1. CHAPTER 4 ENVIRONMENTAL FATE
Narration: This chapter will discuss methods of evaluation environmental fate using approaches based on chemical structure. This evaluation will serve as a preliminary screening of potential substances used in a process.

2. Introduction
This chapter serves as a basis to identify the hazards associated with different substances used and produced in the chemical process, including raw materials, products and or byproducts.
It would then be possible to do follow up with an exposure assessment and a dose-response assessment which are needed to perform risk characterization
Narration : The exposure assessment, dose-response assessment and risk characterization steps will be discussed in further chapters. Once these assessments are performed, risk management can be applied. This would be the step where pollution prevention techniques would be applied.

3. Substance Classification Tree
What Substances?
Physical + Chemical
Properties
Old Analyses
Estimating Exposure
And
Environmental Effects
Classifying the
Substances based on risk
Performing P2 on the
substances...
Narration : This diagram demonstrates potential steps that can be taken in order to perform a preliminary classification of the risk associated with a substance. The first step in classifying the risk substances may pose is to identify all possible materials, products and by-products in the given process. The second step then consists of analyzing the substance’s properties to estimate it’s effect on the environment and on people. This can be done many ways. Old analyses exist where one can find information on given substances. Another option would be to estimate the physical and chemical properties in order to estimate the effects of long and short term exposure, as well as the environmental setbacks. In this chapter, we will estimate these physical and chemical properties by using the compound’s chemical structure. This is only one of many different available methods. Once the exposure and environmental effects have been estimated for all the different substances, they can be classified. There does not exist a set of guidelines used to classify substances, but the main criterias include perisistance, bioaccumulation and toxicity. After classifying the compounds, it is possible to perform correction measures, by applying pollution prevention tools, in order to eliminate harmfull substances.

4. Chemical Properties Used to Perform Environmental Risk Screenings
Environmental Process
Relevant Properties
Dispersion and Fate
Volatility, density, melting point,
water solubility, effectiveness of waste, water treatment.
Persistence in the
Environment
Atmospheric oxidation rate, aqueous hydrolysis rate, photolysis rate, rate of microbial degradation,
and adsorption.
Narration : This list shows the different properties that help assess the environmental and health risks associated with each substance. In this chapter, we will discuss different models used to estimate these properties using the specific compound’s chemical structure. There also exist many other ways to estimate these properties. (continued on the following slide)
Continued on the following slide

5. Environmental Process Relevant Properties
Chemical Properties Used to Perform Environmental Risk Screenings
Environmental Process
Relevant Properties
Uptake by Organisms
Volatility, Lipophilicity,
Molecular Size,
Degradation Rate in Organism.
Human Uptake
Transport Across Dermal Layers, Transport Rates Across Lung Membrane, Degradation Rates
within the Human Body.
Toxicity and other
Health Effects
Dose-Response Relationships.

6. Boiling Point Distinguishes gas and liquid partitioning
Using the substance’s structure, it can be estimated by :
Where:
Tb: normal boiling point (at 1 atm) (K)
ni : number of groups of type i in the molecule,
gi : contribution of each functional group to the boiling point
Corrected using :
Tb= Σ nigi (4.1)
Narration : The boiling point is influenced by molecular weight and intermolecular attractions. In addition to characterizing the partitioning between gas and liquid phases, it can be used to calculate many other properties as will be seen later on in the chapter. The first equation was developped by Joback and Reid (1987) and then corrected by Stein and Brown (1994). The factors required for these equations can be found in table (Green Engineering, Allen and Shonnard, pp.98).
. It is also important to note that this is not the only way to calculate the boiling point.
Tb (corrected) = Tb – *Tb *(Tb)2 (Tb  700K) (4.2)
Tb (corrected) = Tb – *Tb (Tb > 700K) (4.3)

7. Example : Boiling Point Estimation
Estimate the Normal Boiling Point for diethyl ether.
Diethyl ether has the molecular structure CH3-CH2-O-CH2-CH3
Solving :
Group
-O-
2(-CH3)
2(-CH2)
gi contribution
25.16
2(21.98)
2(24.22)
Narration: Each functional group in a molecule is assumed to make a welldefined contribution (in this case, gi)to the property. The group contributions may be simply added together, as in equation 4.1. The application of the method is illustrated in this example.
The actual boiling point for diethyl ether is K

8. Example : Boiling Point Estimation (Continued)
Using equation 4.1 :
Tb (K)= Σ nigi
Tb (K)= (21.98) + 2(24.22)
Tb =
b) Using equation 4.2 :
Tb (corrected) = Tb – *Tb *(Tb)2
Tb (corr) = – (315.76) (315.76)2
Tb (corrected) = K
Narration : using the group values, it is possible to estimate the boiling point using the two given equations. In this case, equation 4.1 lends an error of 2.5% whereas equation 4.2 lends an error of 3.3%.

9. Melting Point Distinguishes solid and liquid partitioning.
Can be estimated using the substance’s boiling point :
Where :
Tm : Melting Point in Kelvins.
Tb : Boiling Point in Kelvins.
Tm (K) = * Tb (K)
(4.4)
Narration : This equation was established by Lyman (1985). There also exist other ways of estimating the melting point of a substance.

10. Example : Melting Point Estimation
Estimate the Melting Point for diethyl ether.
Solving :
Using equation 4.4 to calculate the Tm :
Tm (K) = * Tb (K)
Tm (K) = * K
Tm = K
Narration: One property, which is occasionally used in estimating the phase partitioning of solids, is melting point. Melting Point is sometimes expressed as a simple fraction of Boiling Point (Lyman,1985). The actual melting point is K leading to an error of 14.3 percent.

11. Vapor Pressure Higher Vapor Pressure = Higher Air Concentrations
Can be estimated using the following equations :
ln Pvp = A + B/(T - C) (4.5)
Where : T = Tb at 1 atm
ln(1 atm) = 0 = A + B/(Tb – C) (4.6)
ln Pvp(atm) ={[A(Tb – C)2] / [0.97*R*Tb]}*{1/(Tb – C)-1/(T – C)}
(4.7)
the parameters A and C can be estimated using :
C = Tb (4.7a)
A = KF*(8.75+ R ln Tb) (4.7b)
Narration : Plays significant role in the environmental partitioning of a given chemical and is used to estimate environmental risks and exposures. A higher vapor pressure normally signifies that the substance will be found in the atmosphere at a higher concentration then a substance with a lower vapor pressure. This property varies with temperature and therefore a substance’s atmospheric concentration can vary significantly over the course of a day. As for the boiling point, there exist many different ways to calculate vapor pressure. This slide covers the correlation between vapor pressure, boiling point and heat of vaporization, because the latter can be estimated using the chemical substance’s structure. It is important to watch the units in these calculations. All parameter definitions are given on the following slide

12. Vapor Pressure (continued)
For solids :
ln P = -(4.4 + lnTb) * {1.803*[(Tb/T)- 1)] - [0.803*ln (Tb/T)]}
- 6.8(Tm/T-1) (4.8)
Where :
Pvp : vaporization pressure (atm).
T : absolute temperature and Tb is the boiling point at 1 atm.
A and C are empirical constants.
B : a parameter related to the heat of vaporization.
KF : a correction factor.
R : gas constant ; L-atm K-1 mol-1
Tm : melting point (K).
Narration: Values of KF are given in tables; for any compound not given in the tables, assume
KF = 1.06.

13. Example : Vapor Pressure Estimation
Estimate the Vapor Pressure for diethyl ether
Using the predicted value of K:
C = Tb = (320.2) =
A = Kf ( R ln Tb) = 1.06 [ ln(320.2)] =
(4.7.a)
(4.7.b)
ln Pvp = {[A(Tb – C)2] / [0.97*R*Tb]}*{1/(Tb – C) - 1/(T – C)}
= {[21.39( )2] / [0.97(1.987)(315.76)]}*{1/(273.76) – 1/(256)}
Ln Pvp = ; Pvp = atm = mm Hg.
(4.7)
Narration: Estimated the vapor pressure at 298 K for Diethyl ether, and in example: normal boiling point, its boiling point was estimated to be K. The experimental value for the boiling point is K. We will estimated the vapor pressure using both the predicted and the experimental value for boiling point.
Repeating the calculation for the experimental boiling point leads to a vapor pressure estimated of Pvp = atm = mm Hg.

14. Octanol-Water Partition Coefficient
Describes partition between an aqueous phase and it’s suspended organic phases.
Can be estimated using the substance’s structure :
log Kow = Σ nifi (4.9)
log Kow (corrected) = Σ nifi + Σ njcj (4.10)
Where:
Kow : Octanol-Water Partition Coefficient.
ni : number of groups i in the compound.
fi : factor associated with the group i
nj : number of groups j in the compound that have correction factors.
cj : correction factor for each group j
Narration : This property is critcal for the estimate of other environmental properties and is often used in environmental and toxicological fate modeling. It is also an important factor used in the estimation of bioaccumulation. This partitioning coefficient is also used to calculate other physical and chemical properties.
The factors required for these estimations can be found in tables and (Green Engineering, Allen and Shonnard, pp ).

15. Example : Octanol-Water Partition Coefficient Estimation
Estimate the Octanol-Water Partition Coefficient for diethyl ether.
Solving :
Using equation 4.9 : log Kow = Σ nifi
log Kow = (0.5473) + 2(0.4911) + (1.2566)
log Kow = ≈ therefore Kow = 11.2
Group
-O-
2(-CH3)
2(-CH2)
fi contribution
2(0.5473)
2(0.4911)
Narration: The octanol-water coefficient for this compound is strongly pH-dependent, but this estimation leads to reasonable estimates for slightly basic solutions. Diethyl ether does not contain any groups that require correction terms.

16. Bioconcentration Factor
Describes partitioning between aqueous and lipid phases in living organisms.
Higher bioconcentration factors = higher quantity of bioaccumulation in living organisms
Can be calculated using :
log BCF = 0.79*(log Kow) – (4.11)
log BCF = 0.77*(log Kow) – Σ jj (4.12)
Narration : Describes the ratio of a chemical’s concentration in the fatty tissue of an aquatic organism to its concentration in water. The first equation is the original one, whereas the second is adjusted for non-ionic compounds.
The jj factor required for these estimations can be found in the table (Green Engineering, Allen and Shonnard, pp.110).
Where :
BCF : Bioconcentration Factor.
Kow : octanol-water partition coefficient.
jj : correction factor for each group.

17. Example : Bioconcentration Factor (BCF) Estimation
Estimate the Bioconcentration Factor for diethyl ether.
Solving :
Using equation 4.9 we obtain log Kow :
log Kow = Σ nifi
log Kow = ≈ 1.05
Using equation 4.11 we can calculate BCF :
log BCF = 0.79*(log Kow) – 0.40
log BCF = 0.79* (1.05) – 0.40
log BCF =
therfore BCF =
Narration: Before estimating the BCF, it is first necessary to estimate Kow, and then, proceed to calculate BCF for Diethyl ether. In this case, the value for Kow is taken from the previous example. Again, Diethyl ether does not contain any groups that require correction terms and equation 4.12 does not apply.

18. Water Solubility Can be calculated using :
Used to assess concentrations in water
Can be calculated using :
Log S = – logKow – (Tm –25) + Σhj (4.13)
Log S = –0.854 logKow – (MW) + Σhj (4.14)
Log S = – 0.96 los Kow – (Tm –25) – (MW) + Σhj (4.15)
Where :
S : water solubility (mol/L).
Kow : octanol-water partition coefficient.
Tm : melting point (ªC).
MW :s the molecular weight of the substance.
hj is the correction factor for each functional group j.
Narration : To estimate saturation/concentration of a particular chemical substance in water. This property is often estimated using the octanol-water partition coefficient. The first 2 equations can be more accurate if all the data is obtainable. Any of the three equations can be used, but generally, if more information is available for the correlation (eqns or 4.15), the estimate is more accurate.
The hj factor required for these estimations can be found in the table (Green Engineering, Allen and Shonnard, pp.112).

19. Example : Water Solubility Estimation
Estimate the Water Solubility for diethyl ether.
Solving :
Equation 4.9 gives the log Kow ≈ 1.05
Using equation 4.14 we can calculate the S :
Log S = –0.854 logKow – (MW) + Σhj Log S = – 0.854(1.05) – (74.12) + 0.0
Log S =
Therfore : S = mol/L. = g/L = 16, mg/ L
Narration: Before estimating water solubility, it is first necessary to estimate Kow,. The octanol-water coefficient is taken from the previous calculations, and again, this substance does not contain any groups that require correction terms

20. Henry’s Law Constant Describes the affinity for air over water.
Can be determined using :
-log H = log (air-water partition coeff) = Σ nihi + Σ njcj (4.19)
Where :
H : dimensionless Henry’s Law Constant.
ni : number of bonds of type i in the compound.
hi : bond contribution to the air-water partition coefficient.
nj : number of groups of type j in the molecule.
cj : correction factor for each group.
Narration : Is commonly used in environmental fate descriptions, particularly those relating atmospheric and water relationships. H can also be described in terms of atm-m3/mol. The difference between this relationship and the others seen in this chapter is that the constant is described in terms of bonds instead of functional groups.
Values for cj and hi for these estimations can be found in tables and (Green Engineering, Allen and Shonnard, pp ).

21. Example : Henry’s Law Constant Estimation
Estimate the Henry’s Law Constant for diethyl ether.
H H H H
H-C-C-O-C-C-H
Expressed as a collection of bonds, diethyl ether consists of 10 C-H, 2 C-C bonds, and 2 C-O bonds. The uncorrected value of log (air to water partition constant) is given by :
Narration: A preliminary estimate of the Henry’s Law Constant is obtained by summing each of the bond contributions. This preliminary estimate is then adjusted by correction factors for selected functional groups (Meylan and Howard,1991).
-log H = log (air-water partition coefficient) =
10( ) + 2(0.1163) + 2(1.0855) =
log H-1 =

22. Soil Sorption Coefficient
Used to describe the Soil-Water Partitioning.
Can be estimated by :
log Koc = (log Kow) (4.16)
log Koc = (log S) (4.17)
log Koc = χ Σ njPj (4.18)
Where :
Koc : Soil Sorption Coefficient (μg/g of organic carbon (to μg/mL of liquid)).
Kow : Octanol-Water Partition Coefficient.
S : Water Solubility.
1χ : first order Molecular Connectivity Index (from literature-appendix ).
nj : number of groups of type j in the compound.
Pj : correction factor for each group j.
Narration : This property describes the ratio of mass of a compound adsorbed per unit weight of organic carbon in a soil to the concentration of the compound in a liquid phase. The first 2 equations given in this slide estimate coefficients for a very limited range of substances and the last equation was developed to fit more substances by using a structural parameter called the molecular connectivity. Calculating the molecular connectivity index will be demonstrated in the following slides.
The factors required for these estimations can be found in the table and appendix B(Green Engineering, Allen and Shonnard, pp.117 and ).

23. Molecular Connectivity Index Calculations
The first step in calculating 1χ is to draw the bond structure of the molecule. For example, isopentane would be drawn as:
CH3
H3C-CH-CH2-CH3
The second step is to count the number of carbon atoms to which each carbon is attached. Each C-C bond is given a value of 1 and δi, is the parameter that defines the quantity of carbon atoms connected to a carbon atom i. The diagram below gives the δi, values for the different carbon atoms.
Narration : the molecular connectivity index is needed to calculate the soil sorption coefficient. In the second step, it is important to count any heteroatom as a carbon, but to ignore bonds to hydrogen.
(1)
CH3
H3C-CH-CH2-CH3
(1)
(3)
(2)
(1)

24. Molecular Connectivity Index Calculations (continued)
The third step is to identify the “connectedness” of the carbons connected by the bond (δi , δj). For isopentane, these pairs are:
(1,3)
CH3
H3C-CH-CH2-CH3
(2,1)
(1,3)
(3,2)
The value of 1χ can then be calculated using the equation :
1χ = Σ(δi* δj) (4.19)
For isopentane,
1χ = (1/√3) + (1/√3) + (1/√6) + (1/√2) = 2.68

25. Example : Soil Sorption Coefficient Estimation
Estimate the Soil Sorption Coefficient for diethyl ether.
Solution :
The molecular structure for diethyl ether is :
CH3-CH2-O-CH2-CH3
Using previously calculated values for log Kow (estimated at ) and log S (estimated at ) we can estimate the soil sorption coefficients using equations 4.16 and 4.17 :
log Koc = (log Kow) =
log Koc = (log S) = 3.99
Narration: On the next slide we will calculate the soil sorption coefficients using the molecular connectivity index.

26. Example : Soil Sorption Coefficient Estimation
Using the molecular connectivity we can also estimate the soil sorption coefficient :
First the molecular connectivity index is calculated using eq :
CH3-CH2-O-CH2-CH3 (molecular structure)
2(C-C), 2(C-O), 2(1, 2) , 2(2, 2) (connection pairs)
therefore : 1χ = 2(1/√2) + 2(1/√4) = 2.414
Using equation 4.18 to calculate the soil sorption coefficient :
log Koc = χ Σ njPj
log Koc = χ Σ njPj = 0.53(2.414) (-1.264)
log Koc =
therefore : Koc = 4.32
Narration : in this case, the molecular connectivity is used to estimate the soil sorption coefficient

27. Where to look up this information...
Narration : These are some of the many references available for finding chemical and physical properties for various substances.

28. What do the different Properties mean?
Narration: This table sums up some of the classification criteria of the various properties.
Adapted from the Green Engineering Textbook

29. Estimating Environmental Persistence and Ecosystem Risks
To be discussed :
Atmospheric Lifetimes
Aquatic Lifetimes
Overall Biodegradability
Ecosystems
Narration : Only a limited quantity of information concerning these topics will be discussed.

30. Estimating Atmospheric Lifetimes
One way to estimate the atmospheric lifetime of a compound is to analyze the rate of oxidation of the substance, specifically the hydroxyl radical reaction rate.
Group contributions is again one of the approaches that can be taken to estimate this property.
Using examples, we will show how to estimate reaction rates and half lives while using the appropriate correction factors.
Narration : This analysis method is very important for organic compounds. Hydroxyl radicals can alter many organic compounds because they are highly reactive. For example, they can remove a hydrogen from a saturated organic compound, add to double bonds or add to aromatic rings. These reactions show that certain organic compounds can have very short lifetimes in the atmospher (and in the presence of hydroxyl radicals).

31. Example : Atmospheric Lifetime Estimation
Dimethylsulfide (DMS, CH3SCH3) produced by phytoplankton degredation is thought to be the major source of the sulfate and methanesulfonate aerosol found in the marine boundary layer.
The primary objective of this research effort is to determine the detailed mechanism of, and final product yields from, the OH initiated gas phase oxidation of DMS.
At the low NOx levels that are characteristic of the remote marine boundary layer, reaction with OH is the initial step in DMS oxidation.
Narration: OBJECTIVES. Our objectives include the determination of a) the effective rate coefficients for the OH initiated oxidation of DMS under atmospheric conditions, and b) the elementary rates for adduct formation and adduct reaction with O2. By comparing these data sets we can evaluate the overall consistency of our proposed oxidation mechanism.
There is considerable evidence that some of the impact of DMS oxidation on new particle formation and marine boundary layer (MBL) visibility occurs via oxidation in the upper troposphere. While DMS oxidation occurs rapidly in the MBL it has been suggested that the oxidation products will primarily condense on existing particles. An accurate value of this rate coefficient is critical in defining the lifetime of DMS in the marine boundary layer and hence the rate at which it is oxidized to products, some of which may act as condensation nuclei. This will allow us to provide a definitive expression for the OH initiated oxidation of DMS that will allow the calculation of accurate values for both the overall reaction rate coefficient and the branching ratio between addition and abstraction at any location in the troposphere.
OH + CH3SCH3 ⇒ Products (1)

32. The OH initiated oxidation of DMS proceeds via a complex, two channel, mechanism involving abstraction (1a) and reversible addition (1b, -1b). This can be described by the reaction sequence:
CH3SCH3 + OH ⇒ CH3SCH2 + H2O (1a)
CH3SCH3 + OH + M ⇔ CH3S(OH)CH3 + M (1b, -1b)
CH3S(OH)CH3 + O2 ⇒ Products (3)
Because of this complex mechanism the effective rate coefficients for reaction (1) and its deuterated analog, reaction (2) depend on the partial pressure of O2 at any total pressure.
OH + CD3SCD3 ⇒ Products (2)
The two channel reaction mechanism implies that in the absence of O2 we measure k1a, the abstraction rate. As we add O2 the effective rate increases until we measure a limiting rate (k1a + k1b).

33. Narration: Fig. 1 shows the O2 enhancement in the effective rate coefficients for the reactions of OH with DMS at 240K and at total pressures of 200 and 600 Torr. The behavior is consistent with the two channel mechanism showing a characteristic “roll-off” at high O2. The limiting value of the enhancement at 200 Torr is ~ 1x10-11 cm3 molecule-1 s-1 and increases to ~2.5x10-11 cm3 molecule-1 s-1 at 600 Torr.
These results show that for the effective rate of the OH initiated oxidation of DMS significantly underestimates both the effective rate and branching ratio between abstraction and addition at low temperatures. Current models of the high latitude oxidation of DMS should be significantly impacted by these results.

34. Estimating Aquatic Lifetimes
One way to estimate the aquatic lifetime of a compound is to analyze the rate of hydrolysis of the substance.
The rate of hydrolysis can be estimated by :
log (hydrolysis rate) = log (hydrolysis rate of a reference compound)
+ Constant * σ
Therefore log (hydrolysis rate) = A + Bσ (4.20)
Narration: An added complexity is that rates of reactions, such as hydrolysis, depend not just on the structure of the reactant, but also on the characteristics, of the receiving waters.
Where :
A is rxn and compound class specific(depends on the reference rxn chosen)
B is rxn and compound class specific (depends on type of rxn considered)
σ is a structural parameter commonly used in linear free energy relationship.

35. Estimating Overall Biodegradability
It is difficult to do an overall biodegradability analysis.
It can be estimated using :
Where :
an is the contribution of the functional group (see table ).
fn is the number of different functional group.
MW is the molecular weight.
I is an indicator of aerobic biodegradation rate.
Different Values (of I) represent different life times :
I = a1f1 + a2f2 + a3f anfn + amMW (4.21)
Narration : the rate at which a compound is metabolized (by living organisms) is also important to consider (in addition to its reaction with other compounds in the environment). An overall biodegradability analysis on a substance is hard, however, it is important to consider the original form of the compound, the initial structural stage, the final form(s) of the compound as well as the aerobic and anaerobic stages of degradation. Only possible to analyze a minimal quantity of compounds because of the lack of data. Although it may be difficult to obtain a quantitative value, it is possible (and important) to get a qualitative idea of the compound’s environmental persistence. This can be achieved by using the given equation. Once the I value is determined, by verifying the chart, it is possible to give an expectance for the compound’s lifetime.
I value 5 4 3 2 1
Expected degradation rate
Hours
Days
Weeks
Months
Years

36. Example : Overall Biodegradability Estimation
Estimate the Biodegradation Index for diethyl ether.
Solution :
Molecular weight of diethyl ether :
MW = g/mol
Using equation 4.21, the index can be calculated :
I = a1f1 + a2f2 + a3f anfn + amMW
I = ( ) (74.12) =
Therefor a lifetime of WEEKS
Narration : Diethyl ether contains an aliphatic ether.

37. Estimating Ecosystem Risks
Compare the Fish, Guppy and Daphnids mortalities for an acrylate with log Kow = 1.22 (e.g. ethyl acrylate).
Guppies
log (1/LC50) = log Kow – (4.22)
log (1/LC50) = 0.871(1.22) – 4.87 =
LC50 = µmol/L.
Daphnids
Narration: In assessing ecosystem hazard, the standard practice is to estimate toxicity for a variety of species. For example, mortality for daphnids, fish, and guppies are frequently used in assessing the ecosystem hazard of chemicals described in premanufacture notices submitted to the US EPA.
The mortality for Guppies can be correlated with the octanol-water partition coefficient using the equation (4.22), where LC50 is the concentration that is lethal to 50% of the population over a 14-day exposure (expressed in µmol/L). Other equations used to estimate ecosystem hazard are specific to certain compound classes. For example, toxicities for daphnids and fish can be estimated using equations (4.23), (Daphnids, mortality after 48 hr exposure) and (4.24),(Fish, mortality after 96 hr exposure); where LC50 is expressed in units of millimoles/L.
log LC50 = – log Kow (4.23)
log LC50 = – (1.22) =
LC50 = millimoles/L = 242 µmol/L.

38. Fish Estimating Ecosystem Risks Continued
log LC50 = – 0.18 log Kow (4.24)
log LC50 = – 0.18(1.22) =
LC50 = millimoles/L = 21 µmol/L.
The concentrations yielding 50% mortality are:
Guppies (14 day): µmol/L.
Daphnids (48 hour): millimoles/L = 242 µmol/L.
Fish (96 hour): millimoles/L = 21 µmol/L.

39. Environmental Fate and Exposures
Example : If chemicals are released into a river upstream of a water treament plant, what factors need to be taken into account to estimate the potential danger to the community. What fraction of the chemicals are:
- Absorbed by river sediments. - Volatilized into the air.
- Taken up by living organisms. - Biodegraded.
- Reacted with other compounds. - Removed in the treatment process.
Narration : environmental fate and exposure calculations use the basic concepts of mass balances

40. Classification of Substances Based on Risk
By examining the table XX, we can use the calculated properties to qualitatively quantify the risk associated with the different substances
Three main criteria are normally considered in the classification of the substances : persistence, bioaccumultion and toxicity.
There do not exist a given set of regulations or guidelines on quantifying risk, but the above parameters are used in the process.
Narration : see the following slide for additional references

41. Available Ressources
EPA (persistent, bioaccumulating and toxic substances) :
Pollution Prevention, Waste Minimization and PBT Chemical Reduction :
Environment canada (existing substances evaluation) :