Environmental Modeling Chapter 3: Quantitative Aspects of Chemistry Copyright © 2006 by DBS.
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Transcript of Environmental Modeling Chapter 3: Quantitative Aspects of Chemistry Copyright © 2006 by DBS.
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Environmental Modeling
Chapter 3:Quantitative Aspects of Chemistry
Copyright © 2006 by DBS
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Quote“The noblest of the elements is water”
-Pindar, 476 B.C.
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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
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Concepts
• Free metal ion concentration using equilibria
• Determining KD and KP
• Kinetics of sorption• Sorption isotherms• Kinetics of transformation• Modeling
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Free Metal Ion ConcentrationChemical Equilibria
General Chemistry(simplified solutions)
Environmental Chemistry(more complex solutions -
activities)
Chemical Fate(computer programs –
GEOCHEM, MINTECH, MINEQL)
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Free Metal Ion ConcentrationChemical Equilibria
• General Chemistry
e.g. Pb2+ + 2Cl- → PbCl2(s)
• But! Lead is also present as:
PbCl+, PbCl2(aq), PbCl3-, and PbCl42-
• 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)
General Chemistry(simplified solutions)
Environmental Chemistry(more complex solutions -
activities)
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Free Metal Ion ConcentrationReview of Chemical Equilibria
Gen. Chem.: aA + bB cC + dD
K = [C]c[D]d
[A]a[B]b
Where K is a constant as [A], [B], [C], and [D] varies
Idealized!
Equilibrium constant K has been found to vary with concentration! So it’s not really constant…
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Free Metal Ion ConcentrationReview of Chemical Equilibria
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Free Metal Ion ConcentrationReview of Chemical Equilibria
Use activities rather than concentrations
K = (AC)c(AD)d = [C]c γc [D]d γd
(AA)a(AB)b [A]a γa [B]b γb
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
For K to remain constant Ax must remain constant
If [X] increases,
γx must decrease
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Free Metal Ion ConcentrationReview 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…
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Free Metal Ion ConcentrationReview 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 = x2, and x = [Ca2+] = [SO42-] = 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!
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Free Metal Ion ConcentrationReview of Chemical Equilibria
Example:
• CaSO4 (Ksp = 2.5 x 10-5) in 0.02 M KNO3
NO3-
SO42-
Ca2+ SO42- Ca2+K+
More interactions
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Free Metal Ion ConcentrationReview 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
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Free Metal Ion ConcentrationReview of Chemical Equilibria
• Use Debye-Hückel equation to find activities:
log γ = 0.512 Z2 (μ)1/2 (at 25 ºC)1 + α (μ)1/2/305
log γCa = 0.512(2)2 (0.02)1/2 = -0.230 γCa = 0.5901 + 600(0.02)1/2/305
log γSO4 = 0.512(2)2 (0.02)1/2 = -0.240 γSO4 = 0.5801 + 450(0.02)1/2/305
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!
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Time consuming!
• More general method…
Uses charge and mass balance
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Free Metal Ion ConcentrationChemical Equilibria
• Charge balance: positive charges on the ions = negative charges on the ions
e.g. K3PO4 in solution
[H+] + [K+] = [OH-] + [H2PO4-] + 2[HPO4
2-] + 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
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Free Metal Ion ConcentrationChemical 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+
Total acetic acid added is 0.0500 M, total mass of acid and associated species:
0.0500 M = [CH3COOH] + [CH3COO-]
HA A- + H+
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Free Metal Ion ConcentrationChemical 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+ + S2-
Total 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]
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Free Metal Ion ConcentrationChemical Equilibria
• General steps for this approach to equilibrium:
Step 1: Write the pertinent reactionsStep 2: Write the charge balanceStep 3: Write the mass balanceStep 4: Write the equilibrium expressions + appropriate constantsStep 5: Count the equations and unknownsStep 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)
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Calculate the concentration of Hg22+ in a saturated solution of Hg2Cl2
(no reaction with H2O)
Step 1: Hg2Cl2 Hg22+ + 2Cl- Ksp = 1.2 x 10-18
H2O H+ + OH- Kw = 1.00 x 10-14
Step 2: Charge balance - [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 constantsKsp = [Hg2
2+][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 = [Hg2
2+][Cl-]2 = [Hg22+][2 x Hg2
2+]2 = 1.2 x 10-
18
[Hg22+] = (Ksp/4)1/3 = 6.7 x 10-7 M
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Wait!
• Why is 2[Hg22+] = [Cl-] ???
Hg2Cl2 Hg22+ + 2Cl-
[0.5] [0.5] [1.0]
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Calculate the concentration of Hg2+ in HgS (reaction with H2O)
Step 1: Hg2S Hg2+ + S2- Ksp = 5 x 10-54
S2- + H2O HS- + OH- Kb1 = 0.80
HS- + H2O H2S + OH- Kb2 = 1.1 x 10-7
H2O H+ + OH- 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
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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 [HS-] = Kb1[S2-] -1[S2-] [OH-]
Kb2 = [H2S][OH-] = 1.1 x 10-7 [H2S] = Kb2[HS-] -2 [HS-] [OH-]
Kw = [H+][OH-] = 1.00 x 10-14
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
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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))
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End
• Review
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Equilibrium Applied to Complex SpeciationCase 1: In Presence of a Solid
• Equilibrium equations from a table (given)
• Pb2+ + I- <--> PbI+ K1 = [PbI+] / [Pb2+][I-] = 1.00E+02
• Pb2+ + 2I- <--> PbI2(aq) 2 = [PbI2(aq)] / [Pb2+][I-]2 = 1.40E+03
• Pb2+ + 3I- <--> PbI3- 3 = [PbI3
-] / [Pb2+][I-]3 = 8.30E+03
• Pb2+ + 4I- <--> PbI42- 4 = [PbI4
2-] / [Pb2+][I-]4 = 3.00E+04
• Solve each in terms of the progressively complexed species
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Equilibrium Applied to Complex SpeciationCase 1: In Presence of a Solid
• Ksp controls Pb2+ conc.
Ksp [Pb2 ] [I- ]2 7.9 x 10-9
[Pb2 ] 7.9 x 10-9
[I- ]2 and for [I- ] 0.0001 M
[Pb2 ] 0.790 M.
log (0.790) - 0.102
For [I- ] 10 M, [Pb2 ] 7.9 x 10-11.
log(7.9 x 10-11) -10.102.
-12.000
-10.000
-8.000
-6.000
-4.000
-2.000
0.000
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
log [I-]
log
[P
b s
pe
cie
s]
[Pb2+]
[PbI+]
[PbI2]aq
[PbI3-]
[PbI4(2-)]
[Pb] total
[Pb2+]
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Equilibrium Applied to Complex Speciation In Presence of a Solid Case I
• For PbI+
An I- concentration of 0.0001 results in a Pb2 concentration of 0.790 M.
Substitution into K1 [PbI ]
[Pb2 ][I- ] 100.
yields [PbI ] K1 [Pb2 ][I- ] 7.90 x 10-3.
An I - concentration of 10 M results in a Pb2
concentration of 7.9 x 10-11.
This results in a [PbI ] concentration of 7.90 x 10-8.
-12.000
-10.000
-8.000
-6.000
-4.000
-2.000
0.000
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
log [I-]
log
[P
b s
pe
cie
s]
[Pb2+]
[PbI+]
[PbI2]aq
[PbI3-]
[PbI4(2-)]
[Pb] total
[Pb2+]
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Equilibrium Applied to Complex Speciation In Presence of a Solid Case I
• For PbI2
Again, an I- concentration of 0.0001 results in a Pb 2 concentration of 0.790 M.
Substitution into 2 [PbI2(aq) ]
[Pb2 ][I- ]2 1.4 x 103
yields [PbI2(aq) ] 2[Pb2 ][I- ]2 1.11 x 10-5.
An I- concentration of 10 M results in a Pb 2 concentration of 7.9 x 10-11.
This results in a [PbI2(aq) ] concentration of 1.11 x 10 -5.
-12.000
-10.000
-8.000
-6.000
-4.000
-2.000
0.000
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
log [I-]
log
[P
b s
pe
cie
s]
[Pb2+]
[PbI+]
[PbI2]aq
[PbI3-]
[PbI4(2-)]
[Pb] total
[Pb2+]
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Equilibrium Applied to Complex Speciation Case IFor PbI3
-
-12.000
-10.000
-8.000
-6.000
-4.000
-2.000
0.000
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
log [I-]
log
[P
b s
pe
cie
s]
[Pb2+]
[PbI+]
[PbI2]aq
[PbI3-]
[PbI4(2-)]
[Pb] total
[Pb2+]
An I- concentration of 0.0001 results in a Pb2 concentration of 0.790 M.
Substitution into 3 [PbI3
- ]
[Pb2 ][I- ]3 8.3 x 103
yields [PbI3- ] 3[Pb2 ][I- ]3 6.56 x 10-9.
An I- concentration of 10 M results in a Pb2 concentration of 7.90 x 10-11.
This results in a [PbI3- ] concentration of 6.56x 10-4.
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Equilibrium Applied to Complex SpeciationFor PbI4
-2
-12.000
-10.000
-8.000
-6.000
-4.000
-2.000
0.000
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1
log [I-]
log
[P
b s
pe
cie
s]
[Pb2+]
[PbI+]
[PbI2]aq
[PbI3-]
[PbI4(2-)]
[Pb] total
[Pb2+]
An I- concentration of 0.0001 results in a Pb2 concentration of 0.790 M.
Substitution into 4 [PbI3
2- ]
[Pb2 ][I- ]4 3.00 x 104
yields [PbI42- ] 4[Pb2 ][I- ]4 2.37 x 10-12.
An I- concentration of 10 M results in a Pb2 concentration of 7.90 x 10-11.
This results in a [PbI42- ] concentration of 2.37x 10-2.
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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
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Equilibrium Applied to Complex Speciation Case 2: Equilibrium the Absence of a Solid
Total cadmium concentration = 0.015 M
CT = [Cd2+] + [CdNTA+] + [CdNTA2]
-16.000
-14.000
-12.000
-10.000
-8.000
-6.000
-4.000
-2.000
0.000
-14 -12 -10 -8 -6 -4 -2
log NTA
log
Cd
Sp
ec
ies
Cd2+
CdNTA
CdNTA2
Cd Total
CdNTA
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Equilibrium Applied to Complex Speciation Case 2: Equilibrium the Absence of a Solid
For Cd2+
-16.000
-14.000
-12.000
-10.000
-8.000
-6.000
-4.000
-2.000
0.000
-14 -12 -10 -8 -6 -4 -2
log NTA
log
Cd
Sp
ec
ies
Cd2+
CdNTA
CdNTA2
Cd Total
CdNTA
CT [Cd2] [CdNTA] [CdNTA2]
CT [Cd2] K1[Cd2][NTA -] 2[Cd2][NTA -]2
CT [Cd2] 1 K1[NTA -] 2[NTA]2 [Cd2]
CT
1 K1[NTA -] 2[NTA -]2
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Equilibrium Applied to Complex Speciation Case 2: Equilibrium the Absence of a Solid
Similarly we can derive equations for the other species:
-16.000
-14.000
-12.000
-10.000
-8.000
-6.000
-4.000
-2.000
0.000
-14 -12 -10 -8 -6 -4 -2
log NTA
log
Cd
Sp
ec
ies
Cd2+
CdNTA
CdNTA2
Cd Total
CdNTA
[CdNTA ] CT
1
K1[NTA - ] 1
2 [NTA - ]
K1
[CdNTA 2 ] CT
1
2[NTA - ]2
K1
2[NTA - ] 1
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Equilibrium Applied to Complex Speciation Case 3: Stability Diagrams
EH-pH Diagram
for dissolved iron
Purpose is to show
the dominant phase
or chemical species
Fe2+ may ppt and form a reactive sorption surface
-Reduces pollutant
-Absorb Mn+ from water
Fate of another metal may be affected by changing pH-E of water containing Fe
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Equilibrium Applied to Complex Speciation Computer Software
• MINEQL+ calculates equilibrium speciation of all cataions and anions based on pH and EH
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End
• Review
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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)
orKp = Conc pollutant on organic particle (mg/kg)
Conc. Pollutant in water (mg/L)
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Methods for Determining KD and KP
• To describe partitioning as it relates to conc. Organic matter
Korganic Carbon = Kd or Kp
Fraction of organic matter in sample
KOC = Kp/fOC
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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
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Kd and Kp Total mass (g) of pollutant added to flask
0.00725
Mass of pollutant recovered in blank (mg)
0.00720
Mass of pollutant measured in water phase (mg)
0.00542
Volume of wate r (L)
0.0300
Conc. Of pollutant in water phase (mg/L)
0.181
Mass of pollutant on solid phase (mg)
0.00178
Mass of solid phase (kg)
3.58 x 10-5
Conc. Of pollutant on solid phase (mg/kg)
49.7
Kd 275
Table 3.2: Cd2+ on sediment
Designates exp. Measurements
Note the other wayof doing this with multiple measurements(Figure 3.12)
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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
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Kd and Kp
Karickhoff et al., 1979
• Hydrophobic pollutants sorb more with greater OM content
Slope here is: Kp/fOC = KOC
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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
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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
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End
• Review
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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)
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Kinetics of Desorption
Figure 3.8 PCB desorption from clay as a function of timeDunnivant, 1988
Rapid
Slow
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Kinetics of Desorption
Figure 3.8 PCB desorption from clay as a function of timeDunnivant, 1988
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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
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Sorption Isotherms
Pollutant concentrations constantly vary in nature
Isotherms establish shape of sorption relationship
Most common
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Sorption Isotherms
• Langmuir isotherm (L-type)S + A ↔ SA
(where SA is the adsorbate on surface sites)
Where ΓA = [SA]/mass absorbent, [A] = conc. of absorbate in solution, K = equilibrium constant
(note assumptions in text)
.[A]K 1
K[A] maxA
Intercept 1/Γmax , slope 1/Γmax K
1/ ΓA
1/[A]
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Sorption Isotherms
• Freundlich equation
• 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
nA [A]K
Does not limit maximum amount adsorbed or sorbed
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Sorption Isotherms
• Comparison of Frendlich and langmuir isotherms
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End
• Review
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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
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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
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Kinetics of Transformation Reactions
• Many chemical reactions are of the form:
• Rate = k[pollutant][some ox or red pollutant]
• Conc. of OX/RED reactants is high and relatively constant compared to low pollutant conc., assume conc. does not change with time
• Reduces to: Rate = k’[pollutant]
(where k’ = k[some ox or red pollutant])
• Greatly simplifies reactions!!!
NOM, mineral surface, photon, microbe
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End
• Review
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Modeling with ChemistryChemical Factor Metals
(+radionuclides)
Ionizable Organics Hydrophobic Organics
Section 2.4
pH
Solubility
Vapor pressure
HLC
√
√
X
X
Potentially √
√
Can be √
Can be √
X
√
Can be √
√
For Inorganics
Acid-base
Redox
Precipitation
√
√
√
Section 2.5
Sorption √ √ √
Section 2.6
Abiotic
Photochemical
Biological
Can be √
Can be √
Can be √
Can be √
Can be √
Can be √
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Modeling with ChemistryCase 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)
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Modeling with ChemistryCase 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
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Modeling with Chemistry
Change of mass sum of sum of internal sum of all sum of allin system with time all inputs sources outputs internal sinks
= + - -
C/t mass of any source or mass of removal from pollutant generation pollutant the system input of the pollutant exiting the by sorption
from within system orthe system degradation
reactions
• Use box model approach
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End
• Review
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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.
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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.
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Exercises
From Stumm and Morgan, 1996
5.
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Exercises
15. 14700 L kg-1
17. 8752 L kg-1