Environmental Modeling Chapter 3: Quantitative Aspects of Chemistry Copyright © 2006 by DBS.

transcript

Environmental Modeling

Chapter 3:Quantitative Aspects of Chemistry

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 ConcentrationChemical Equilibria

General Chemistry(simplified solutions)

Environmental Chemistry(more complex solutions -

activities)

Chemical Fate(computer programs –

GEOCHEM, MINTECH, MINEQL)

• 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)

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…

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

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…

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!

Example:

• CaSO4 (Ksp = 2.5 x 10-5) in 0.02 M KNO3

Ca2+ SO42- Ca2+K+

More interactions

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

• 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!

Time consuming!

• More general method…

Uses charge and mass balance

• 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

• 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+

• 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]

• 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)

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-

[Hg22+] = (Ksp/4)1/3 = 6.7 x 10-7 M

• 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 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

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

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))

• Review

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

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

-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1

log [I-]

[Pb2+]

[PbI+]

[PbI2]aq

[PbI3-]

[PbI4(2-)]

[Pb] total

[Pb2+]

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

-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1

log [I-]

[Pb2+]

[PbI+]

[PbI2]aq

[PbI3-]

[PbI4(2-)]

[Pb] total

[Pb2+]

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

-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1

log [I-]

[Pb2+]

[PbI+]

[PbI2]aq

[PbI3-]

[PbI4(2-)]

[Pb] total

[Pb2+]

Equilibrium Applied to Complex Speciation Case IFor PbI3

-12.000

-10.000

-8.000

-6.000

-4.000

-2.000

-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1

log [I-]

[Pb2+]

[PbI+]

[PbI2]aq

[PbI3-]

[PbI4(2-)]

[Pb] total

[Pb2+]

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.

Equilibrium Applied to Complex SpeciationFor PbI4

-12.000

-10.000

-8.000

-6.000

-4.000

-2.000

-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1

log [I-]

[Pb2+]

[PbI+]

[PbI2]aq

[PbI3-]

[PbI4(2-)]

[Pb] total

[Pb2+]

Substitution into 4 [PbI3

[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.

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

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

-14 -12 -10 -8 -6 -4 -2

log NTA

CdNTA2

Cd Total

For Cd2+

-16.000

-14.000

-12.000

-10.000

-8.000

-6.000

-4.000

-2.000

-14 -12 -10 -8 -6 -4 -2

log NTA

CdNTA2

Cd Total

CT [Cd2] [CdNTA] [CdNTA2]

CT [Cd2] K1[Cd2][NTA -] 2[Cd2][NTA -]2

CT [Cd2] 1 K1[NTA -] 2[NTA]2 [Cd2]

1 K1[NTA -] 2[NTA -]2

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

-14 -12 -10 -8 -6 -4 -2

log NTA

CdNTA2

Cd Total

[CdNTA ] CT

K1[NTA - ] 1

2 [NTA - ]

[CdNTA 2 ] CT

2[NTA - ]2

2[NTA - ] 1

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

Equilibrium Applied to Complex Speciation Computer Software

• MINEQL+ calculates equilibrium speciation of all cataions and anions based on pH and EH

• 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)

orKp = Conc pollutant on organic particle (mg/kg)

Conc. Pollutant in water (mg/L)

• To describe partitioning as it relates to conc. Organic matter

Korganic Carbon = Kd or Kp

Fraction of organic matter in sample

KOC = Kp/fOC

• 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 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)

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)

Kd 275

Table 3.2: Cd2+ on sediment

Designates exp. Measurements

Note the other wayof 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

Karickhoff et al., 1979

• Hydrophobic pollutants sorb more with greater OM content

Slope here is: Kp/fOC = KOC

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

• 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

Figure 3.8 PCB desorption from clay as a function of timeDunnivant, 1988

Kinetics of Desorption

Figure 3.8 PCB desorption from clay as a function of timeDunnivant, 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

Pollutant concentrations constantly vary in nature

Isotherms establish shape of sorption relationship

Most common

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

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

Sorption Isotherms

• Comparison of Frendlich and langmuir isotherms

• 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

Solve using calculus:

ln (Ct/C0) = -kt

Where Ct = pollutant concenatration at time t, C0 = pollutant concentration at time = 0

• 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

• 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

• Review

Modeling with ChemistryChemical Factor Metals

(+radionuclides)

Ionizable Organics Hydrophobic Organics

Section 2.4

Solubility

Vapor pressure

Potentially √

Can be √

For Inorganics

Acid-base

Precipitation

Section 2.5

Sorption √ √ √

Section 2.6

Abiotic

Photochemical

Biological

Can be √

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)

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

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

• Review

Environmental Modeling Chapter 3: Quantitative Aspects of Chemistry Copyright © 2006 by DBS.

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