 79-01-6 Trichloroethylene (TCE) 2 1100 0.0103 0.422 2.71 2.22 166 183 -85 75-01-4 Vinyl chloride 2 2760 0.027 1.11 1.5 1.27 18.6 126 -154 108-38-3 m-Xylene 2 161 0.00734 0.301 3.2 2.61 407 211 -48 95-47-6 o-Xylene 2 178 0.00519 0.213 3.13 2.56 363 214 -25 106-42-3 p-Xylene 2 185 0.00766 0.314 3.17 2.59 389 211 13

a Adapted from: U.S. EPA, 1996, Soil Screening Guidance: Technical Background Document, Office of Emergency and Remedial Response, Washington, D.C., EPA/540/R95/128. b Group 1: Semi-volatile nonionizing organic compounds. Fitted to: log Koc = 0.983 log Kow + 0.00028.

Group 2: Volatile organic compounds, chlorobenzenes, and certain chlorinated pesticides. Fitted to: log Koc = 0.7919 log Kow + 0.0784. c Compounds solid at soil temperature are defined as those with a melting point > 20°C. Compounds liquid at soil temperature are defined as those with a melting point < 20°C.

sorption strengths), and moving back upgradient through the plume, there are progressively slower moving contaminants with progressively larger values of Kd.

Rules of Thumb

The mobility of dissolved organic compounds in groundwater depends on Kd.

1. If Kd = 0, the organic does not sorb to the soils it passes through and moves at the groundwater velocity.

2. If Kd > 0, movement of the organic is retarded.

### Example 4.10

Consider the case where groundwater contaminated with benzene and toluene is moving through subsurface soils containing 1.6% organic carbon. Compare benzene and toluene qualitatively with respect to their mobility in the subsurface.

Koc(benzene) = 58.9 L/kg, and Koc (toluene) = 182 L/kg.

Using Kd = Kofoc,

Kd (benzene) = 58.9 x 0.016 = 0.942 and Kd (toluene) = 182 x 0.016 = 2.91.

Kd (toluene) is approximately 3 times larger than Kd(benzene), indicating that toluene sorbs more strongly to the soil and will move significantly more slowly than benzene through the subsurface.

Retardation Factor

A retardation factor for contaminant movement can be calculated using Kd.34 The retardation factor, R, for a contaminant is defined as

R _ average linear velocity of groundwater (4 27)

average linear velocity of contaminant

Thus, a retardation factor of 10 means that the contaminant moves at one-tenth of the average velocity of the groundwater. The average linear velocity of the contaminant is measured at the point where its concentration is equal to 1/2 of its concentration at the contaminant source.

Assuming a linear partition coefficient, Kd = —^, the retardation factor becomes3

or for a Freundlich partition coefficient, Kd _ ^ ,

h pY Kd

n where p = soil bulk density (g/cm3), typically 1.5 to 1.9 g/cm3.

h = effective soil porosity,* typically 0.35 to 0.55.

Some retardation factors and qualitative mobility classifications determined from data in Table 4.6 are in Table 4.7. The table was developed using the linear Equation 4.28 with typical values for soil foc, porosity, and bulk density.

TABLE 4.7

Calculated Retardation Factors and Mobility Classifications

TABLE 4.7

Calculated Retardation Factors and Mobility Classifications

 Mobility R Examples Kd Classification <3 Methylene Chloride, MTBE, 1,2-DCA <0.03 Very Mobile 3-9 Benzene, 1,1-DCA, Chloroform 0.03-1.2 Mobile 9-30 Ethylbenzene, Toluene, Xylenes 1.2-4.3 Intermediate 30-100 Styrene, Pyrene, Lindane 4.3-15 Low Mobility >100 Naphthalene, Dioxin, Heptachlor >15 Immobile

Note: Assumed values: foc = 0.01; p = 2.0 g/cm3; h = 0.3.

Note: Assumed values: foc = 0.01; p = 2.0 g/cm3; h = 0.3.

Effect of Biodégradation on Effective Retardation Factor

Equations 4.28 and 4.29 are based only on sorption to organic carbon and do not consider the influence of processes like biodegradation, ion-exchange, precipitation, and chemical changes. As shown in Figure 4.6, biodegradation causes additional retardation to plume movement.

### Rules of Thumb

1. Dissolved contaminants generally move more slowly than groundwater because of sorption, ionexchange, precipitation, and biodegradation, but the calculated retardation factor considers sorption only.

2. The movement rate of biodegradable compounds is overestimated by the sorption retardation factor, especially over long periods of time.

3. Sorption retardation factors not only slow the growth of a plume but also the rate of cleanup by pump-and-treat, and they increase the water volume that must be extracted.

Example 4.11

### Calculation of a Sorption Retardation Factor

Calculate the retardation factor for 1,2-dichloroethane (1,2-DCA) in a soil with 2.7% TOC (total organic carbon), dry bulk density of 1.7 g/cm3, and soil porosity of 40%.

* Soil porosity is the ratio of the volume of empty space (pore volume) to total volume of soil. It can be expressed as a simple ratio or as a percentage. Thus, for a soil sample in which 1/4 of the total volume is empty, or void, space, the porosity may be expressed as 0.25 or 25%. Effective porosity is the ratio of the volume of effective void space to the total volume of soil, expressed as a percentage. Effective porosity accounts for the fact that pore spaces that are not connected, or are too small for water to overcome capillarity, do not contribute to fluid movement. The difference between true porosity and effective porosity is most noticeable with clay, where true porosity is very high, 34—60%. Effective porosity, however, is very low, 1-2%, making clay relatively impermeable.

(a) Horizontal cross section through plumes from a continuous point source.

(b) Decrease in concentration with distance from the source.

FIGURE 4.6 Positions of contaminant plume front as affected by different flow conditions. A represents a nonretarded plume, B represents a plume retarded by sorption, and C represents a plume retarded by sorption and biodegradation.

(a) Horizontal cross section through plumes from a continuous point source.

(b) Decrease in concentration with distance from the source.

FIGURE 4.6 Positions of contaminant plume front as affected by different flow conditions. A represents a nonretarded plume, B represents a plume retarded by sorption, and C represents a plume retarded by sorption and biodegradation.

Answer: Kd = Kocfoc where Koc = 17.4 (from Table 4.6).

R = 1 + —^. h r = dry bulk soil density = 1.7 g/cm3. h = effective porosity of soil = 0.40. foc = 0.025.

Divide the groundwater velocity by 3.0 to get the velocity with which 1,2-DCA moves through this particular soil.

### Example 4.12

Calculate the sorption retardation factor for tetrachloroethene (PERC) in a silty clayey loam soil, using data from Table 4.6. Soil density is 2.5 g/cm3 and porosity is 31%.

Answer: For tetrachloroethene, Koc = 155. Since no measured value for foc is available, a value from Table 4.5 can be used. For a silty clayey loam soil, foc « 0.03.

Kd = 155 x 0.03 = 4.7, and R = 1 + 2'5 X 4'7 = 38. d 0.31

Tetrachloroethene has a Koc nearly 10 times that of 1,2-DCA, which means that it is much less water-soluble. Thus, it is less mobile and has a higher retardation factor.

### A Model for Sorption and Retardation

Consider the result of Example 4.11, where the retardation factor for 1,2-DCA was found to be 3.0 in a particular soil. Since the contaminant is slowed by a factor of 3 relative to the groundwater velocity, it might appear that flushing 3 pore volumes of water through the impacted soil would completely desorb 1,2-DCA from that part of the subsurface. In other words, the retardation factor might be erroneously interpreted as the number of groundwater pore volumes that must be flushed through the contaminated zone to desorb a contaminant from the impacted soil.

In fact, however, most low solubility organic contaminants can never be completely flushed from the soil because normally there is some fraction of the contaminant that becomes almost irreversibly bound to the organic matter in soil. This part cannot be removed in any reasonable time by water flushing. It can remain in the soil as a slowly diminishing source of groundwater contamination for tens, or even hundreds, of years.

There is an aging process that causes the fraction of a contaminant that is desorbed by flushing to decrease with time. This is why pump-and-treat remediation methods are seldom successful at removing the last part of contamination from the soil. When pump-and-treat methods become ineffective, other approaches, such as bioremediation, are needed to achieve required cleanup levels.

A conceptual model for sorption has been proposed1 that is consistent with time-dependent irreversible sorption and other observations, such as a decrease in biodegradation rates with time.

Model of Sorption

• A compound with large Kd is initially adsorbed rapidly from water to the external surfaces of soil particles.

• The initially adsorbed fraction can be rapidly desorbed again to water, if the elapsed time is not too long. It is also available to microorganisms for biodegradation, and to organisms that might be susceptible to toxic effects such as fish and humans.

• As time passes, a portion of the surface-adsorbed compound begins to diffuse into micropores in the solid surface, moving away from the surface into the particle interior. Here, the compound becomes sequestered within the soil in locations remote from the surface where the compound is less available for desorption.

• Thus, aging appears to be associated with continuous diffusion into more remote sites on the solid particle where the molecules are retained and rendered less accessible to biological, chemical, and physical changes. After 1 to 10 years, depending on site-specific soil and pollutant conditions, a large fraction of the remaining sorbed pollutant will not desorb from the soil.

Another limitation of the retardation factor is that the equations unrealistically assume that the soil matrix is homogeneous, which is rarely the case. However, the results are still useful for rough estimates and for estimating relative mobilities of dissolved contaminants.

### Soil Properties

The subsurface environment contains inorganic minerals, organic humic materials, air, and water. Also found are plant roots, microorganisms, and burrowing animals, not to mention building

TABLE 4.8

Representative Values of Effective Porosity for Some Soil Types

Soil Type

Well-sorted sand or gravel

Sand and gravel, mixed

Medium sand

Glacial sediments

Silt

Clay

Effective Porosity (%)

TABLE 4.9

Soil Particle Size Range for Some Soil Types

Soil type Particle size range (mm)

Clay <0.002

Silt 0.002-0.04

Very fine sand 0.04-0.10

Fine sand 0.10-0.20

Medium sand 0.20-0.40

Coarse sand 0.40-0.9 Very coarse sand 0.9-2.0 Fine gravel 2.0-10.0

Medium gravel 10.0-20 Coarse gravel 20-40 Very coarse gravel 40-80

foundations, utility service lines, and other man-made structures. All these can affect the movement of contaminants through the subsurface.

The physical properties of soil that have the greatest effect on water and contaminant movement are effective porosity, particle size range, and hydraulic conductivity. Representative values for these properties are given in Table 4.8, 4.9, and 4.10.

• Effective porosity is the ratio of the volume of effective void space to the total volume of material, expressed as a percentage. Effective porosity accounts for the fact that pore spaces that are not connected, or are too small for water to overcome capillarity, do not contribute to fluid movement. The difference between true porosity and effective porosity is most noticeable with clay, where true porosity is very high, 34-60%. Effective porosity, however, is very low, 1-2%, making clay relatively impermeable.

• Soil particle range determines the average soil pore size and strongly influences the capillary attraction of soil particles to moving liquids. The smaller the pore size, the stronger is capillary attraction and the greater is the total particle surface area within a given volume of soil. The larger the surface area, the larger is the volume of liquid immobilized by sorption to soil surfaces. Where pore size is small, the distance between adjacent particle surfaces also is small and capillary attraction can extend across significant fractions of the pore volume. Capillary attractions retard the movement of liquids through soil and can also immobilize a fraction of the liquid. Thus, silt retards the movement of liquids more than coarse sand. Silt also will immobilize a larger quantity of liquid than will an equal volume of coarse sand.

• Hydraulic conductivity indicates the ability of subsurface materials to transmit a particular fluid. For example, the hydraulic conductivity of a given soil to transmit water is greater than it would be to transmit more viscous fluids, such as diesel fuel. The hydraulic conductivity of a very porous soil, such as coarse sand, is greater than for a less porous soil, such as fine silt.

TABLE 4.10

Representative Values of Water Hydraulic Conductivity for Some Soils

Hydraulic Conductivity for Water: Soil Type Typical Range (cm/sec)

Clay 10-6 to 10-9

Silt 10-3 to 10-7

Fine sand 10-2 to 10-5

Medium sand 10-1 to 10-4

Coarse sand 1 to 10-4

Gravel 102 to 10-1

4.8 PARTICULATE TRANSPORT IN GROUNDWATER: COLLOIDS

Contaminants move in groundwater systems as dissolved species in the water, as flowing free-phase liquids, or combined with moving particulates. Particulates that can move through soils with groundwater must be small enough to move through the soil pore spaces. Such particulates are generally less than 2.0 |jm (micron) in diameter and are called colloids. Colloids are a special class of matter with properties that lie between those of the dissolved state and the solid or immiscible liquid states.

Colloids have a high surface area to mass ratio due to their very small size. Groundwater concentrations of colloidal materials can be as high as 75 mg/L, corresponding to as many as 1012 particles/L. This represents a large surface area available for transporting sorbed contaminants.

There are many sources of colloidal material in groundwater. Colloids are formed in soil when fragments of soil, mineral, or contaminant particles become detached from their parent solid because of weathering. Then they are carried to the groundwater when water from irrigation or precipitation percolates downward through the soil. Colloids also form as fine precipitates when dissolved minerals in groundwater undergo pH or redox potential changes. Colloids are often introduced directly into groundwater from landfills, and they can form as emulsions of small droplets from free-phase hydrocarbons or other immiscible liquids.

### Colloid Particle Size and Surface Area

When particle size is reduced to 1-2 |jm or smaller, surface forces arising from surface charge or London force attractions begin to exert a significant influence on particle behavior. Consider the effect of reducing the particle size of a given mass of solid. A cube that is 10 mm on a side has a surface area of 6.0 x 10-4 m2. Cut it in half in each of the 3 directions to get 8 cubes, which are each 5 mm on a side. The surface area is now 12.0 x 10-4 m2. Continue subdividing until the cubes are 1 |Ltm on an edge. The total surface area is now 6.0 m2 — an increase of 10,000 times over the original cube.

Montmorillonite, a clay mineral, in the dispersed state may break down into platy particles only one unit cell in thickness, about 10-9 meters. Its specific surface area then is about 800 m2/g. A monolayer of 10 g of this material would cover a football field.

### Particle Transport Properties

Contaminants of low solubility can be transported as colloids or associated by adsorption or occlusion with colloids, resulting in the unexpected mobility of a low-solubility material. When contaminants are sorbed to colloids, their transport behavior is determined by the properties of the colloid, not the sorbed contaminant. In the low velocity flow conditions of groundwater, particles larger than 2 |jm tend to settle by gravity. Particles smaller than 0.1 |jm tend to sorb readily to larger soil particles, becoming retarded or immobilized. Thus, particles in the range 0.1 to 2.0 |jm are the most mobile in groundwater.

Colloid adsorptive behavior is influenced by

1. Forces acting on colloidal particles a) Electrostatic attraction and repulsion b) London attractions c) Brownian motion

2. Properties of the groundwater a) Ionic strength (related to TDS and conductivity)

b) Ionic composition c) Flow velocity

3. Properties of the colloids a) Size b) Chemical nature c) Concentration

4. Properties of the soil matrix a) Geologic composition b) Particle size distribution c) Soil surface area

Rules of Thumb

1. Usually, the most important factor governing colloid behavior is groundwater chemistry, and the least important is flow velocity.

2. Generally, colloids are more mobile when the dissolved ion concentration is low, and less mobile in solutions with high salt content.

### Electrical Charges on Colloids and Soil Surfaces

The explanation for how water chemistry affects colloidal behavior lies in the behavior of charged particles in water solutions. Soil surfaces in an aquifer generally have a net negative charge due to the dominance of silicates in the minerals which have exposed electronegative oxygen atoms. Colloidal particles also are usually charged.

• Metal oxide colloids tend to be positively charged.

• Sulfur and the noble metals tend to be negatively charged.

• Organic macromolecules, such as humic materials, proteins, or flocculating resins, acquire a charge that depends on the pH. At a low pH, protons bind to the molecules and make the colloid positive. At a high pH, the binding of hydroxyl ions makes the charge negative.

### The Electrical Double Layer

The charge on colloids and on soil surfaces attracts dissolved ions into a configuration known as an electric double layer. Consider the double layer that forms adjacent to a negatively charged surface. The first layer forms when dissolved cations (positive ions) are attracted to the oppositely charged surfaces. Then anions (negative ions) are attracted to the positive ion region of the first layer to form a second, more diffuse, oppositely charged layer surrounding the first. Positively charged surfaces acquire an inner layer of anions and an outer layer of cations. The inner layer effectively neutralizes the surface charge, and the second layer effectively neutralizes the inner layer. Subsequent layers can form in principle, but they are too diffuse to have any effect.

If two colloid particles can come close enough together, London attractive forces will pull them together into a larger particle starting the process of coagulation. The same thing happens if a colloidal particle comes close enough to a soil grain. London attractive forces will pull them together and the colloid will become sorbed to the soil surface.

Two colloid particles of the same material have the same sign of charge in the outer part of their double layer and, therefore, repel one another. For two colloid particles to coagulate, they must collide with enough energy to force past their repulsive double layers and approach close enough for London attractive forces to be effective. The same is true for adsorption to soil particle surfaces.

Brownian motion provides the collisional energy for overcoming the electrical repulsion of the double layer. In high-energy collisions, particle momentum overcomes charge repulsion and allows particles to approach close enough to enter the zone of London attraction where adsorption and coagulation can occur.

• At high ionic strength (high dissolved ion concentrations; high TDS), the double layer is thin because the ion atmosphere is dense. The higher ion charge density neutralizes the net particle charge with a thinner layer.

• At low ionic strength (low TDS), a thicker ionic charge layer is required to neutralize the charge on colloidal particles.

Thus, colloidal particles can approach surfaces and other particles more closely at high TDS than at low TDS, allowing London attraction to be more effective. There are several familiar illustrations of this principle:

• When river water carrying colloidal clay reaches the ocean, the salt water induces coagulation. This is a major cause of silting in estuaries.

• Applying a styptic pencil stops bleeding from small cuts. The styptic pencil contains aluminum salts. Dissolving high concentrations of Al3+ into the wound initiates coagulation of colloidal proteins in the blood.

• Colloidal material can be "salted out" by dissolving salts into the solution.

• The high TDS of landfill leachate provides optimal conditions for immobilizing entrained colloids by sorption and coagulation.

Rule of Thumb

Coagulation and adsorption of colloids are more efficient under high TDS conditions.

For organic contaminants of low solubility and low volatility, biodégradation is the ultimate form of contaminant removal and soil and groundwater cleanup. Because some fraction of these contaminants always becomes nearly irreversibly sorbed to soil and cannot readily be removed mechanically, it can remain in place for many years continually serving as a source of groundwater contamination. There are only three practical approaches to achieving a complete remediation in such cases:

1. Excavation of the contaminated soil and treating it on the surface (e.g., by incineration), or sending it to a landfill.

2. Isolating the contaminated soil by capping its surface with a clay or membrane liner, and diverting surface and groundwater away from the area.

3. Allowing biodegradation to transform the contaminants into nonhazardous substances.

Biodegradation is often the most economical and practical approach. If the rate of natural biodegradation (intrinsic bioremediation) is too slow, adding nutrients, oxygen, and appropriate microbes can often accelerate it (engineered bioremediation). Much progress has been made in recent years in understanding the many different processes of biodegradation and adapting them to successfully degrading types of organic compounds once thought resistant to biodegradation such as chlorinated hydrocarbons. The EPA issues many bulletins related to this new technology. The EPA's Web site is a good place to find the latest references. An extensive and excellent reference for the application of intrinsic bioremediation to fuel hydrocarbon contaminants is Wiedemeier, et al., 1995 (or Reference 13).

Biodegradation is an oxidation-reduction process. This means that energy for biodegradation arises from electron-transfer reactions. There are six basic requirements that must be present for biodegradation to occur:

1. Suitable degrading organisms, generally bacteria or fungi.

2. An energy source, generally organic carbon, which is mostly oxidized to CO2. The energy source is the electron-donor.

3. Electron acceptors, generally O2, NO3-, SO42-, Fe3+, and CO2.

4. Carbon for cell growth, which also comes from organic carbon. About 50% of bacterial dry mass is carbon.

5. Nutrients, including nitrogen, phosphorus, calcium, magnesium, iron, and trace elements. Nitrogen and phosphorus are needed in the largest quantities. Bacterial dry mass is about 12% nitrogen and about 2-3% phosphorus.

6. Acceptable environmental conditions: pH, salinity, hydrostatic pressure, solar radiation, toxic substances, oxygen, etc., must all be within acceptable limits for the particular bioprocesses.

Upon accepting electrons from an energy source, electron acceptors are converted to the products indicated in Equations 4.30-4.34.

Microbes capable of degrading petroleum hydrocarbons generally prefer a pH between 6 and 8, and temperatures between 5° C and 25° C. Temperatures lower than 5° C tend to inhibit biodegradation in general. Microbial reactions can be grouped into two classes, aerobic and anaerobic. Aerobic microbial reactions require the presence of oxygen as an electron acceptor. Anaerobic microbial reactions utilize electron acceptors other than oxygen. When sufficient oxygen is present, it will be utilized in preference to alternative electron acceptors. Aerobic and anaerobic reactions require different kinds of bacteria, with aerobic reactions being considerably faster than anaerobic reactions.

### Natural Aerobic Biodegradation of NAPL Hydrocarbons

Hydrocarbons of low solubility are called nonaqueous phase liquids, or NAPL. NAPL is further divided into hydrocarbons that are less dense than water, light nonaqueous phase liquids, or LNAPL, and hydrocarbons that are denser than water, dense nonaqueous phase liquids, or DNAPL. If nonaqueous phase liquids are mixed with water, they separate from water into a separate immiscible liquid phase. LNAPL floats on the water surface and DNAPL sinks to the bottom of the water. Generally, most of the NAPL (>90%) is in the immiscible phase and only a small fraction dissolves into the water (<10%).

When natural biodegradation of NAPL (oils, many solvents, gasoline, etc.) occurs in the saturated zone, indigenous aerobic bacteria react with dissolved oxygen to consume some of the immiscible-phase NAPL directly. These bacteria also form a biosurfactant that helps to increase the rate of NAPL dissolution into groundwater.

### Rules of Thumb

1. Within the unsaturated zone, in situ aerobic biodegradation of NAPL is often nutrient-limited by available nitrogen. Addition of nitrogen, as nitrate or ammonia, usually enhances biodegradation.

2. Within the saturated zone, dissolved oxygen is usually the limiting factor for in situ aerobic biodegradation of NAPL.

3. For fuels, the contaminants of regulatory interest are usually the most toxic and soluble components known as BTEX: benzene, toluene, ethylbenzene, and the xylene isomers.

4. It takes about 1 mg of O2 to biodegrade 0.32 mg of BTEX.

5. About 95% of the easily biodegradable hydrocarbons are converted to CO2 and water in a few months. These are low molecular weight unbranched alkanes (smaller than C-30 to C-40) and aromatics (smaller than C-10).

6. The remainder — unbranched alkanes larger than C-40, branched alkanes, alkenes, and aromatics larger than C-10 — can last for many years.

7. Polar hydrocarbons containing S, O, N, Cl, or Br may be resistant to biodegradation.

8. Chemicals that are highly water soluble biodegrade more readily than those with low water solubility.

10. Chemicals with low Kow values biodegrade more readily than those with high Kow values.

11. Chemicals that leach easily from soils biodegrade more readily than those that are not easily leached.

12. Rates of hydrocarbon biodegradation roughly double for every 10°C increase in groundwater temperature, over the range 5 to 25° C.

### Example 4.13

How long will it take to naturally biodegrade BTEX from a 100-gallon gasoline (an LNAPL) release (about 250 kg) in the saturated zone, given the following conditions:

Depth of LNAPL penetration into saturated zone = 2 m. Width of LNAPL release in saturated zone = 10 m. Groundwater Darcy velocity = 1 m/day. Background DO concentration = 5 mg/L.

Oxygen-hydrocarbon consumption ratio = 1 mg 02/0.32 mg BTEX (from rules of thumb).

Assume the gasoline LNAPL is immobilized by sorption in the soil matrix and that BTEX is 25% of the LNAPL weight. Also assume that aerobic biodegradation is instantaneous compared to normal groundwater movement. In the presence of excess oxygen, aerobic bacteria can degrade 1 mg/L of benzene in about 8 days, essentially instantaneous compared to the years required for flowing groundwater to replenish a plume area with oxygen. Under these conditions, the rate of biodegradation is equal to the rate at which sufficient dissolved oxygen can be brought into the residual gasoline LNAPL plume by groundwater flow. Approach the problem by calculating how long it will take enough water to pass through the plume cross-sectional area to supply 1 mg O2 per 0.32 mg of BTEX.

1 g NAPL

1 mg O2

0.32 mg BTEX y

1 m3 103 L

This approach of calculating how rapidly dissolved oxygen can be supplied to the plume area does not work for DNAPL composed of chlorinated compounds because they are resistant to aerobic biodegradation.