Transcript ENVS4450 - Robert Morris University

Environmental Modeling
Chapter 3:
Quantitative Aspects of Chemistry
Copyright © 2006 by DBS
Quote
“The noblest of the elements is water”
-Pindar, 476 B.C.
Where Next?
Next step:
Convert chemical process into a ‘mathematical
form’ that can be integrated into an
environmental model
Cannot fit all chemistry into models
…has to be simplified
Concepts
•
•
•
•
•
•
Free metal ion concentration using equilibria
Determining KD and KP
Kinetics of sorption
Sorption isotherms
Kinetics of transformation
Modeling
Free Metal Ion Concentration
Chemical Equilibria
General Chemistry
(simplified solutions)
Environmental Chemistry
(more complex solutions activities)
Chemical Fate
(computer programs –
GEOCHEM, MINTECH,
MINEQL)
Free Metal Ion Concentration
Chemical Equilibria
• General Chemistry
e.g. Pb2+ + 2Cl- → PbCl2(s)
• But! Lead is also present as:
PbCl+, PbCl2(aq), PbCl3-, and PbCl42-
General Chemistry
(simplified solutions)
Environmental Chemistry
(more complex solutions activities)
• Whys is this important?
– Speciation determines transport
– Variable toxicity/bioavailability
• Which ion is most toxic?
– Usually hydrated free metal e.g. Pb2+ not PbCl+ (except Hg)
Free Metal Ion Concentration
Review of Chemical Equilibria
Gen. Chem.:
aA + bB
K=
cC + dD
[C]c[D]d
[A]a[B]b
Idealized!
Where K is a constant as [A], [B], [C], and [D] varies
Equilibrium constant K has been found to vary with
concentration! So it’s not really constant…
Free Metal Ion Concentration
Review of Chemical Equilibria
Free Metal Ion Concentration
Review of Chemical Equilibria
Use activities rather than concentrations
K=
(AC)c(AD)d =
(AA)a(AB)b
[C]c γc [D]d γd
[A]a γa [B]b γb
For K to remain
constant Ax must
remain constant
If [X] increases,
γx must decrease
Where A is the activity and γ is the activity coefficient and γ C = A
For ions in DI water:
[X] = AX , γX = 1.00
Free Metal Ion Concentration
Review of Chemical Equilibria
General Rule
If you add an inert salt to a solution you increase the solubility of
another salt when the 2 do not share a common ion (exception:
anions ligate to form insoluble complexes)
We will prove this with the following 2 examples…
Free Metal Ion Concentration
Review of Chemical Equilibria
Example:
• CaSO4 (Ksp = 2.5 x 10-5) in DI water, C = A
Ksp = ACa2+ ASO42- = [Ca2+][SO42-]
Let x = cation and anion concentrations:
Ksp =
x 2,
and x =
[Ca2+]
= [SO4
2-]
= 4.9 x
10-3
M
Solubility of CaSO4
is limited by:
attraction between
+ve and –ve ions
Decrease interactions between Ca2+ and SO42- by adding
another electrolyte…ion concentrations should increase!
Free Metal Ion Concentration
Review of Chemical Equilibria
Example:
• CaSO4 (Ksp = 2.5 x 10-5) in 0.02 M KNO3
SO42-
K+
Ca2+
NO3-
More
interactions
SO42-
Ca2+
Free Metal Ion Concentration
Review of Chemical Equilibria
Example:
• CaSO4 (Ksp = 2.5 x 10-5) in 0.02 M KNO3
Since concentration CaSO4 << KNO3 ionic strength (μ) is dominated by KNO3
First calculate μ
= 0.500  CiZi
= 0.500([K+] + [NO3-]) = 0.500(0.02+0.02) = 0.02 M
Free Metal Ion Concentration
Review of Chemical Equilibria
•
Use Debye-Hückel equation to find activities:
γ=
0.512 Z2 (μ)1/2
1 + α (μ)1/2/305
(at 25 ºC)
log γCa =
0.512(2)2 (0.02)1/2 =
1 + 600(0.02)1/2/305
-0.230

γCa = 0.590
log γSO4 =
0.512(2)2 (0.02)1/2 =
1 + 450(0.02)1/2/305
-0.240

γSO4 = 0.580
log
Ksp = ACa2+ ASO42- = [Ca2+] γCa2+ [SO42-] γSO42- = x2 γCa2+ γSO42- = 2.4 x 10-5
x2 (0.590)(0.580) = 2.4 x 10-5
x = 8.37 x 10-3 M
 [Ca2+] = [SO42-] = 8.37 x 10-3 M
Concentration of ions
has increased!
Time consuming!
• More general method…
Uses charge and mass balance
Free Metal Ion Concentration
Chemical Equilibria
• Charge balance:
 positive charges on the ions =  negative charges on the ions
e.g. K3PO4 in solution
[H+] + [K+] = [OH-] + [H2PO4-] + 2[HPO42-] + 3[PO43-]
Conc. Is multiplied by Z
e.g. PO43- at 0.300 M, the charge concentration
(3 charges per ion) is 3 x 0.300 M = 0.900 M
Free Metal Ion Concentration
Chemical Equilibria
•
Mass balance:
The total amount of A put into the system must equal the sum of all the
various species in which A exists:
[A]T = [HA] + [A-]
e.g. CH3COOH
CH3COO- + H+
HA
A- + H+
Total acetic acid added is 0.0500 M, total mass of acid and associated
species:
0.0500 M = [CH3COOH] + [CH3COO-]
Free Metal Ion Concentration
Chemical Equilibria
• Mass balance:
Quantity of element in all species put into a solution must equal
the amount delivered to the solution
e.g. Na2S → 2Na+ + S2Total sodium sulfide added is 0.105 M but S2- reacts with water to
form HS- and H2S
Total S = 0.105 M = [S2-] + [HS-] + [H2S]
Free Metal Ion Concentration
Chemical Equilibria
• General steps for this approach to equilibrium:
Step 1: Write the pertinent reactions
Step 2: Write the charge balance
Step 3: Write the mass balance
Step 4: Write the equilibrium expressions + appropriate constants
Step 5: Count the equations and unknowns
Step 6: If the number of equations is equal to or greater than the number of
unknowns, solve
• Example on pages 102-105 for Hg2Cl2 (no reaction with H2O)
and HgS (reacts)
Calculate the concentration of Hg22+ in a saturated solution of Hg2Cl2
(no reaction with H2O)
Step 1:
Hg2Cl2
H2O
Step 2: Charge balance -
Hg22+ + 2ClH+ + OH-
Ksp = 1.2 x 10-18
Kw = 1.00 x 10-14
[H+] + 2[Hg22+] = [Cl-] + [OH-]
Step 3: Mass balance – neither Hg22+ or Cl- reacts with water
[H+] = [OH-]
and
2[Hg22+] = [Cl-]
Step 4: Equilibrium constants
Ksp = [Hg22+][Cl-]2 = 1.2 x 10-18
Kw = [H+][OH-] = 1.00 x 10-14
Step 5: Count equations and unknowns
Step 6: Solve – For pure water:
[H+] = [OH-] = 1.00 x 10-7
For Hg2Cl2: 2[Hg22+] = [Cl-]; Ksp = [Hg22+][Cl-]2 = [Hg22+][2 x Hg22+]2 = 1.2 x 10-18
[Hg22+] = (Ksp/4)1/3 = 6.7 x 10-7 M
Wait!
•
Why is 2[Hg22+] = [Cl-] ???
Hg2Cl2
Hg22+ + 2Cl-
[0.5]
[0.5]
[1.0]
Calculate the concentration of Hg2+ in HgS
(reaction with H2O)
Step 1:
Hg2S
S2- + H2O
HS- + H2O
H2O
Hg2+ + S2HS- + OHH2S + OHH+ + OH-
Ksp = 5 x 10-54
Kb1 = 0.80
Kb2 = 1.1 x 10-7
Kw = 1.00 x 10-14
S2- is a strong base, so [H+] will not equal [OH-]
Step 2: Charge balance
2[Hg2+] + [H+] = 2[S2-] + [HS-] + [OH-]
Step 3: Mass balance
[Hg2+] = [S2-] + [HS-] + [H2S]
Decomposition of HgS is 1:1, total Hg
= total S species
Calculate the concentration of Hg2+ in HgS
(reaction with H2O)
Step 4: Equilibrium constants
Ksp = [Hg2+][S2-] = 5 x 10-54
Kb1 = [HS-][OH-] = 0.80
[S2-]
Kb2 = [H2S][OH-] = 1.1 x 10-7
[HS-]
Kw = [H+][OH-] = 1.00 x 10-14
[HS-] = Kb1[S2-] -1
[OH-]
[H2S] = Kb2[HS-] -2
[OH-]
Step 5: 6 equations with 6 unknowns
Step 6: Simplify by letting pH = 8.00, [H+] = 10-8.00, [OH-] = 1.00 x 10-6
Substitute 1 into 2:
[H2S] = Kb1Kb2[S2-]
[OH-]2
Calculate the concentration of Hg2+ in HgS
(reaction with H2O)
From mass balance: [Hg2+] = [S2-] + [HS-] + [H2S]
[Hg2+] = [S2-] + Kb1[S2-] + Kb1Kb2[S2-]
[OH-]
[OH-]2
Rearrange:
[Hg2+] = [S2-] (1 + Kb1/[OH-] + Kb1Kb2/[OH-]2)
[S2-] =
[Hg2+]
(1 + Kb1/[OH-] + Kb1Kb2/[OH-]2)
Substitute [S2-] into Ksp
Ksp = [Hg2+] x [Hg2+]
(1 + Kb1/[OH-] + Kb1Kb2/[OH-]2)
[Hg2+] = √(Ksp(1 + Kb1/[OH-] + Kb1Kb2/[OH-]2))
End
• Review
Equilibrium Applied to Complex Speciation
Case 1: In Presence of a Solid
• Equilibrium equations from a table (given)
• Pb2+ + I- <--> PbI+
• Pb2+ + 2I- <--> PbI2(aq)
1.40E+03
• Pb2+ + 3I- <--> PbI38.30E+03
• Pb2+ + 4I- <--> PbI423.00E+04
K1 = [PbI+] / [Pb2+][I-] = 1.00E+02
2 = [PbI2(aq)] / [Pb2+][I-]2 =
 3 = [PbI3-] / [Pb2+][I-]3 =
 4 = [PbI42-] / [Pb2+][I-]4 =
• Solve each in terms of the progressively complexed species
Equilibrium Applied to Complex Speciation
Case 1: In Presence of a Solid
• Ksp controls Pb2+ conc.
0.000
[Pb2+]
-2.000
Ksp  [Pb2 ] [I- ]2  7.9 x 10-9
7.9 x 10-9
and for [I- ]  0.0001-4.000
M
- 2
[I ]
[P b2 ]  0.790 M.
log (0.790)  - 0.102
2
For [I- ]  10 M, [Pb
]  7.9 x 10-11.
log [Pb species]
[P b2 ] 
[PbI+]
[Pb] total
[PbI2]aq
-6.000
[PbI3-]
-8.000
log(7.9 x 10-11 )  -10.102.
[Pb2+]
-10.000
[PbI4(2-)]

-12.000
-4
-3.5
-3
-2.5
-2
-1.5
log [I-]
-1
-0.5
0
0.5
1
Equilibrium Applied to Complex Speciation
In Presence of a Solid Case I
• For PbI+
An I- concentration of 0.0001 results in a2Pb
concentration of 0.790 M.
Substit ution into K
1 
[P bI ]
 100.
[P b2 ][I- ]
0.000
yields [PbI ]  K1 [P b2 ][I- ]  7.90 x 10-3.
[Pb2+]

An I- concentration of 10 M results in a2 Pconcentration
b
of -2.000
7.9 x -11
10
.

T his results in a [PbI
] concentration of 7.90 x -810
.
log [Pb species]

[PbI+]
[Pb] total
-4.000
[PbI2]aq
-6.000
[PbI3-]
-8.000
[Pb2+]
-10.000
[PbI4(2-)]
-12.000
-4
-3.5
-3
-2.5
-2
-1.5
log [I-]
-1
-0.5
0
0.5
1
In Presence of a Solid Case I
• For PbI2
Again, an I- concent rat ion of 0.0001 results in a2Pb
concent rat ion of 0.790 M.
Subst itut ion into2 
[P bI2(aq) ]
[P b2 ][I- ]2
 1.4 x 103
0.000
yields [P bI2(aq) ]  2[P b ][I ]  1.11 x 10.
2
- 2
-5
[Pb2+]
2
An I- concent rat ion of 10 M results in a Pconcent
b
rat ion of 7.9 x -11
10
.
-2.000
[PbI+]
-5
T his results in a [P bI
.
2(aq) ] concentration of 1.11 x 10
log [Pb species]

Equilibrium Applied to Complex Speciation
[Pb] total
-4.000
[PbI2]aq
-6.000
[PbI3-]
-8.000
[Pb2+]
-10.000
[PbI4(2-)]
-12.000
-4
-3.5
-3
-2.5
-2
-1.5
log [I-]
-1
-0.5
0
0.5
1
Equilibrium Applied to Complex Speciation
Case I
For PbI3-
An I- concentration of 0.0001 results in a2Pconcentration
b
of 0.790 M.
Subst itution into3 
-
[P bI3 ]
 8.3 x 103
[P b2 ][I- ]3
yields [PbI3 ]  3 [P b2 ][I- ]3  6.56 x 10-9.
-

An I- concentration of 10 M result s in a2 Pb
concentration of 7.90 x -1110
.
-
-4
T his results in a [P bI
.
3 ] concentration of 6.56x 10
0.000
[Pb2+]
-2.000
log [Pb species]
[PbI+]
[Pb] total
-4.000
[PbI2]aq
-6.000
[PbI3-]
-8.000
[Pb2+]
-10.000
[PbI4(2-)]
-12.000
-4
-3.5
-3
-2.5
-2
-1.5
log [I-]
-1
-0.5
0
0.5
1
Equilibrium Applied to Complex Speciation
For PbI4-2
An I- concentration of 0.0001 results in a2Pconcentration
b
of 0.790 M.
Subst itution into4 
2-
[P bI3 ]
 3.00 x 104
[P b2 ][I- ]4
yields [PbI4 ]  4[P b2 ][I- ]4  2.37 x 10-12.
2-

An I- concentration of 10 M result s in a2 Pb
concentration of 7.90 x -1110
.
2-
-2
T his results in a [P bI
.
4 ] concentration of 2.37x 10
0.000
[Pb2+]
-2.000
log [Pb species]
[PbI+]
[Pb] total
-4.000
[PbI2]aq
-6.000
[PbI3-]
-8.000
[Pb2+]
-10.000
[PbI4(2-)]
-12.000
-4
-3.5
-3
-2.5
-2
-1.5
log [I-]
-1
-0.5
0
0.5
1
Equilibrium Applied to Complex Speciation
Case 2: Equilibrium the Absence of a Solid
• Problem statement: Nitrilotetraacetic (NTA) acid was a common
component in detergents and was a chemical of concern (COC)
in sewage effluent in the past. It is known for its high metal
complexing power; thus it keeps metals in solution, increasing
their mobility, and it may reduce their toxicity. Draw a pCd- pNTA
diagram for a cadmium system (total cadmium concentration =
0.015 M) using NTA concentrations ranging from 1.00 x 10-1 to
1.00 x 10-14 M. Use the following data:
Cd2+ + NTA- <--> CdNTA+ K1 = [CdNTA+] / [Cd2+][NTA-] = 6.31 x 109
Cd2+ + 2NTA- <--> Cd(NTA)2(aq)
2 = [CdNTA(aq)] / [Cd2+][NTA-]2 = 1.58 x
1015
Equilibrium Applied to Complex Speciation
Case 2: Equilibrium the Absence of a Solid
Total cadmium concentration = 0.015 M
CT = [Cd2+] + [CdNTA+] + [CdNTA2]
0.000
Cd Total
-2.000
-4.000
log Cd Species
CdNTA
CdNTA
-6.000
CdNTA2
-8.000
-10.000
Cd2+
-12.000
-14.000
-16.000
-14
-12
-10
-8
log NTA
-6
-4
-2
Equilibrium Applied to Complex Speciation
Case 2: Equilibrium the Absence of a Solid
For Cd2+
C T  [Cd2 ]  [CdNT A ]  [CdNT A2 ]
C T  [Cd2 ]  K1[Cd2 ][NT A- ]   2 [Cd2 ][NT A- ]2
C T  [Cd2 ] 1  K1[NT A- ]   2[NT A]2 
CT
1  K1[NT A- ]  2[NT A- ]2 
0.000
Cd Total
-2.000
-4.000
CdNTA
log Cd Species
[Cd 2 ] 
CdNTA
-6.000
CdNTA2
-8.000
-10.000
Cd2+
-12.000
-14.000
-16.000
-14
-12
-10
-8
log NTA
-6
-4
-2
Equilibrium Applied to Complex Speciation
Case 2: Equilibrium the Absence of a Solid
Similarly we can derive equations for the other species:
[CdNT A2 ] 
CT
1
2 [NT A- ]

1

K1[NT A- ]
K1
CT
1
K1

 1
- 2
2 [NT A ]
2[NT A- ]
0.000
Cd Total
-2.000
-4.000
CdNTA
log Cd Species
[CdNT A ] 
CdNTA
-6.000
CdNTA2
-8.000
-10.000
Cd2+
-12.000
-14.000
-16.000
-14
-12
-10
-8
log NTA
-6
-4
-2
Equilibrium Applied to Complex Speciation
Case 3: Stability Diagrams
EH-pH Diagram
for dissolved iron
Fe2+ may ppt and form a
reactive sorption surface
-Reduces pollutant
-Absorb Mn+ from water
Purpose is to show
the dominant phase
or chemical species
Fate of another
metal may be
affected by
changing pH-E of
water containing Fe
Equilibrium Applied to Complex Speciation
Computer Software
•
MINEQL+ calculates equilibrium
speciation of all cataions and
anions based on pH and EH
End
• Review
Methods for Determining KD and KP
•
Aqueous phase / dissolved organic matter
Pollutant(aq) ↔ PollutantDOM
KDOM =
•
Conc pollutant in DOM (mg/kg)
Conc. Pollutant in water (mg/L)
Aqueous phase and particles in solution:
Pollutant(aq) ↔ Pollutantparticle
Kd =
Conc pollutant on particle (mg/kg)
Conc. Pollutant in water (mg/L)
Kp =
Conc pollutant on organic particle (mg/kg)
Conc. Pollutant in water (mg/L)
or
Methods for Determining KD and KP
• To describe partitioning as it relates to conc. Organic matter
Korganic Carbon =
KOC = Kp/fOC
Kd or Kp
Fraction of organic matter in sample
Methods for Determining KD and KP
• Lab work: solution of known pollutant mass and water,
soil/sediment, or DOM is mixed together for 3 days
• Solid and aqueous phases are separated by 0.45 μm filtration
Measurement in lab is an approximation!
Kd depends on pH, type of cations present, ionic
strength, surface charge, solids-to-water ratio
Assumes reversible sorption
Kd and Kp
T ot al mass (g
) of
pollut ant added
t o flask
Mass of pollut ant
recovered in
blank (mg)
Mass of pollut ant
measured in
wat er phase (mg)
0.00725
Volume of water
(L)
Conc. Of
pollut ant in
wat er phase
(mg/L)
Mass of pollut ant
on solid phase
(mg)
Mass of solid
phase (kg)
Conc. Of
pollut ant on soli
d
phase (mg/kg)
Kd
0.0300
Table 3.2: Cd2+ on
sediment
0.00720
0.00542
0.181
0.00178
3.58 x 10-5
49.7
275
Designates exp.
Measurements
Note the other way
of doing this with
multiple measurements
(Figure 3.12)
Kd and Kp
•
•
•
Kp usually more constant
Methoxychlor (pesticide) on clay
Hydrophobic organic pollutant
Kp is independent
of water phase
concentration
Karickhoff et al., 1979
Kd and Kp
•
Hydrophobic pollutants sorb
more with greater OM content
Slope here is:
Kp/fOC = KOC
Karickhoff et al., 1979
Kd and Kp
•
•
Measuring Kp or KOC for every
pollutant takes time
KOC is estimated based on
correlation with octanol-water
partition coefficients (KOW)
e.g. log KOC = 1.00 KOW – 0.21
Kd and Kp
•
•
•
In theory Kd and Kp should be independent of suspended solids
concentrations
Investigations indicate strong dependence on solids concentration
K and concentration of pollutant sorbed decreases
End
• Review
Kinetics of Sorption
•
•
•
Consider time scale to reach equilibrium
3 days is used for lab studies
– particles are not uniform size, may take days - months
Particles are in ever changing state of sorption equilibrium
Cd2+ desorption from clay as a function of time (takes longer than sorption)
Kinetics of Desorption
Slow
Rapid
Figure 3.8 PCB desorption from clay as a function of time
Dunnivant, 1988
Kinetics of Desorption
Figure 3.8 PCB desorption from clay as a function of time
Dunnivant, 1988
Sorption Isotherms
Sorption isotherms (sorption studies conducted at constant
temperature)
Three types of sorption mechanisms:
(1) Chemical reactions at surfaces such as
surface hydrolysis
surface complexation
surface-ligand exchange, and
hydrogen bond formation
(2) Electrostatic interactions, and
(3) Hydrophobic expulsion
Sorption Isotherms
Most common
Pollutant concentrations
constantly vary in nature
Isotherms establish
shape of sorption
relationship
Sorption Isotherms
•
Langmuir isotherm (L-type)
S + A ↔ SA
(where SA is the adsorbate on
surface sites)
1/ ΓA
K[A]
A  max
.
1  K [A]
Where ΓA = [SA]/mass absorbent,
[A] = conc. of absorbate in solution,
K = equilibrium constant
1/[A]
(note assumptions in text)
Intercept 1/Γmax , slope 1/Γmax K
Sorption Isotherms
• Freundlich equation
A  K [A]n
Does not limit
maximum amount
adsorbed or sorbed
• Where ΓA = [SA]/mass absorbent, [A] = conc. of absorbate in
solution, K = equilibrium constant, n = constant
• In general due to the low concentrations of pollutants under
consideration we rely on Kd and Kp values and assume linear
isotherm
Sorption Isotherms
• Comparison of Frendlich and langmuir isotherms
End
• Review
Kinetics of Transformation Reactions
•
When ever possible we use first-order or pseudo first-order
kinetics. Why?
Rate = change in concentration with time = - ΔC = kC
Δt
Solve using calculus:
ln (Ct/C0) = -kt
Where Ct = pollutant concenatration at time t,
C0 = pollutant concentration at time = 0
Kinetics of Transformation Reactions
•
Half-life (t1/2) is the time when half of the initial pollutant
concentration has been removed:
Ct = C0/2,
substitute into above, ln(1/2) = - kt1/2
ln(1/2) = - kt1/2
t1/2 = ln 2
k
Kinetics of Transformation Reactions
•
•
Many chemical reactions are of the form:
Rate = k[pollutant][some ox or red pollutant]
NOM, mineral surface, photon, microbe
•
Conc. of OX/RED reactants is high and relatively constant compared to
low pollutant conc., assume conc. does not change with time
•
Reduces to:
•
Greatly simplifies reactions!!!
Rate = k’[pollutant]
(where k’ = k[some ox or red pollutant])
End
• Review
Chemical Factor
Metals
Ionizable Organics
Hydrophobic
Organics
Potentially √
√
Can be √
Can be √
X
√
Can be √
√
√
√
Can be √
Can be √
Can be √
Can be √
Can be √
Can be √
Modeling with Chemistry
(+radionuclides)
Section 2.4
pH
Solubility
Vapor pressure
HLC
√
√
X
X
For Inorganics
Acid-base
Redox
Precipitation
√
√
√
Section 2.5
Sorption
√
Section 2.6
Abiotic
Photochemical
Biological
Modeling with Chemistry
Case I - Metal
• Controlled by Ksp, VP andand HLC not useful
• Sorption greatly influences transport, adsorbed metals are less
bioavailable
• Settle out and incorporated into sediments
• Metals adsorbed to DOM are generally transported out of the
system
• Transformation reactions are of little consequence (except
radionuclide decay)
Modeling with Chemistry
Case II – Hydrophobic Pollutants
• Controlled by HLC
• For atmospheric systems vapor pressure determines input
• EH influences biotic and abiotic degradation reactions
• Sorption to NOM or minerals is important
• Kp >>> Kd values
Modeling with Chemistry
•
Use box model approach
Change of mass
= sum of
in system with time
all inputs
C/t
mass of
pollutant
input
+
sum of internal
sources
any source or
generation
of the pollutant
from within
the system
-
sum of all
outputs
mass of
pollutant
exiting the
system
-
sum of all
internal sinks
removal from
the system
by sorption
or
degradation
reactions
End
• Review
Further Reading
Journals
• Karickhoff, S.W., Brown, D.S., and Scott, T.A. (1979) Sorption of
hydrophobic pollutants on natural sediments. Water Research, Vol. 13,
pp. 241-248.
• Karickhoff, S.W. (1984) Organic pollutant sorption in aquatic systems.
Journal of Hydraulic Engineering, Vol. 110, No. 6, pp. 707-735.
• Karickhoff, S.W. and Morris, K.W. (1985) Sorption dynamics of
hydrophobic pollutants in sediment suspensions. Environmental
Toxicology and Chemistry, Vol. 4, pp. 467-479.
• O’Connor, D.J. and Conolly, J.P. (1980) the effect of concentration of
absorbing solids on the partition coefficient. Water Research, Vol. 14,
pp. 1517-1523.
• Wu, S. and Gschwend P.M. (1986) Sorption kinetics of hydrophobic
organic compounds to natural sediments and soils. Environmental
Science and Technology, Vol. 20, pp. 1213-1217.
Books
•
Dunnivant, F.M. (1988) Congener-Specific PCB Chemical and Physical
Parameters for Evaluation of Environmental Weathing of Arochlors.
Ph.D. Dissertation, Environmental Systems Engineering, Clemson
University, Clemson, SC.
5.
Exercises
From Stumm and Morgan, 1996
Exercises
15. 14700 L kg-1
17. 8752 L kg-1