Transcript CE 510 Hazardous Waste Engineering

CE 510
Hazardous Waste Engineering
Department of Civil Engineering
Southern Illinois University Carbondale
Instructors: Jemil Yesuf and
Dr. L.R. Chevalier
Lecture Series 3:
Hazardous Waste Properties and Classification
Course Goals
 Review the history and impact of environmental laws in the






United States
Understand the terminology, nomenclature, and
significance of properties of hazardous wastes and
hazardous materials
Develop strategies to find physical and chemical
properties, i.e., water solubility, density, flammability,
and toxicity for hazardous compounds
Elucidate procedures for describing, assessing, and
sampling hazardous wastes at industrial facilities and
contaminated sites
Predict the behavior of hazardous chemicals in surface
impoundments, soils, groundwater and treatment systems
Assess the toxicity and risk associated with exposure to
hazardous chemicals
Apply scientific principles and process designs of hazardous
wastes management, remediation and treatment
Concentration - mg/L to ppm
In environmental engineering, we are
sometimes dealing with dilute aqueous
solutions. In such a case, we assume that the
substance does not change the density of
water.
Based on this assumption, we can relate mg/L
to ppm.
Concentration - mg/L to ppm
From the density of water, we know that
10-3 ml of water weighs 1 mg.
w 1
g
ml
 1000 m g / m l
Concentration - mg/L to ppm
Substituting the volume of water for the mass
of water in 1 mg/L we obtain the following:
1
mg
L

1 mg
1 kg

1 mg
6
10 mg
 1 part per million
 1 ppm
THEREFORE, IN OUR DILUTE AQUEOUS SOLUTIONS
WE COMMONLY INTERCHANGE mg/L WITH ppm
Class Example
A 500-L tank contains 255 mg/L 2,4,6trichlorophenol dissolved in hexane. What is
the trichlorophenol concentration in ppm?
Assume the temperature of the system is
20°C.
Specific gravity of Hexane is 0.6603
(Appendix C)
Solution
Mass of trichlorophenol dissolved in hexane
is:
(255 mg/L)(500L) = 127,500 mg = 0.128 kg
Mass of hexane as a solvent is:
(500 L)(0.6603)(1000kg/1000L) = 330 kg
Therefore, concentration of trichlorophenol
in hexane is:
(127,500 mg)/(330 kg) = 386 mg/kg
= 386 ppm
......end of example
Concentration - air
 In dilute aqueous solutions, ppm is a mass-




to-mass ratio
In air, ppm is a volume-to-volume ratio
Changes in temperature and pressure do not
change the volume of the pollutant gas to
the volume of the air that contains it
i.e. we can compare ppm from Washington
D.C. to Denver without further conversion
We will define the Ideal Gas Law to further
understand this…
Ideal Gas Law
P V  nR T
w here
P  absolute pressure  atm  or  kP a 
 
V  volum e  L  or m
3
n  m ass [ m ol ]
T  absolute tem perature  K 
R  proportion ality const . or ideal gas const .
 0 .0821 L  atm / K  m ol 
 8 .3143 J / K  m ol 
Example
How much volume in liters will one
mole of any ideal gas occupy at
standard temperature and pressure
(STP).
Note: STP is 273.13°K and
101.325 kPa (0°C and 1 atm).
Solution
Use the ideal gas law to solve for volume.
Note: J = N. m Pa = N/m2
 N m 
 m ole  
K
 K  m ole 
V 
N
m

1
 m
3
2
m ole   8 .3143
  273.16
101.325 kP a  1000 P a kP a 
J
K  m ole
K
1000 L m 
3
 22 .414 L
......end of example
Solution
Similarly, if we consider the volume at 25°C
 N m 
 mole 
K 
 K  mole 
3
V 
m
N
m

2
1mole 8 . 3143 J K  mole 273  25  K 
1000
Pa
101 . 325 kPa 1000 kPa 
L
m
3

 24 . 45 L
......end of example
Example
The chemical para-dichlorobenzene (p-DCB)
is used in an enclosed area. At 20C (68F)
the saturated vapor pressure of 1,4-DCB is 5.3
x 10-4 atm. What would be the concentration
in the air of the enclosed area (units of g/m3)
at 20C ?
Solution
Rearrange the Ideal Gas Law to solve for the
concentration of 1,4-DCB in the air
n
V

P
RT

5 . 3  10
4
atm
L  atm 

o
0
.
08205
293
K


mol  K 



g 

 5 mol   1000 L  
  2 . 2  10

  147

3
L  m
mol 


 3 .2
g
m
3
Example
Anaerobic microorganisms metabolize
organic matter to carbon dioxide and
methane gases. Estimate the volume of gas
produced (at atmospheric pressure and 25°
C) from the anaerobic decomposition of 2
moles of glucose. The reaction is:
C 6 H 12 O 6  3CH 4  3CO 2
Solution
Each mole of glucose produces 3 moles of methane and
3 moles of carbon dioxide gases, for a total of 6 moles.
Therefore, 2 moles of glucose produces a total of 12
moles. The entire volume is then
L  atm 


12 mol  0 . 0821
 298 K
nRT
K  mol 

V 

1 atm 
P


 294 L
Note: The volume of 1 mole of any gas is the same. Thus, 1 mole
of carbon dioxide gas is the same volume of 1 mole of methane
gas.
Concentration - Air
Correction factor for conditions other than STP
L  T2

V   22 . 414


mole
273
K


ppm 
volume
  101 . 325 kPa
 
P2





3
conta min ant
volume
(m )
3
air
 10
6
(m )
Incorporate this
factor in order to
report as ppm
Concentration - Air
Based on these two equations, we can convert mg/L (g/m3) to
ppm using
C ppm
 RT 
6
 C
  10
 PM 
M is the molecular weight in g/mole
Concentration is g/m3
Example
Convert 80 mg/m3 of SO2 in 1 m3 of
air, 25° C, 101.325 kPa to ppm
Solution
C ppm
 RT 
6
 C
  10
 PM 

 80  10
3
 30 . 6 ppm

J


8
.
314

  273  25  K 
K  mol 

6
 10
Pa  
g 

101 . 325 kPa  1000
  64

kPa  
mol 

Water Solubility
 the maximum concentration of a substance that will
dissolve in water at equilibrium at a given temperature
and pressure
 Controls



Concentration of a chemical in groundwater
Proportion that exists as a free product
Partitioning into solids
 Inversely related to sorptivity, bioaccumulation, and
volatilization
 Affects a number of other pathways such as
biodegradation, photolysis, chemical oxidation
 Together with another chemical parameter, Kow,
determine the fate of chemicals in the environment
Equilibrium
Most chemical reactions are to some extent
reversible when the rates of reaction are the
same, i.e. the products are formed on the left
at the same rate at which they are formed on
the right.
In such a case, the reaction is said to have
reached equilibrium.
Generalized Reversible Reaction
aA  bB  cC  dD
where
a , b , c , d # of
molecules
A , B , C , D chemical
 reaction
same time
species
in both directions
at the
Generalized Reversible Reaction
aA  bB  cC  dD
C   D 
a
b
 A  B 
c
d
 K
Here, K is the equilibrium constant
Provides the ratio of the concentration of
individual reactants and products for any
reaction at equilibrium
Use this equation with a degree
of caution
 Valid only when chemical equilibrium
is established
 Natural systems are often subject to
constantly changing inputs
 Some reactions occur very slowly
 Equilibrium may never be established
Important Equilibrium Processes in
Environmental Systems
 Acids and bases
 Dissolved and precipitated chemicals
 Pure compounds and air (volatilization)
 Chemicals dissolved in water and air
(Henry’s Law)
 Chemicals dissolved in water and
adsorbed on a solid (adsorption and ion
exchange)
Acid-Base Chemistry
 Among the most important in
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

environmental engineering
Waste must be neutralized
Aquatic life sensitive to changes
The concentration of acid/base is given by
the pH
By controlling pH, unwanted substances
may be driven out of solution as gases or
precipitates
Acid-Base Chemistry

HA  H  A

We’ll evaluate this equation further later.
First, lets consider the dissociation of water
and the definition of pH
Acid-Base Chemistry
Water dissociates slightly into
Hydrogen Ions (H+)
Hydroxide Ions (OH-)
H 2O  H

H



OH
H 2 O
 OH



 K
Acid-Base Chemistry
 H  O H   K


pH= -log [H+]
[H+] = 10-pH
w
 10
 14

@ 25 C
Example
What percentage of total ammonia
(i.e. NH3 + NH4) is present as NH3 at a
pH of 7? The pKa for NH4+ is 9.3.
NH

4
 H
K a  10
 9 .3


 NH
3
 NH 3 H
NH 

4


Solution
The problem is asking:
 NH 3 
 NH 3   NH

4

 100
Solution
The problem is asking:
 NH 3 
 NH 3   NH

4

 100
However, this expression has two unknowns.
Therefore, we need a second equation.
?????
Solution
The problem is asking:
 NH 3 
 NH 3   NH

4

 100
K a  10
 9 .3

 NH 3 H
NH 

4


Solution
K a  10
 9 .3

 NH 3 H

NH 
 NH 10 

NH 

4
7
K a  10
 9 .3

3

4
Recall, pH=7
means
[H] = 10-7
Solution
K a  10
Therefore:
 9 .3

 NH 3 10
7
NH 


4
NH   200  NH 

4
3
Solution
 NH 3 
 100

 NH 3   NH 4 
 NH 3 

 100 %
 NH 3   200  NH 3 
 0 .5 %
Acid-Base Chemistry
Relationship between pH and pKa
Consider a generalized acid reaction
HA  H

 A

Acid-Base Chemistry

H  A 


Ka

Take the logarithm
 HA 
log K a  log H

  log
A 

 HA 
or
A 

pH  pK a  log
 HA 
Henderson-Hasselbalch equation
Example
How is pH related to pKa when the acid
is 50% ionized? (i.e. [A-] = [HA])
Solution
A 

pH  pK a  log
 HA 
 pK a  log 1 
or
pH  pK a
......end of example
Solubility Product
Solid Chemical  Dissolved Chemical
Equilibrium Constant Notation:
Solubility Product (Ksp)
Solubility Product
 All complexes are soluble in water to a
certain extent
 Likewise, all complexes are limited by
how much can be dissolved in water
 Example


NaCl very soluble
AgCl only a small amount will go into
solution
Solubility Product
 Visualize a solid compound being
placed in distilled water
 Some of the compound will go into
solution
 At some time, no more of the
compound will dissolve
 At this point, equilibrium is reached


may take seconds
may take centuries
Solubility Product
Consider the following
NaCl  s   Na

 Cl

The solid form, NaCl, may be dissociating into
its ionic components (dissolution) OR
The ionic components may be recombining into
the solid form (precipitation)
Solubility Product
The conditions of equilibrium can be
expressed in an equilibrium or mass
action equation for a generalized
reaction.
A x B y  xA  yB
Solid
Compound
 A x  B  y
A
x
By

 K
Ionic
Components
Solubility Product
 The brackets indicate molar concentrations
 K is an equilibrium constant for a given substance in
pure water at a given temperature
 At equilibrium, the solid phase does not change
concentration because dissolution and precipitation
are equal.
Solubility Product
 The brackets indicate molar concentrations
 K is an equilibrium constant for a given substance in
pure water at a given temperature
 At equilibrium, the solid phase does not change
concentration because dissolution and precipitation
are equal.
A
x
B y   K s  const .
 A   B   KK s  K sp
x
y
Solubility Product
 Supersaturated Solution



Created by dissolving a solid at an
elevated temperature and allowing it to
cool
Once cooled, precipitation may not occur,
although the solution is not at equilibrium
Precipitation will occur if the reaction
vessel is shaken or otherwise disturbed
Solubility Product
 Unsaturated Solution


Not at equilibrium
Can dissolve more solid
 Saturated Solution


At equilibrium
Cannot dissolve more solid unless the
temperature or pressure is increased
Solubility Product
 A  B 
x
y
 K sp
If this product is > Ksp, we have a supersaturated solution
If this product is < Ksp, we have an unsaturated solution
If this product is = Ksp we have a saturated solution
Example
The solubility product for the dissociation of
Mg(OH)2 is 9 x 10-12. Determine the
concentration of Mg2+ and OH- at
equilibrium.
Solution
1. Write the equation for the reaction.
Mg  OH  2  Mg
2
 2 OH

2. The solubility product equation is:
 Mg  OH 
2

2
 9  10
12
Solution
3. If x is the number of moles of Mg2+ resulting from
the dissociation, then the number of moles of OHis equal to 2x.
 x  2 x   9  10 12
2
4 x  9  10
3
x  1 . 3  10
12
4
2 x  2 . 6  10
moles / L  Mg
4
moles / L  OH
.....end of example
Solubility Product
 A   B   K sp
x
y
 Ksp is the solubility product for the ion pair
 If the concentration of either or both of the
ions is increased (without a change in
temperature or pressure), the product of the
ionic concentration will exceed Ksp
 Precipitation will occur to maintain
equilibrium conditions
Solubility Product
 Precipitation can occur when a
chemical reaction transforms a solute
to a much less soluble form, typically
mixing a precipitant with the solution
 Precipitation depends greatly upon pH
 Most metals precipitate at high pH
levels as hydroxides
Example
Magnesium is removed from an
industrial waste stream by hydroxide
precipitation at a pH = 10. Determine
the solubility of Mg2+ in pure water at
25° C and pKsp of 10.74.
Mg  OH 2   s   Mg
2
 2 OH

Solution
1. What are your two governing equations?
 Mg  OH   10
 OH  H   10
2



2
10 . 74
14
2. Two unknowns, and two equations.
Solution
3. Given the pH, we know [H+].
 H   10

 pH
 10
10
4. Solve for [OH-]2
 OH  

 OH 

2
10
14
H 
 10


10
10
 28  2 pH
14
 pH
 10
14  pH
Solution
5. Substitute into 1st governing equation, and
solve for [Mg2+].
 Mg
2


10
10
10 . 74
 28  2 pH
 10
17 . 26  2 pH
Solution
6. Substitute value of pH given in the
problem statement, then convert to mg/L.
NOTE: units in [ ] are moles per liter!
 Mg
10
2
  10
  24 . 3
 2 . 74 mol
 2 . 74 mol
L
L
g
3

10
mol 
mg
g

 44 . 2 mg
L
Solution
7. For a pH of 11, the solubility is 0.442 mg/L.
For a pH of 12 the solubility is 0.004 mg/L.
Work these solutions on your own.
..... end of example.
General Comments on Ksp
 Ksp is dependent on



pH
temperature
pressure
 The higher the Ksp, the more soluble
the compound
LNAPL
LNAPL
saturated
C.F.
unsaturated
Generalized Schematic: LNAPL Spill
water table
impermeable boundary
LNAPL
LNAPL
water table
dissolved phase transport
impermeable boundary
saturated
C.F.
unsaturated
Generalized Schematic: LNAPL Spill
LNAPL
LNAPL
water table
dissolved phase transport
impermeable boundary
saturated
C.F.
residual NAPL
vapor and
dissolved phase
transport
unsaturated
Generalized Schematic: LNAPL Spill
water table
DNAPL
DNAPL
impermeable boundary
saturated
C.F.
unsaturated
Generalized Schematic: DNAPL Spill
water table
DNAPL
DNAPL
impermeable boundary
dissolved phase
transport
saturated
C.F.
unsaturated
Generalized Schematic: DNAPL Spill
C.F.
residual NAPL
water table
DNAPL
DNAPL
impermeable boundary
dissolved phase
transport
saturated
vapor and
dissolved phase
transport
unsaturated
Generalized Schematic: DNAPL Spill
vapor and
dissolved phase
transport
LNAPL
trapped residual NAPL
water table
DNAPL
DNAPL
impermeable boundary
dissolved phase
transport
saturated
C.F.
residual NAPL
unsaturated
Generalized Schematic: DNAPL Spill
Dissolution of NAPL
 NAPL may have various compounds
 Dissolution is a function of





solubility of compound(s)
groundwater velocity
distribution of the NAPL
pore distribution
molecular diffusion of compound(s)
Based on Pankow, J.F. and Cherry, J.A. 1996. Dense Chlorinated
Solvents and other DNAPLs in Groundwater. Waterloo Press.
Solubility Data
Compound
Chloroform
Carbon Tetrachloride
1,1,1-Trichloroethane
Trichloroethylene
Source
1
8700
780
1250
1400
2
7920
793
1495
1100
3
7925
1000
720
1050
It is not uncommon to find a wide range of
values for solubility across different
reference. Advice: go with the most current
one using a good analytical method.
Solubility Data
Compound
Chloroform
Carbon Tetrachloride
1,1,1-Trichloroethane
Trichloroethylene
Source
1
8700
780
1250
1400
2
7920
793
1495
1100
1. Broholm and Feenstra (1995) 23-24° C
2. Horvath (1982) 25° C
3. Mackay and Shiu (1981) 25° C
3
7925
1000
720
1050
Solubility of NAPL Mixture
For an organic NAPL mixture, the aqueous
phase concentration of each component
that is in equilibrium with the mixture can
be approximated using a solubility analog
of Raoult’s Law for vapor pressure.
C sat , m  X m C
o
sat
Solubility of NAPL Mixture
For an organic NAPL mixture, the aqueous
phase concentration of each component
that is in equilibrium with the mixture can
be approximated using a solubility analog
of Raoult’s Law for vapor pressure.
C sat , m  X m C
o
sat
Aqueous solubility of component m
from the mixture, also referred to as
the “effective solubility”
Solubility of NAPL Mixture
For an organic NAPL mixture, the aqueous
phase concentration of each component
that is in equilibrium with the mixture can
be approximated using a solubility analog
of Raoult’s Law for vapor pressure.
C sat , m  X m C
o
sat
mole fraction of component m
Solubility of NAPL Mixture
For an organic NAPL mixture, the aqueous
phase concentration of each component
that is in equilibrium with the mixture can
be approximated using a solubility analog
of Raoult’s Law for vapor pressure.
C sat , m  X m C
o
sat
solubility of the pure compound
Solubility of NAPL Mixture
C sat , m  X m C
o
sat
 Laboratory experimental studies suggest that this
equation is a reasonable approximation for
mixtures of structurally similarly hydrophobic
organics
 Other studies have shown that with most dissimilar
compounds that deviations from ideal behavior are
small enough to be safely ignored in environmental
systems with much larger uncertainties
Solubility of NAPL Mixture
A comparison of observed groundwater
concentrations with calculated effective
solubilities values have been reported for a
Pennsylvania Superfund site. DNAPL was
found in numerous wells in a fractured
sandstone aquifer, though at different
concentrations. This type of variability is
likely to be common at many multicomponent DNAPL sites. Samples from two
wells are presented on the next slide for
discussion.
Solubility of NAPL Mixture
1 ,2,3TrichloloroProperty
propane
Toluene
Pure compound solubility (ppm) 1 900
580
Molecular weight (g/mol)
1 46
92
Xy lene
200
1 06
Weight %, DNAPL Sample 1
Effectiv e Solubility (ppm)
23
400
4.2
35
17
43
3.8
9.1
Weight %, DNAPL Sample 2
Effectiv e Solubility (ppm)
73
1 400
0.9
8.1
5.8
16
0.9
2.3
Observ ed Groundwater Conc.
Lower V alue (mg/L)
Upper V alue (mg/L)
220
1 200
2.1
53
7 .3
74
1 .1
12
Note: Assumed molecular weight of unidentified
portions of the samples is 150 g/mol
Ethy lbenzene
1 90
1 06
Example Applied Problem
Consider a liter of weathered gasoline spilled
into the soil. Based on the table below,
determine the effective solubilities of each
compound in the water phase due to this spill.
The density of the gasoline is 0.87 g/cm3.
Solution
Compound
% of Total Grams
Mass
Moles
Mole
Fraction
Benzene
Toluene
Ethylbenzene
Xylene
30
50
15
5
3.341
4.721
1.229
0.410
0.344
0.487
0.127
0.042
261
435
130.5
43.5
Total Moles = 9.701
Solution
Compound
% of Total Grams
Mass
Moles
Mole
Fraction
Benzene
Toluene
Ethylbenzene
Xylene
30
50
15
5
3.341
4.721
1.229
0.410
0.344
0.487
0.127
0.042
261
435
130.5
43.5
Total Moles = 9.701
At this stage, you would use the literature to
determine the solubility of each compound of
BTEX and apply
o
C sat , m  X m C sat
Example Applied Problem
A 37,900 L UST containing tolune has leaked and
the tolune is floating at the top of the shallow
aquifer as a free product. Its water solubility is
controlling the concentration of tolune in the
groundwater. If 37.9 x 106 L ( 10 million gallons)
of groundwater have been polluted with tolune
at a concentration equal to its water solubility,
what volume of tolune remains as floating
product?
Water solubility of tolune is 546 mg/L.
Density of tolune is 0.867 kg/L
Solution
Mass of tolune that can be dissolved based on its
water solubility:
= (3.79X106 L)(546 mg/L) = 2.07X1010 mg
= 20,700 kg
Actual mass of tolune spilled:
= (37,900 L)(0.867 kg/L) = 32,900 kg
Mass of tolune not dissolved in water of aquifer
= 32,000 kg – 20,700 kg = 12,200 kg
Therefore, volume of tolune as a free product:
= (12,200 kg)/(0.867 kg/L) = 14,100 L.
..... end of example.
Flammability
 Concern that vapors may ignite from spark or
open flame if at a certain concentration

Lower flammability limit, LFL
 Insufficient mass (conc.) is available for flame
propagation
 Mixture is “too lean” to burn or explode

Upper flammability limit, UFL
 Threshold concentration that limits combustion
 “too rich”
 See Table 3.8 p. 178
Flammability Limits of Mixtures
Most hazardous waste disposal and storage
areas contain mixtures of chemicals
LFL mixture 
1
n

i 1
UFL mixture 
yi
LFL i
1
n
yi
 UFL
i 1
i
where
yi = mole fraction of
compound i
n = the number of
compounds in the mixtures
Approach to Problem Solving
 Determine the v/v % of compounds
 Calculate the mole fraction of each
compound

Number of moles of each gas is directly
proportional to its volume (ideal gas law)
 Use data to determine UFL and LFL
Class Problem
Determine the LFL an UFL of a gaseous mixture 1.30%
benzene, 0.4% pentane and 0.6% ethanol
(volume/volume).
Solution
Compound
v/v
Mole fraction
LFL
UFL
Benzene
1.3
1.13
1.3
7.9
Pentane
0.4
0.35
1.5
7.8
Ethanol
0.6
0.52
3.3
19
Total
2.3
LFLmix
1.59
UGLmix
9.30
Spreadsheet
Flash Point
 Definitions



Flash Point: Minimum temperature at which a
compound emits sufficient vapor to form an
ignitable mixture with air
Ignitable mixture: Vapor-air combination within
flammability limits capable of propagating a
flame away from the source of ignition
Ignition temperature: minimum temperature
required to initiate direct combustion of a
material, whether it is a solid, liquid or gas
Review Table 3.9 p. 183
Labels and Placards
 Two systems

Transportation
 Developed by US DOT

Storage
 Developed by National Fire Protection
Association NFPA
Transportation
 Provides critical data to first
responders
 Placard must be placed on all four
sides of the vehicle
 9 Classes of hazardous materials (DOT)


Use universal icons for each hazard code
See p. 192 Table 3.13
Nine Classes of Haz. Mat. Used
by the DOT
Labels and Placards
DOT hazard symbol
Hazard Code
1375
Office of Hazardous Materials Safety
(HAZMAT)
HAZMAT Guidebook
UN/NA
ID #
Storage
 Regulated by NFPA
 Focuses on three categories



Health (for firefighting, as
opposed to occupational or
public health
Flammability
Reactivity
Fire
Health
 Each category has five levels
– Table 3.15 p.197


0 No hazard
4 Severe hazard
NFPA
1
4
Reactivity
2
W
Hazard Class
Incompatible with Water
Priority Pollutants
 Result of a law suit filed by the Natural
Resources Defense Council (NRDC)
against the EPA
 Identification of the chemicals that
would be regulated under the 1977
Amendment to the Clean Water Act

See Table. 3.38 p. 203
Chemical Abstracts Service
 Unique classification

Ambiguity is found in numerous IUPAC, common
and trade names
 Multi-digit numbers derived from computer
language description of chemical’s molecular
structure
 Equivalent to a “social security number” for
chemicals
 See Table 3.17 p. 200
Resources
 Cross Reference Index of Hazardous
Chemicals, Synonyms and CAS Registry
Numbers
 Handbook of Environmental Data on Organic
Chemicals

Solubility and density
 CRC Handbook of Chemistry and Physics
 Fire Protection Guide on Hazardous Materials

Flash point, ignition temperature and hazardous
labeling
 MSDS
Summary of Important Points
and Concepts
 Water solubility significantly influences
environmental transport and transformation
 Hazardous contaminants may not mix with water
(LNAPL, DNAPL)
 Parameters that aid in determining explosivity and
flammability

Flammability
 Range of flammable concentration in air

Flash Point
 Temperature that results in sufficient volatilization for
compound to reach lower flammability limit in the air above
the liquid

Ignition temperature
 Minimum temperature for combustion
Summary of Important Points and
Concepts
 Two label of placard systems are
universal


DOT and UN/NA
NFPA
 CAS provides a definitive means of
identifying hazardous chemicals
 Priority Pollutants identified