toxic metals in the environment: thermodynamic ... · kirk g. scheckel, christopher a....

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TJ1228-02 EST.cls September 1, 2004 19:29

Critical Reviews in Environmental Science and Technology, 34:495–604, 2004Copyright © Taylor & Francis Inc.ISSN: 1064-3389 print / 1547-6537 onlineDOI: 10.1080/10643380490492412

Toxic Metals in the Environment:Thermodynamic Considerations for Possible

Immobilization Strategies for Pb, Cd, As, and Hg

SPENCER K. PORTERNational Council on the Aging, Washington, DC, USA

KIRK G. SCHECKEL, CHRISTOPHER A. IMPELLITTERI,AND JAMES A. RYAN

United States Environmental Protection Agency, Cincinnati, OH, USA

The contamination of soils by toxic metals is a widespread, seriousproblem that demands immediate action either by removal or im-mobilization, which is defined as a process which puts the metalinto a chemical form, probably as a mineral, which will be inertand highly insoluble under conditions that will exist in the soil. Ifmetals are to be immobilized, this might be achieved by the additionof sufficient amounts of the anion or anions which can form the in-ert mineral. A serious complication arises from the fact that all soilshave several other cations that can and do react with the anions.

This paper is a review of the equilibrium-state chemistry forthe possible immobilizations of four metals: lead, cadmium, ar-senic, and mercury. The anions which might precipitate these met-als include: oxide, hydroxide, chloride, sulfate, sulfide, phosphates,molybdate, and carbonate. The metal ions which can interfere withthese precipitation reactions are calcium, magnesium, iron, alu-minum, and manganese. All of the probable combinations are re-viewed, and possible immobilization strategies are evaluated fromthe point of view of thermodynamic stability using free energies offormation scattered throughout the literature. The systems are ex-amined from the point of view of the phase rule and stoichiometricconsideration over the 2–12 pH range.

KEY WORDS: equilibrium, precipitation, soil, solubility, remedia-tion, modeling

Address correspondence to Kirk G. Scheckel, United States Environmental ProtectionAgency, 5995 Center Hill Avenue, Cincinnati, OH 45224. E-mail: [email protected]

495

SDMS DOCID# 1127640

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496 S. K. Porter et al.

I. INTRODUCTION

Many soils in the industrial world are contaminated with toxic metals forwhich the health hazards are well documented.1–7 Metals are also known toseriously disrupt the life cycles of flora and fauna and the health of entireecosystems.8–12 Regulatory efforts over the past few decades have succeededin cutting off some of the sources of this pollution, but the neglect of previousyears as well as more recent pollution has given us a difficult problem withoutobvious solution at any price.

One possibility is surely removal, but experience has shown that treat-ments, such as concentrated nitric acid or ethylenediamminetetraacetic acid(EDTA), which will remove such metals, do severe damage to the soil itself,often rendering it sterile and useless.3,13–15 As a consequence considerableresearch has been focused on techniques for immobilization, which is de-fined to be a treatment which will put the toxic metal into a salt or mineralwhich is highly insoluble and stable over wide ranges of pH and oxidizingconditions (pe).16–21 It is also desirable that the salt or mineral be inert in theface of possible future manipulations of the soil’s chemistry by organisms,agriculture, industry, etc.

While there is a large collection of information available on the thermo-dynamics of possible crystalline phases and the aqueous solutions in equi-librium with them,22–25 especially Gibbs free energies of formation, much ofthe research done to date must be described as Edisonian—the experimentalapproach of trying everything, relentlessly, until a solution is found. The goalof this paper is an extensive review of the thermodynamics of minerals andcrystals relevant to the possible immobilization and long-term stabilities ofPb, Cd, As, and Hg. This review cannot tell us what treatments will work;like all results from thermodynamics it will tell us only what is impossible orpossible.

We may determine which salts or minerals of a particular metal will meetthe test of being highly insoluble by calculating the solubility as a functionof pH and pe, and this will be done for several compounds of each toxicmetal by modeling methods to be described in the next section. We will thendiscuss possible treatments which could make the desired compounds, andit will be necessary to consider the possible interactions of these treatmentswith other constituents of a soil. For example certain lead phosphate min-erals may result in the immobilization of lead, but the phosphate treatmentscan themselves react with a number of common minerals and salts in soils,and they do. The modeling methods will be used, therefore, to understandsystems which contain phosphate minerals of several metals. The number ofmetallic elements in a soil which might precipitate phosphate is large, andthis number includes Ca, Mg, Fe, Mn, Al, Zn, Cu, as well as Pb, Hg(II), andCd. Thermodynamics can answer the question of which of the several phos-phates is the most stable, even if the question of which will form the most

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Toxic Metals in the Environment 497

rapidly requires experiment. This paper will present thermodynamic results,and these have proved to be quite useful in the design of experiments andtreatments.

Some treatments, such as phosphate, are benign or even beneficent inmoderate amounts, but others are not. Lead may be precipitated as wulfenite,PbMoO4, but molybdate is in itself harmful. A similar case is that of makingarsenic into Ag3AsO4, which is known to be highly insoluble, but the additionof silver ion to a soil is not likely to benefit its fertility. Such facts will limitpossible remediation schemes.

Minerals in soils may undergo redox reactions over time, and these maybe quite significant to our purposes even if they are slow. For examplegalena, PbS, is quite insoluble but subject to slow oxidation to anglesite,PbSO4, which is orders of magnitude more soluble.22 Mercury is a particulartroublesome case in this regard because it has three oxidation states, and allthree can exist in soils. It will be necessary to consider, therefore, the possibleredox reactions of each element and each treatment as well as the conditionsof pH and pe under which each form might be stable. These reactions willfurther limit our choices or perhaps the long-term efficacy of our methods.

II. MODELING

1. Systems of One Mineral in Equilibrium with Water

The object of each model made for this study is a description of a particularchemical system at equilibrium. We know of course that not all chemicalsystems will come to equilibrium, but we do know that no chemical sys-tem can move by itself away from its equilibrium state.22,25 We also wish toknow how a particular system will change as the pH and pe vary. For ex-ample we would like to know how the solubility of hydroxypyromorphite,Pb5(PO4)3OH, will change with pH (pe is not an issue in this case). It is cer-tainly true that if solid hydroxypyromorphite is stirred with pure water untilequilibrium is obtained, there will be a definite pH determined by the relativeamounts of the different ions, but in soil systems the small concentrations oflead and phosphate ions will not determine pH. There will be many otherions of higher concentrations which will.

The problem of describing a heterogeneous chemical system like hy-droxypyromorphite in equilibrium with an aqueous solution is straight-forward even if the algebra can become complex. First the system mustbe defined with precision and the number of components determined. Inthe model we allow the mineral to come to equilibrium with a solution con-taining a fixed amount of NaCl, because chloro-complexes are likely to besignificant, and an activity of CO2 which would be in equilibrium with theatmosphere. The system has, therefore, three phases and six components(PbO, P2O5, CO2, H2O, Na2O, and HCl). This gives five degrees of freedom,

P1: GIM



and two of these will be satisfied by using standard temperature and pres-sure. Thus three tests on the stoichiometry of the solution will be required,and these will be tests on the total amounts of sodium and chlorine and onthe ratio of phosphorus to lead which must be three to five at equilibrium.Finally the charge of the model system must be determined. If the systemis in a beaker, the definite pH described in the previous paragraph will beobtained. This pH is labeled the natural pH here. If the Pb5(PO4)3OH is inthe environment, the aqueous phase and what that contains will determinethe pH. For this reason the models done for this paper are calculated overa range of pH. If redox chemistry is possible, calculations are done over arange of pe as well.

A review of the thermochemical tables shows that there are forty-one solute species in the hydroxypyromorphite system, beside H+ andOH−, whose activities must be determined by equilibrium-constant equa-tions found from Gibbs free energies of formation. All of this gives a few toomany simultaneous equations to be solved conveniently, so the techniquesdeveloped by the authors (described below) were used. A spreadsheet ofseveral columns was calculated (Table 1), and each column has a fixed pH.Each of the forty-four solute species, other than the hydrogen and hydroxideions, was given a row, and the first three rows (21–23 in the example) werethose that needed to be determined by trial and error in order to meet thethree tests on the stoichiometry (rows 70–72). The three trial-and-error quan-tities must be one chlorine species, one sodium species, and one of eitherlead or phosphate. In this work these choices were always neutral speciesincluding the one from the anion of the mineral, and in this case these werepHCl0, pNaOH0, and pH3PO0

4. To be sure these choices were arbitrary, butalways beginning with neutral species made the process consistent and theformulas for calculating the other species easy to debug. A sample is given inthe next paragraph. The equilibrium-constant equations were always done inlogarithmic form, and all activities are given as pa’s. (Since the spreadsheetprogram [LOTUS 1-2-3, Release 4] does not allow superscripts, it should benoted that pX[n] = pXn in the first column.)

The activity of dissolved CO2 (row 24) was taken to be constant on theassumption of an atmosphere with the gas at 270 ppm. The activities of thespecies on rows 25 through 42 were calculated from the activities on rows20 through 24 and equilibrium-constant equations derived from free energiesof formation. Examples of the equations used:

pCl[−] = pHCl[0] − pH − 2.999

pH2PO4[−] = pH3PO4[0] − pH + 2.148

pNa[+] = pNaOH[0] + pH − 13.994

pNaCO3[−] = pNaOH[0] + pCO2[0] − pH + 1.261, and so forth.

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TAB

LE1

.A

spre

adsh

eet

of

the

hyd

roxy

pyr

om

orp

hite

(Pb

5(P

O4) 3

OH

)sy

stem

ove

rth

epH

range

0.00

to12

.00

atin

terv

als

of

0.50

.The

aqueo

us

phas

eco

nta

ins

NaC

lw

ithan

activ

itysu

chth

atpN

a(t)

=pCl(t)

=3.

00.

Ther

eis

dis

solv

edca

rbon

dio

xide

ineq

uili

brium

with

the

atm

osp

her

e(2

70ppm

)gi

ving

pCO

0 2=

5.00

2.The

mak

ing

ofth

issp

read

shee

tis

des

crib

edin

the

text

.

1H

ydro

xypyr

om

orp

hite

2in

aqueo

us

solu

tion

pK

sp0

Com

ponen

tsPhas

es

3Pb5(

PO

4)3O

H17

.289

PbO

gas

4pCO

2[0]

5.00

2P2O

5so

l’n5

pCl(t)

3.00

0pH

inc.

CO

2Pb5(

PO

4)3O

H6

pN

a(t)

3.00

00.

50H

2O7

Poss

ible

carb

onat

es.et

c.pK

sp0

Na2

OP

=3

8PbCO

314

.533

HCl

9Pb(O

H)2

9.59

5F

=5

10Pb4O

(PO

4)2

16.1

16C

=6

file

nam

e:PY

RM

ORPH

11PbH

PO

418

.570

12Pb(H

2PO

4)2

23.3

0513

PbCl2

28.4

3614

Pb2C

O3C

l222

.361

15Pb3(

CO

3)2(

OH

)212

.865

16PbO

PbCO

39.

772

17Pb3(

PO

4)2

18.0

6918

Pb5(

PO

4)3O

H17

.289

19Pb5(

PO

4)3C

l22

.069

20p

H0

.00

0.5

01

.00

1.5

02

.00

2.5

03

.00

3.5

04

.00

4.5

05

.00

5.5

0

21pH

Cl[0

]8.

065

7.94

47.

866

7.90

78.

149

8.54

99.

017

9.50

610

.002

10.5

0011

.000

11.5

0022

pN

aOH

[0]

16.9

9416

.494

15.9

9415

.494

14.9

9414

.494

13.9

9413

.494

12.9

9412

.494

11.9

9411

.494

23pH

3PO

4[0]

0.00

80.

621

1.24

41.

891

2.59

33.

382

4.25

05.

162

6.09

07.

025

7.96

29.

902

24pCO

2[0]

5.00

25.

002

5.00

25.

002

5.00

25.

002

5.00

25.

002

5.00

25.

002

5.00

25.

002

25pH

2CO

3[0]

4.99

54.

995

4.99

54.

995

4.99

54.

995

4.99

54.

995

4.99

54.

995

4.99

54.

995

26pH

CO

3[−]

11.3

6510

.865

10.3

659.

865

9.36

58.

865

8.36

57.

865

7.36

56.

865

6.36

55.

865

27pCO

3[2−

]21

.729

20.7

2919

.729

18.7

2917

.729

16.7

2915

.729

14.7

2913

.729

12.7

2911

.729

10.7

2928

pCl[−

]5.

066

4.44

53.

867

3.40

83.

150

3.05

03.

018

3.00

73.

003

3.00

13.

001

3.00

129

pH

2PO

4[−]

2.15

62.

269

2.39

22.

539

2.74

13.

030

3.39

83.

810

4.23

84.

673

5.11

05.

550

30pH

PO

4[2−

]9.

354

8.96

78.

590

8.23

77.

939

7.72

87.

596

7.50

87.

436

7.37

17.

308

7.24

831

pPO

4[3−

]21

.705

20.8

1819

.941

19.0

8818

.290

17.5

7916

.947

16.3

5915

.787

15.2

2214

.659

14.0

9932

pH

4P2O

7[0]

6.94

58.

171

9.41

810

.711

12.1

1513

.693

15.4

2917

.252

19.1

1020

.979

22.8

5424

.732

33pH

3P2O

7[−]

7.74

48.

470

9.21

710

.010

10.9

1411

.992

13.2

2814

.551

15.9

0917

.278

18.6

5320

.031

34pH

2P2O

7[2−

]10

.024

10.2

5010

.497

10.7

5011

.194

11.7

7212

.508

13.3

3114

.189

15.0

5815

.933

16.8

11

(Con

tin

ued

onn

ext

page

)

499

P1: GIM


TAB

LE1

.(C

onti

nu

ed)

35pH

P2O

7[3−

]16

.723

16.4

4916

.196

15.9

6915

.893

15.9

7116

.207

16.5

3016

.888

17.2

5717

.632

18.0

1036

pP2O

7[4−

]26

.135

25.3

6124

.608

23.9

0123

.305

22.8

8322

.619

22.4

4222

.300

22.1

6922

.044

21.9

2237

pN

a[+]

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

38pN

aCl[0

]8.

081

7.46

06.

882

6.42

36.

165

6.06

66.

033

6.02

26.

018

6.01

76.

016

6.01

639

pN

aCO

3[−]

23.2

5722

.257

21.2

5720

.257

19.2

5718

.257

17.2

5716

.257

15.2

5714

.257

13.2

5712

.257

40pN

a2CO

3[0]

27.3

0826

.308

25.3

0824

.308

23.3

0922

.309

21.3

0920

.309

19.3

0918

.309

17.3

0916

.309

41pN

aHCO

3[0]

13.9

4813

.448

12.9

4812

.448

11.9

4811

.448

10.9

4810

.448

9.94

89.

448

8.94

88.

448

42pN

aHPO

4[−]

11.1

6110

.774

10.3

9710

.044

9.74

69.

535

9.40

49.

315

9.24

49.

179

9.11

69.

055

43pPb(O

H)2

[0]

17.2

8416

.916

16.5

4216

.154

15.7

3315

.260

14.7

3914

.192

13.6

3513

.074

12.5

1211

.948

44pPb[2

+]−0

.318

0.31

40.

940

1.55

22.

131

2.65

83.

137

3.59

04.

033

4.47

24.

910

5.34

645

pPbO

H[+

]7.

230

7.36

27.

488

7.60

07.

679

7.70

67.

685

7.63

87.

581

7.52

07.

458

7.39

446

pPb(O

H)3

[−]

27.6

2426

.756

25.8

8224

.994

24.0

7323

.100

22.0

7921

.032

19.9

7518

.914

17.8

5216

.788

47pPb(O

H)4

[2−]

39.0

2037

.652

36.2

7834

.890

33.4

6931

.996

30.4

7528

.928

27.3

7125

.810

24.2

4822

.684

48pPb2O

H[3

+]5.

467

6.23

26.

984

7.70

88.

366

8.91

99.

377

9.78

310

.199

10.5

4710

.922

11.2

9549

pPb3(

OH

)4[2

+]22

.495

22.3

9122

.269

22.1

0521

.842

21.4

2120

.859

20.2

1819

.546

18.8

9418

.177

17.4

8650

pPb4(

OH

)4[4

+]19

.033

19.5

6220

.066

20.5

1320

.829

20.9

3520

.851

20.6

6420

.435

20.1

9219

.942

19.6

8851

pPb6(

OH

)8[4

+]40

.789

40.5

8340

.339

40.0

1039

.484

38.6

4337

.517

36.2

3634

.893

33.5

2832

.154

30.7

7252

pPbH

PO

4[0]

5.78

86.

033

6.28

36.

541

6.88

27.

138

7.48

57.

850

8.22

18.

595

8.97

09.

346

53pPbH

2PO

4[+]

0.18

80.

933

1.68

32.

441

3.22

24.

038

4.88

55.

750

6.62

17.

495

8.37

09.

246

54pPbP2O

7[2−

]14

.375

14.2

3314

.106

14.0

1213

.994

14.0

9914

.314

14.5

9014

.890

15.1

9915

.511

15.8

2655

pPbPO

4[−]

12.1

4011

.885

11.6

3511

.393

11.1

7410

.990

10.8

3710

.702

10.5

7310

.447

10.3

2210

.198

56pPb(P

O4)

2[4−

]28

.445

27.3

0326

.176

25.0

8224

.064

23.1

6922

.384

21.6

6020

.960

20.2

6919

.581

18.8

9657

pPb(P

2O7)

2[6−

]43

.435

42.5

1941

.639

40.8

3840

.224

39.9

0739

.859

39.9

5740

.115

40.2

9340

.480

40.6

7458

pPb(H

PO

4)2[

2−]

15.7

4515

.603

15.4

7615

.382

15.3

6415

.469

15.6

8415

.960

16.2

6016

.569

16.8

8117

.196

59pPbCl[+

]3.

004

3.01

63.

064

3.21

63.

537

3.96

44.

411

4.85

35.

291

5.72

96.

166

6.60

360

pPbCl2

[0]

7.88

77.

279

6.74

96.

442

6.50

56.

833

7.24

67.

677

8.11

28.

548

6.98

59.

421

61pPbCl3

[−]

13.0

5011

.822

10.7

139.

947

9.75

29.

980

10.3

6110

.781

11.2

1211

.647

12.0

8312

.519

500

P1: GIM


62pPbCl4

[2−]

18.4

1716

.568

14.8

8113

.656

13.2

0213

.332

13.6

8014

.088

14.5

1514

.949

15.3

8415

.820

63pPbCO

3[0]

14.1

6213

.794

13.4

2013

.032

12.6

1112

.138

11.6

1711

.070

10.5

139.

952

9.39

08.

826

64pPb(C

O3)

2[2−

]32

.645

31.2

7729

.903

28.5

1527

.094

25.6

2124

.100

22.5

5320

.996

19.4

3517

.873

16.3

0965

pCl(t)

4.69

74.

697

4.69

74.

697

4.69

74.

697

4.69

74.

697

4.69

74.

695

4.68

84.

669

66pC(t)

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

67pN

a(t)

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

68pP(t)

−0.2

140.

442

1.08

71.

713

2.30

42.

842

3.32

93.

786

4.23

04.

670

5.10

75.

541

69pPb(t)

−0.4

360.

220

0.86

51.

491

2.08

22.

620

3.10

73.

564

4.00

84.

448

4.88

55.

319

70TEST

1:pCl(t)-p

Cl(*)

6.7E

-16

5.8E

-16

9.5E

-16

8.5E

-16

6.7E

-16

9.1E

-16

8.0E

-16

9.0E

-16

9.4E

-16

9.6E

-16

8.2E

-16

7.7E

-16

71Te

st2:

pN

a(t)-p

Na(

*)−8

.3E-1

6−7

.4E-1

6−9

.5E-1

6−8

.2E-1

6−5

.3E-1

6−8

.5E-1

6−7

.7E-1

6−5

.5E-1

6−5

.8E-1

6−7

.5E-1

6−7

.0E-1

6−7

.0E-1

672

TEST

3:pP(t)-

pPb(t)-

log(

5/3)

9.6E

-18

4.2E

-16

9.0E

-16

7.8E

-16

6.7E

-16

4.9E

-16

4.4E

-16

6.1E

-16

5.4E

-16

2.2E

-16

9.8E

-16

3.9E

-16

73pQ

(+)

−0.7

64−0

.147

0.45

31.

032

1.57

42.

058

2.45

52.

734

2.88

92.

959

2.88

52.

995

74p(a

bs(

Q(−

))2.

156

2.26

62.

378

2.48

42.

598

2.73

92.

866

2.94

32.

978

2.99

22.

997

2.99

975

p(a

bs(

Q(t))

−0.7

64−0

.148

0.45

91.

048

1.61

72.

159

2.66

73.

152

3.62

34.

089

4.55

75.

075

76p

H0

.00

0.5

01

.00

1.5

02

.00

2.5

03

.00

3.5

04

.00

4.5

05

.00

5.5

077

Poss

ible

pre

cipita

tes:

78pQ

sp0-

pK

sp0

79PbCO

37.

753

7.38

57.

011

6.62

36.

202

5.72

95.

208

4.66

14.

104

3.54

32.

981

2.41

780

Pb(O

H)2

7.68

97.

321

6.94

76.

559

6.13

85.

665

5.14

44.

597

4.04

03.

479

2.91

72.

353

81Pb4O

(PO

4)2

1.17

21.

111

1.04

90.

984

0.91

40.

835

0.74

80.

657

0.56

40.

470

0.37

70.

283

82PbH

PO

4−1

.278

−1.0

33−0

.783

−0.5

25−0

.244

0.07

20.

419

0.78

41.

155

1.52

91.

904

2.28

083

Pb(H

2PO

4)2

−6.0

05−5

.147

−4.2

74−3

.368

−2.3

86−1

.281

−0.0

661.

210

2.51

03.

819

5.13

16.

446

84PbCl2

4.97

74.

369

3.83

93.

532

3.59

53.

923

4.33

64.

767

5.20

25.

638

6.07

56.

511

85Pb2C

O3C

l25.

489

5.00

14.

549

4.20

14.

022

3.94

93.

896

3.83

83.

776

3.71

43.

651

3.58

886

Pb3(

CO

3)2(

OH

)27.

754

7.38

67.

012

6.62

46.

203

5.72

95.

209

4.66

24.

104

3.54

42.

981

2.41

887

PbO

·PbCO

310

.013

9.64

59.

271

8.88

38.

462

7.98

97.

468

6.92

16.

364

5.80

35.

241

4.67

788

Pb3(

PO

4)2

−0.7

79−0

.739

−0.6

97−0

.654

−0.6

07−0

.555

−0.4

97−0

.436

−0.3

74−0

.312

−0.2

49−0

.187

89Pb5(

PO

4)3O

H0.

000

0.00

00.

000

0.00

00.

000

0.00

00.

000

0.00

00.

000

0.00

00.

000

0.00

090

Pb5(

PO

4)3C

l−3

.167

−3.1

91−3

.207

−3.1

99−3

.150

−3.0

70−2

.977

−2.8

79−2

.780

−2.6

80−2

.580

−2.4

80(C

onti

nu

edon

nex

tpa

ge)

501

P1: GIM


TAB

LE1

.(C

onti

nu

ed)

206

.00

6.5

07

.00

7.5

08

.00

8.5

09

.00

9.5

01

0.0

01

0.5

01

1.0

01

1.5

01

2.0

0p

H

2111

.999

12.4

9912

.999

13.4

9913

.999

14.4

9914

.999

15.4

9915

.999

16.4

9916

.999

17.4

9917

.999

pH

Cl[0

]22

10.9

9410

.494

9.99

49.

495

8.99

58.

496

8.00

27.

532

7.21

67.

333

7.75

48.

242

8.73

9pN

aOH

[0]

239.

845

10.7

9411

.721

12.5

1413

.164

13.7

7714

.331

14.6

3714

.700

14.7

0714

.716

14.7

4314

.828

pH

3PO

4[0]

245.

002

5.00

25.

002

5.00

25.

002

5.00

25.

002

5.00

25.

002

5.00

25.

002

5.00

25.

002

pCO

2[0]

254.

995

4.99

54.

995

4.99

54.

995

4.99

54.

995

4.99

54.

995

4.99

54.

995

4.99

54.

995

pH

2CO

3[0]

265.

365

4.86

54.

365

3.86

53.

365

2.86

52.

365

1.86

51.

365

0.86

50.

365

−0.1

35−0

.635

pH

CO

3[−]

279.

729

8.72

97.

729

6.72

95.

729

4.72

93.

729

2.72

91.

729

0.72

9−0

.271

−1.2

71−2

.271

pCO

3[2−

]28

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

0pCl[−

]29

5.99

36.

442

6.86

97.

162

7.31

27.

425

7.47

97.

285

6.84

86.

355

5.86

45.

391

4.97

6pH

2PO

4[−]

307.

191

7.14

07.

067

6.86

06.

510

6.12

35.

677

4.98

34.

046

3.05

32.

062

1.08

90.

174

pH

PO

4[2−

]31

13.5

4212

.991

12.4

1811

.711

10.8

619.

974

9.02

87.

834

6.39

74.

904

3.41

31.

940

0.52

5pPO

4[3−

]32

26.6

1928

.517

30.3

7231

.956

33.2

5734

.484

35.5

9136

.203

36.3

2836

.344

36.3

6136

.415

36.5

85pH

4P2O

7[0]

3321

.418

22.8

1624

.171

25.2

5526

.056

26.7

8327

.390

27.5

0227

.127

26.6

4326

.160

25.7

1425

.384

pH

3P2O

7[−]

3417

.698

18.5

9619

.451

20.0

3520

.336

20.5

6320

.670

20.2

8219

.407

18.4

2317

.440

16.4

9415

.664

pH

2P2O

7[2−

]35

18.3

9718

.795

19.1

5019

.234

19.0

3518

.762

18.3

6917

.481

16.1

0614

.622

13.1

3911

.693

10.3

63pH

P2O

7[3−

]36

21.8

0921

.707

21.5

6221

.146

20.4

4719

.674

18.7

8117

.393

15.5

1813

.534

11.5

519.

605

7.77

5pP2O

7[4−

]37

3.00

03.

000

3.00

03.

001

3.00

13.

002

3.00

83.

038

3.22

23.

839

4.76

05.

748

6.74

5pN

a[+]

386.

016

6.01

66.

016

6.01

66.

016

6.01

86.

023

6.05

46.

238

6.85

47.

775

8.76

39.

760

pN

aCl[0

]39

11.2

5710

.257

9.25

78.

258

7.25

86.

259

5.26

54.

295

3.47

93.

096

3.01

73.

005

3.00

2pN

aCO

3[−]

4015

.309

14.3

0913

.309

12.3

0911

.310

10.3

129.

323

8.38

57.

753

7.98

68.

829

9.80

310

.798

pN

a2CO

3[0]

417.

948

7.44

86.

948

6.44

95.

949

5.45

04.

956

4.48

64.

170

4.28

74.

708

5.19

65.

693

pN

aHCO

3[0]

428.

998

8.94

78.

875

8.66

78.

318

7.93

37.

492

6.82

86.

075

5.70

05.

629

5.64

35.

726

pN

aHPO

4[−]

4311

.382

10.8

1310

.256

9.78

19.

390

9.02

38.

690

8.50

78.

469

8.46

58.

459

8.44

38.

392

pPb(O

H)2

[0]

445.

780

6.21

16.

654

7.17

97.

788

8.42

19.

088

9.90

510

.867

11.8

6312

.857

13.6

4114

.790

pPb[2

+]45

7.32

87.

259

7.20

27.

227

7.33

67.

469

7.63

67.

953

8.41

58.

911

9.40

59.

889

10.3

38pPbO

H[+

]46

15.7

2214

.653

13.5

9612

.621

11.7

3010

.863

10.0

309.

347

8.80

98.

305

7.79

97.

283

6.73

2pPb(O

H)3

[−]

4721

.118

19.5

4917

.992

16.5

1715

.126

13.7

5912

.426

11.2

4310

.205

9.20

18.

195

7.17

96.

128

pPb(O

H)4

[2−]

4811

.663

12.0

2412

.411

12.9

6113

.680

14.4

4415

.280

16.4

1317

.837

19.3

2820

.618

22.2

8623

.684

pPb2O

H[3

+]49

16.7

8816

.080

15.4

1114

.984

14.8

1314

.710

14.7

1315

.162

16.0

5017

.036

18.0

2018

.972

19.8

19pPb3(

OH

)4[2

+]50

19.4

2419

.147

18.9

2119

.019

19.4

5819

.986

20.6

5721

.923

23.7

7325

.754

27.7

3429

.669

31.4

65pPb4(

OH

)4[4

+]51

29.3

7627

.960

26.6

2125

.769

25.4

2725

.220

25.2

2626

.125

27.8

9929

.871

31.8

4133

.744

35.4

38pPb6(

OH

)8[4

+]52

9.72

310

.103

10.4

7410

.790

11.0

5111

.296

11.5

1711

.640

11.6

6511

.668

11.6

7111

.682

11.7

16pPbH

PO

4[0]

5310

.123

11.0

0311

.874

12.6

9013

.451

14.1

9614

.917

15.5

4016

.065

16.5

6917

.071

17.5

8218

.118

pPbH

2PO

4[+]

5416

.147

16.4

7516

.774

16.8

8316

.794

16.6

5216

.428

15.8

5614

.944

13.9

5412

.966

12.0

0411

.123

pPbP2O

7[2−

]55

10.0

759.

955

9.82

69.

642

9.40

39.

148

8.86

98.

492

8.01

77.

520

7.02

36.

534

6.06

8pPbPO

4[−]

502

P1: GIM


5618

.217

17.5

4516

.844

15.9

5314

.864

13.7

2212

.498

10.9

269.

014

7.02

45.

036

3.07

41.

193

pPb(P

O4)

2[4−

]57

40.8

8141

.107

41.2

6040

.954

40.1

6639

.251

38.1

3436

.174

33.3

8730

.413

27.4

4224

.534

21.8

22pPb(P

2O7)

2[6−

]58

17.5

1717

.845

18.1

4418

.253

18.1

6418

.022

17.7

9817

.226

16.3

1415

.324

14.3

3613

.374

12.4

93pPb(H

PO

4)2[

2−]

597.

037

7.46

77.

911

8.43

59.

045

9.67

710

.345

11.1

6112

.123

13.1

1914

.113

15.0

9716

.046

pPbCl[+

]60

9.85

510

.286

10.7

2911

.254

11.8

6312

.495

13.1

6313

.980

14.9

4215

.937

16.9

3117

.915

18.8

64pPbCl2

[0]

6112

.952

13.3

8313

.826

14.3

5114

.961

15.5

9316

.261

17.0

7718

.039

19.0

3420

.028

21.0

1221

.961

pPbCl3

[−]

6216

.254

16.6

8417

.128

17.6

5218

.262

18.8

9419

.562

20.3

7821

.340

22.3

3523

.329

24.3

1325

.262

pPbCl4

[2−]

638.

260

7.69

17.

134

6.65

96.

268

5.90

15.

568

5.38

55.

347

5.34

35.

337

5.32

15.

270

pPbCO

3[0]

6414

.743

13.1

7411

.617

10.1

428.

751

7.38

46.

051

4.86

83.

830

2.82

61.

820

0.80

4−0

.247

pPb(C

O3)

2[2−

]65

4.61

34.

471

4.19

83.

803

3.34

22.

851

2.34

31.

806

1.20

40.

486

−0.3

67−1

.308

−2.2

89pC(t)

663.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

pCl(t)

673.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

3.00

03.

000

pN

a(t)

685.

966

6.36

16.

653

6.67

96.

440

6.09

55.

663

4.97

44.

039

3.04

62.

042

1.02

4−0

.040

pP(t)

695.

744

6.13

96.

431

6.45

76.

218

5.87

35.

442

4.75

23.

817

2.62

41.

620

0.80

2−0

.262

pPb(t)

705.

2E-1

65.

8E-1

69.

2E-1

67.

8E-1

65.

1E-1

67.

5E-1

66.

4E-1

66.

8E-1

68.

6E-1

68.

8E-1

68.

6E-1

68.

7E-1

66.

2E-1

6TEST

1:pCl(t)-p

Cl(*)

71−5

.2E-1

6−7

.0E-1

6−6

.1E-1

6−8

.9E-1

6−9

.9E-1

6−5

.8E-1

6−6

.9E-1

6−6

.5E-1

6−5

.2E-1

6−9

.8E-1

6−8

.9E-1

6−7

.4E-1

6−8

.4E-1

6TEST

2:pN

a(t)-p

Na(

*)72

4.2E

-16

2.5E

-16

2.8E

-16

−6.7

E-1

6−2

.3E-1

6−3

.8E-1

69.

9E-1

6−3

.2E-1

6−8

.2E-1

6−4

.7E-1

6−9

.5E-1

6−5

.7E-1

6−8

.0E-1

6TEST

3:pP(t)-

pPb(t)-

log(

5/3)

732.

998

3.00

03.

000

3.00

03.

001

3.00

23.

008

3.03

83.

222

3.83

94.

760

5.74

86.

745

pQ

(+)

742.

998

2.99

42.

982

2.94

42.

843

2.61

92.

243

1.73

31.

084

0.26

7−0

.625

−1.5

93−2

.584

p(a

bs(

Q(−

))75

6.02

94.

901

4.36

63.

861

3.35

82.

850

2.32

51.

755

1.08

70.

287

−0.6

25−1

.593

−2.5

84p(a

bs(

Q(t))

766

.00

6.5

07

.00

7.5

08

.00

8.5

09

.00

9.5

01

0.0

01

0.5

01

1.0

01

1.5

01

2.0

0p

H77 78

pQ

sp0

−pK

sp0

791.

851

1.28

20.

725

0.25

0−0

.141

−0.5

08−0

.841

−1.0

24−1

.062

−1.0

66−1

.072

−1.0

88−1

.139

PbCO

380

1.78

71.

218

0.66

10.

186

−0.2

05−0

.572

−0.9

05−1

.080

−1.1

26−1

.130

−1.1

36−1

.152

−1.2

03Pb(O

H)2

810.

189

0.09

40.

001

−0.0

78−0

.143

−0.2

05−0

.260

−0.2

91−0

.297

−0.2

98−0

.299

−0.3

01−0

.310

Pb4O

(PO

4)2

822.

657

3.03

73.

408

3.72

43.

985

4.23

04.

451

4.57

44.

599

4.60

24.

605

4.61

64.

650

PbH

PO

483

7.76

79.

095

10.3

9411

.503

12.4

1413

.272

14.0

4814

.476

14.5

6414

.574

14.5

8614

.624

14.7

43Pb(H

2PO

4)2

846.

945

7.37

67.

619

8.34

48.

953

9.58

510

.253

11.0

7012

.032

13.0

2714

.021

15.0

0515

.954

PbCl2

853.

522

3.45

23.

396

3.42

03.

530

3.66

23.

830

4.14

64.

608

5.10

45.

598

6.08

26.

531

Pb2C

O3C

l286

1.85

21.

282

0.72

60.

250

−0.1

40−0

.508

−0.8

40−1

.024

−1.0

61−1

.066

−1.0

71−1

.087

−1.1

38Pb3(

CO

3)2(

OH

)287

4.11

13.

542

2.98

52.

510

2.11

91.

752

1.41

91.

236

1.19

81.

194

1.18

81.

172

1.12

1PbO

.PbCO

388

−0.1

24−0

.060

0.00

10.

054

0.09

80.

138

0.17

50.

196

0.20

00.

200

0.20

10.

203

0.20

9Pb3(

PO

4)2

890.

000

0.00

00.

000

0.00

00.

000

0.00

00.

000

0.00

00.

000

0.00

00.

000

0.00

00.

000

Pb5(

PO

4)3O

H90

−2.3

80−2

.280

−2.1

80−2

.080

−1.9

80−1

.880

−1.7

80−1

.680

−1.5

80−1

.480

−1.3

80−1

.280

−1.1

80Pb5[

PO

4]3C

l

503

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In this work these equations were always written in terms of the neutralspecies, and this made the form of each equation easy to check because thesign and magnitude of the pH term equaled the ion’s charge.

Row 43 has pPb(OH)2[0], and this activity was found as follows: A quan-tity named pK0

sp was calculated for the mineral, and this is the negative log-arithm of the equilibrium constant for the reaction shown by equation [1].

9/5 H2O + 1/5 Pb5(PO4)3OH = Pb(OH)02 + 3/5 H3PO0

4 [1]

The conventions used are (1) the mineral is on the left-hand side, (2) the num-ber of moles of the metal is one, and (3) the species on the right-hand sideare uncharged. Free energies of formation were then used to find pK0

sp, and itis 17.289 for hydroxypyromorphite. It is shown on row 3 of the spreadsheet,and the pK0

sp’s for other possible precipitates from the system’s componentsare shown on rows 8 through 19. These conventions allowed the calculationof pPb(OH)02 on row 43 from

pPb(OH)02 = pK0

sp − 3/5p H3PO04

The activities of the other lead species were then calculated in terms ofpH, pHCl0, pH3PO0

4, pCO02, and pPb(OH)02 and put on rows 44 through 64.

The total activities of each of the five elements in solution, C, Cl, Na,P, and Pb were found by summing the activities of each of the species con-taining them, and these are given on rows 65 through 69 as pX(t)’s. Thesetotals were then used to make the tests on stoichiometry which determinedthe values for the variables on rows 21 through 23. A macro, written in thelanguage supplied with Lotus 1-2-3, was used to make, by trial and error,each of the three agreements to be better than 1 × 10−15. These are shownon rows 70 through 72. (Experience has shown that the agreement must bethis close to get with consistency smooth curves on the figures.)

Thirty of the species in solution are charged, and ten of these are cations.The negative log of the sum of the cation charges in moles/liter is given online 73. The negative log of the absolute value of the anion charges is shownon line 74, and the negative log of the absolute value of the total charge is online 75. Figure 1 is a graph of rows 73 and 74, and it shows the lines crossingat pH 6. This pH is what would occur if the system were made exactly asdescribed, and this is the natural pH. In a soil environment or in an analyticaltesting procedure the pH will usually be determined for the system ratherthan by it, and the charges of the system described by the spreadsheet will beunbalanced even as no actual system ever could be.

Rows 79 through 90 are devoted to calculations to see if other mineralsor salts of the system’s components might precipitate. Further, pQ0

sp has thesame form as pK0

sp but uses the actual activities instead of the ones neededfor equilibrium. Thus, row 79, for cerrusite, is row 43 plus row 24 minus the

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FIGURE 1. The hydroxypyromorphite (Pb5(PO4)3OH) system: total cation charges as pQ(+)and total anion charges as p(abs(Q(−))). The natural pH of this system is shown by thecrossing of these lines.

pK0sp for cerrusite, which is 14.533 (row 8). Positive results for the difference

show that the mineral cannot form, while negative results would show thatit might. As can be seen, a number of other minerals may precipitate here,and the most prominent is chloropyromorphite, Pb5(PO4)3Cl.

The finished spreadsheet was then used to make graphs of the mineral’ssolubility as a function of pH. The principal solute species of lead and phos-phorus for such a system are shown in Figures 2 and 3. These graphs have anumber of interesting features. The solubility of the mineral reaches a min-imum at a pH of about 7.5, about 1.5 above the natural pH, and increasessharply on either side of that. The principal lead species is the dipositiveion at low pH, but the complexes of chloride and phosphate are seen to beclose behind in activity. At high pH the carbonate complexes accentuate thetendency of the mineral to be somewhat amphoteric. Surely a system withhigh phosphate or chloride concentration, as in a treatment or soil high inthese elements, might result in a strong increase in soil-lead solubility leadingto precipitation in more stable Pb forms. The chloride concentration in thismodel was arbitrarily fixed at one millimole per kilogram of water.

Some of the systems that will be discussed in this paper have thepossibility of redox chemistry with sulfur, arsenic, and mercury. The en-vironment that each system finds itself in will determine the redox chem-istry of these three elements, and it will be necessary to represent in aquantitative way the conditions just as is done with the pH. One wayto do this is to increase the number of components in the system by

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FIGURE 2. The hydroxypyromorphite (Pb5(PO4)3OH) system: the activities as functions ofpH of the important solute species which contain lead.

noting that water is made of H2 and O2. This increases the degrees offreedom as well and does allow for changes in redox conditions. A sim-pler method which allows the methods of calculation described aboveto be used is to make the electron a reactant with an activity given bype, i.e., the negative log of the virtual activity of the free electron insolution.

FIGURE 3. The hydroxypyromorphite (Pb5(PO4)3OH) system: the activities as functions ofpH of the important solute species which contain phosphate.

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In what follows, when one of these three elements is part of the system,the spreadsheets and graphs are calculated across the pH range at someconstant value of (pH + pe) and with H2O as one component rather thantwo or calculations are made across an area in pH-pe space. Using constantvalues of (pH + pe) gives some interesting results. We find, for example,that PbS and PbSO4 can co-exist at equilibrium at a constant value of (pH +pe), which we can calculate as 5.055. Incidentally, virtually all of the sulfurin solution over these two minerals at equilibrium is sulfate rather sulfide.If (pH + pe) > 5.055, then only lead sulfate will be stable, while PbS willremain unoxidized below 5.055.

pK0sp for each mineral containing sulfur is calculated using pH2S0, re-

gardless of the actual oxidation state in the mineral. So for the two leadminerals:

{PbS, galena} pK0sp = pPb(OH)0

2 + pH2S0 = 25.329

{PbSO4, anglesite} pK0sp = pPb(OH)0

2 + pH2S0 − 8(pH + pe) = −15.111

Subtracting the second equation from the first gives (pe + pH) = 5.055. pK0sp

values for minerals of iron are calculated using Fe(III), while arsenic mineralsuse As(V). Calculations for Hg will use Hg0, that is the atom as a solute.

2. Systems with Several Minerals

The methods described above are easy to extend to systems with severalminerals, and two examples will be given in this section. If the system hastwo minerals rather than one without an increase in the number of compo-nents, the number of degrees of freedom falls to four, and only two testson the stoichiometry are needed. For example we may consider the systemhydroxypyromorphite-cerrusite in contact with the same solution of NaCland dissolved carbon dioxide. Now F = 4, and we need test only pCl(t) andpNa(t). There are two minerals and, therefore, two pK0

sp equations:

hydroxypyromorphite {Pb5(PO4)3OH}: pK0sp = pPb(OH)0

2 + 3/5 pH3PO04

= 17.289

cerrusite {PbCO3}: pK0sp = pPb(OH)0

2 + pCO02

= 14.533

Since pCO02 is fixed by the atmosphere at 5.002, pPb(OH)02 = 9.531, and

pH3PO04 = 12.930. These quantities are constant over the entire pH range, and

we may find the activities of all 46 solute species as before. Figure 4 showsthe total element concentrations with the changes in lead concentrations fromthe hydroxypyromorphite-only system, shown as pPb(*).

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FIGURE 4. Hydroxypyromorphite (Pb5(PO4)3OH) with cerrusite (PbCO3): elemental activitieswith pPb(t) compared to lead in the Pb5(PO4)3OH-only system, Figure 2.

Since pyromorphite will never be the only source of phosphate in a soilsystem, it is useful to examine its interaction with lead carbonate in a slightlydifferent way. Consider the chemical equilibrium [2] below which also showsthe chemistry under discussion.

4/5 H2O + pCO02 + 1/5 Pb5(PO4)3OH = PbCO3 + 3/5 H3PO0

4 [2]

The equilibrium constant of this reaction may be found from the differencebetween the pK0

sp’s as in

pKeq = 17.289 − 14.533 = 2.756

At equilibrium

2.756 = 3/5 pH3PO04 − pCO0

2

If pCO02 is fixed at 5.002 from the atmosphere, then pH3PO0

4 is 12.930 asbefore, but if pH3PO0

4 is not 12.930, as is usually the case; the equilibriumwill shift one way or the other. Thus we see that if the total phosphate con-centration is below the pP(t) line of Figure 4, the carbonate will be changedto hydroxypyromorphite. Since this line represents a very small concentra-tion of phosphate at all pH’s below 9, we see that the conversion of cerrusiteto hydroxypyromorphite may be easily done by adding phosphate unlesssomething else is precipitating it.

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A different sort of problem arises if we have two minerals rather thanone and there is one more component. An example of this sort of problemis anglesite (PbSO4), hydroxypyromorphite, a solution, and the atmosphere.The new component is SO3 (used in this example instead of H2S by assum-ing oxidizing conditions), and there are now seven of them. There are fivedegrees of freedom, and there will be three tests on the stoichiometry. Thereare two pK0

sp equations:

anglesite {PbSO4}, pK0sp = pPb(OH)0

2 + pH2SO04 = 25.540

pyromorphite {Pb5(PO4)3OH}, pK0sp = pPb(OH)0

2 + 3/5 pH3PO04 = 17.289

The three quantities in these equations cannot be constant, because twoequations are not sufficient to determine three variables. There will be a teston the stoichiometry, and it will be given by the following equation, providedthat the only sources of these elements are the two minerals.

Pb(t) = 5/3 P(t) + S(t)

The result is Figure 5 which implies that pyromorphite is less solublethan anglesite at pH’s over 4 and more soluble below 4. The spreadsheetand this graph were constructed on the assumption that the only sourcesof phosphate and sulfate in the solution were the minerals themselves, butthis situation is unlikely in a soil environment. If the two minerals are inequilibrium with a solution and each other, the shift of lead from one to

FIGURE 5. Hydroxypyromorphite (Pb5(PO4)3OH) with anglesite (PbSO4): elemental activitiesvs. pH.

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FIGURE 6. Hydroxypyromorphite (Pb5(PO4)3OH) with anglesite (PbSO4): the negative logof the phosphate-to-sulfate ratio vs. pH in the system of Figure 5.

the other is always possible, and Figure 6 shows what the P:S ratio must beto make either reaction happen. If the P:S ratio is actually above the curve,hydroxypyromorphite will spontaneously change to anglesite, while if theratio is below the curve, the opposite reaction may occur. And as can be seen,the pH dependence is quite strong. At pH’s close to neutral, the conversionof anglesite to pyromorphite is easy in most cases, but there are plenty ofpolluted sites, such as those close to old mining and smelting operations,where the concentration of sulfate is very high. These sites may be quiteacidic as well, and this will make conversion to the phosphate difficult. Thelesson from these observations would seem to be that we must know thechemistry of the soil system quite well, far beyond knowing how much leadis present and in what form.

A further illustration of the complexity of the chemistry of these systemsmay be seen in Figure 7, which shows the principal lead-containing speciesas a function of pH. There are sixteen such species in the system, and thirteenare present in sufficient quantities to appear on the graph. Five of these arecomplex ions of phosphate and chloride, and higher concentrations than wehave here of either of these anions might increase the solubility substantially–even when very insoluble lead minerals are present. The two carbonatecomplexes are quite important at high pH and strengthen the tendency oflead minerals to be amphoteric.

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FIGURE 7. Hydroxypyromorphite (Pb5(PO4)3OH) with anglesite (PbSO4): activities of theimportant solute species of lead vs. pH.

III. STRATEGIES FOR LEAD IMMOBILIZATION

1. Slightly Soluble Lead Minerals

If lead immobilization is going to work, the metal will have to be put intoa form which is highly insoluble over a large pH range including that foundin the stomach after ingestion.21,26,27 To see what forms might qualify, wecalculated the solubility of several lead minerals as a function of pH by themethods described above. In each case the solution in contact had carbondioxide from the atmosphere, assumed to be 270 ppm, and a concentrationof sodium chloride equal to one millimole per liter. The results are given inFigure 8. Surely the best candidates are galena (PbS), chloropyromorphite(Pb5(PO4)3Cl), and wulfenite (PbMoO4). Galena is a common form for theelement in nature, and it is quite insoluble. Unfortunately, it is subject tooxidation in the air, and it slowly goes to anglesite (PbSO4), which is severalorders of magnitude more soluble. Wulfenite is a desirable form, but makingit requires the addition of molybdates to soils, and such a treatment couldcause more problems than it might solve. Chloropyromorphite leads to nosuch difficulty, as phosphate is a constituent of all living cells and a wellknown and useful fertilizer.

Possible problems were set aside, and the modeling techniques de-scribed above were used to discover how easily these minerals might beconverted, one to another. The first model done was to study to the conver-sion of hydroxypyromorphite to chloropyromorphite. The calculations on

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FIGURE 8. The solubilities as pPb(t) of several lead minerals as functions of pH.

row 83 of Table 1 show that the chemical potential for the conversion isnegative under the conditions of the model, and another was built to findthe minimum concentration of chloride necessary for the conversion. Thismodel assumed the presence of both pyromorphites and had four phasesand six components. There were four degrees of freedom, and, therefore,two tests on stoichiometry: The ratio of lead to phosphate must be five tothree, and the total concentration of sodium must be determined. Here wefix it at the same level as the total chloride, and in this system pHCl0 wasfixed by the two pK0

sp equations:

{Pb5(PO4)3Cl}, pK0sp = 22.069 = pPb(OH)0

2 + 3/5pH3PO04 + 1/5pHCl0

{Pb5(PO4)3OH}, pK0sp = 17.289 = pPb(OH)0

2 + 3/5pH3PO04

so pHCl0 = 23.900.

The trial-and-error variables were, therefore, pH3PO04 and pNaOH0. The re-

sults, as total element activities as functions of pH, are shown in Figure 9.This graph shows that the amount of chloride needed to sustain the equi-librium is quite small at all pH’s, and we should, therefore, expect thatchloropyromorphite will be the mineral which will form, if chloride is presentat all.

Next four separate models were made of pairs of minerals with one halfof each pair being chloropyromorphite. The other minerals were cerrusite(PbCO3), anglesite (PbSO4), galena (PbS), and wulfenite (PbMoO4). Thesemodels gave the ratios of phosphate to the other anions at equilibrium suchas was done with the hydroxypyromorphite-anglesite equilibrium in Figure 6.

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FIGURE 9. Hydroxypyromorphite (Pb5(PO4)3OH) with chloropyromorphite (Pb5(PO4)3Cl):elemental activities as functions of pH. All the phosphate in solution is from the minerals, andpNa(t) is forced to be equal to pCl(t).

The results are given as Figures 10 through 13, and these show that convertingcerrusite and anglesite to chloropyromorphite will be relatively easy at mostpH’s. The chloropyromorphite-wulfenite system could go either way, buteither form is probably an excellent result for immobilization. The conversion

FIGURE 10. Chloropyromorphite (Pb5(PO4)3Cl) with cerrusite (PbCO3): elemental activitiesvs. pH.

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FIGURE 11. Chloropyromorphite (Pb5(PO4)3Cl) with anglesite (PbSO4) [oxidizing conditions]:elemental activities as functions of pH.

of galena to chloropyromorphite is another matter, and the phosphate-sulfideratio in the solution is going to have to be quite large.

Figure 14 shows calculations of p{P(t)/X(t)} for each of the four equi-libria discussed in the previous paragraph, where X is carbonate, sulfate,sulfide, or molybdate. The larger this number the easier it is to make

FIGURE 12. Chloropyromorphite (Pb5(PO4)3Cl) with galena (PbS) [reducing conditions]:elemental activities as functions of pH.

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FIGURE 13. Chloropyromorphite (Pb5(PO4)3Cl) with wulfenite (PbMoO4): elemental activi-ties vs. pH.

chloropyromorphite. So the ratio of phosphate to carbonate can be quitesmall, and cerrusite will still become chloropyromorphite. But with galenathe phosphate-to-sulfide ratio will have to be very large, on the order of 104

at least. Of course PbS is by itself quite insoluble, and the concentration of

FIGURE 14. The negative logs of the ratios of phosphate to carbonate, sulfate, sulfide, andmolybdate as functions of pH in the systems of Figures 10 through 13.

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sulfide over it in solution is not large, as shown in Figure 8. Whether anyfeasible phosphate treatment will give sufficient phosphate in solution to re-act with galena remains to be seen. Other soil constituents, such as metaloxides and carbonates, are known to rapidly precipitate soluble phosphateand would probably limit the conversion of galena to pyromorphite.

2. The Fates of Phosphate Additions to Soils

This section will discuss possible phosphate treatments and their likely fateswhen added to soils. There are several elements in soils present in largeamounts as oxides and carbonates which have the capacity to precipitatephosphate and lower its activity in ground water. This chemistry must beunderstood before we can determine whether any phosphate treatment couldever change lead in contaminated soils to chloropyromorphite.

The elements which must be considered are aluminum, iron, calcium,magnesium, and manganese. Soil compositions vary widely, but if we use theaverage compositions from Lindsay’s table {p7},22 we can get a rough ideaof how much phosphate might react with each. The numbers which followare in micromoles of the element per gram of soil with the content in partsper million following: P (20 µmol/g, 600 ppm); Al (2600, 70,000); Fe (700,40,000); Ca (350, 14,000); Mg (200, 5000); and Mn (11, 600). With the excep-tion of manganese all the amounts of the metallic elements are much largerthan the amount of phosphate in an “average” soil. All five of these elementsform multiple insoluble phosphates, and we must consider them all if we aregoing to understand how the activity of phosphate in the water is controlledby them. Fortunately, thermodynamic data are available for these salts andminerals, and it was possible to calculate pK0

sp’s for them. The Appendixcontains these numbers as well as values for all the minerals used in thispaper hereafter. All of these compounds can and do occur in soils, and thereare as well cations of the five elements either free in the water or adsorbedon clays or in combination with a variety of organic molecules. These factsmake for systems of staggering complexity, but modeling is possible simplybecause the phosphate minerals will be in the thermodynamic sense the moststable forms of phosphate. Ultimately, as the systems go to equilibrium, thephosphate will be found in the least soluble phosphate minerals, and we canmake useful models by discovering which these are.

We will consider the five most common metallic elements in order oftheir abundances beginning with aluminum. The chemistry of this elementin soils is complex, and it is found in many minerals, not just the hydroxideor oxide.22 (There is no stable Al carbonate.) Aluminum is most commonlyfound in alumino-silicate minerals such as pyrophyllite and albite.28 Aluminaor gibbsite is usually found only in weathered soils as silica tends to dissolvemore rapidly over time than alumina.29,30 In spite of the complexity, we can

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do a rather simple analysis to understand the effect on phosphate. Of the twophosphates of aluminum, variscite is the less soluble. Its pK0

sp equation is

{variscite, AlPO4·2H2O}pK0sp = pAl(OH)0

3 + pH3PO04 = 19.250

If pAl(OH)03 is determined by the alumino-silicate minerals, and it would

seem that it is, we can find the value of pH3PO04 and determine the solubility

of phosphate as a function of pH. If we have gibbsite, then

pAl(OH)03 = pK0sp = 6.956, and

pH3PO04 = 12.294

If we have an alumino-silicate mineral such as dickite, we will also havesilica and the following equations:

{silica, SiO2}pK0sp = pH4SiO0

4 = 3.096

{dickite, Al2Si2O5(OH)4}pK0sp = 15.474 = pAl(OH)0

3 + pH4SiO04

giving pAl(OH)03 = 12.378.

(There are in fact several forms of silica, and each has its own pK0sp. Here we

have used an active form, which Lindsay22 labels silica, “soil,” and such a sub-stance will suppress the level of aluminum. Thus we get the range discussedimmediately hereafter.) If we have sillimanite with silica, the same sort ofcalculation gives pH3PO0

4 = 9.673, and these two numbers give a range forthis activity for all the alumino-silicate minerals which we have found in theliterature. Figures 15 and 16 show the total element concentrations for thesystems of variscite, silica, and either dickite or sillimanite. It is reasonable tosuppose that the pP(t) line for any similar system would be between the twolines shown. Comparing either of these to Figure 12 shows that the phos-phate activity should be sufficient to convert galena to chloropyromorphiteabove pH six. But we have not yet considered the other abundant metallicelements, and they may impact the activity of phosphate in solution.

The second most abundant element on our list is iron (700 µmol/gsoil). The most likely and stable combination is iron(III) oxide and strengite,FePO4·2H2O, but the situation is complicated by the fact that iron formsseveral oxides from Fe(III), several more from Fe(II), and a number withmixed oxidation states. Furthermore, iron(II) forms two phosphates, and themore stable of these is vivianite, Fe3(PO4)2·8H2O. Using the same techniqueas was used to find the equilibrium between galena and anglesite (p 5),(pe + pH) was set at 8.763 so that strengite and vivianite might co-exist. Ofthe several iron oxides and hydroxides, the one described as “crystalline”24

with a pK0sp of 8.083 was used because this was the smallest pK0

sp of the

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FIGURE 15. Variscite (AlPO4·2H2O), silica (SiO2), and dickite (Al2Si2O5(OH)4): elementalactivities vs. pH.

several calculated from the data in the literature, leading to the lowest activityof phosphate in the solution. (See the Appendix.)

If we assume sodium chloride in solution at one mmol/kg and carbondioxide from the atmosphere as before, we have a system of four phases withsix components and four degrees of freedom. Figure 17 shows the results,and the amount of phosphate in solution is similar to that of Figure 16, that

FIGURE 16. Variscite (AlPO4·2H2O), silica (SiO2), and sillimanite (Al2SiO5): elemental activi-ties vs. pH.

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FIGURE 17. Iron(III) hydroxide “crystalline” (Fe(OH)3), strengite (FePO4·2H2O), and vivianite(Fe3(PO4)2·8H2O): elemental activities and the activities of Fe(II) and Fe(III) as functions ofpH when (pH + pe) = 8.763, so that the two phosphates will be in equilibrium.

is to say sufficient to convert galena to chloropyromorphite. The situationdescribed by this model is what might be obtained by adding freshly madeiron(III) hydroxide to a soil. Over time and with weathering the more sta-ble oxides are likely to form, and these will give lower levels of iron in thewater along with higher levels of phosphate. Consequently, we may say thatthe combination of iron(III) oxide and iron phosphate will give a level ofdissolved phosphate sufficient to the transformation of galena to chloropyro-morphite, at least as long as the conditions are not highly reducing. Anothermodel was made with iron(III) hydroxide and vivianite at (pe + pH) equalto four (results not shown). Compared to Figure 17, it was seen that theiron activity increased by two or three orders, with much more Fe(II), andthat the phosphate activity decreased by two or three orders. This combina-tion would probably not change galena. It should also be noted that chang-ing iron(III) hydroxide to iron(II) oxide or carbonate takes severely reduc-ing conditions. “Crystalline” Fe(OH)3 will co-exist with Fe(OH)3 (siderite) at(pe + pH) = 3.638, and it will take conditions much more reducing than thisfor hematite, Fe2O3 (pK0

sp = 11.820) to co-exist with siderite.The third most abundant element is calcium (350 µmol/g soil), and

the situation is complicated by the existence of an oxide, a hydroxide, acarbonate, and at least eight minerals of the system CaO-P2O5-H2O-CO2.The solubilities of six of these as functions of pH are shown in Figure 18.Over much of the range the least soluble, and therefore the most likely tobe stable is hydroxyapatite, Ca10(PO4)6(OH)2.

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FIGURE 18. The solubilities vs. pH of six minerals and crystals of the system CaO-P2O5-H2O-CO2.

In a thought experiment calcite, CaCO3, was added to the system shownas Figure 17. The results are shown in Figure 19, and the most important resultis that the activities of calcium and phosphorus are both high. The naturalpH of this system is close to 7.5, and the system might exist as shown. Othercalculations show, however, that the chemical potentials for the formations ofthe minerals of the system CaO-P2O5-H2O (Figure 18) are almost all negative.

FIGURE 19. The system of Figure 17 plus calcite (CaCO3): elemental activities vs. pH.

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FIGURE 20. Iron(III) hydroxide “crystalline” (Fe(OH)3), apatite (Ca10(PO4)6(OH)2), and cal-cite (CaCO3): elemental activities vs. pH. (pH + pe) = 8.763 as in Figure 17, but the phosphatesof iron will not form.

These results tell us that the strengite will likely disappear and that a calciumphosphate will form by a metathesis reaction like [3].

CaCO3 + FePO4·2H2O + H2O = Fe(OH)3 + 1/10Ca10(PO4)6(OH)2

+ CO02 + 3/5 H3PO0

4 [3]

In a second thought experiment we have all the variscite being changedto iron(III) oxide and apatite, which occurs as Equation [3] goes to the right-hand side. We also assume that since calcium is much more abundant in soilsthan phosphate, some calcite will remain. This system is shown in Figure 20,and the most important result here is that the excess calcium suppresses thelevel of phosphate to the point where the conversion of galena to chloropy-romorphite is either not possible or marginal (compare the pP(t) curves ofFigures 20 and 12). Figure 20 also shows that the calcium concentrationbecomes very large below pH 6 indicating that calcite does in fact dissolve,and that this model does not hold at low pH.

Figure 21 shows the same components under more reducing conditionswith (pH + pe) = 4.417. This is the boundary between apatite and vivianite,with the latter being the single stable phosphate below 4.417. The levelof dissolved phosphate is still quite low here and controlled by apatite. If(pH + pe) < 4.417, phosphate is controlled by iron (vivianite) and is evenlower.

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FIGURE 21. The system of Figure 20 at (pH + pe) = 4.417, the condition under whichvivianite (Fe3(PO4)2·8H2O) will be in equilibrium with apatite (Ca10(PO4)6(OH)2).

If we assume a total calcium content in the soil of 350 µmol/g soil,a soil-to-water ratio of 10:1, and 10% of the calcium in solution; we haveFigure 22. The pP(t) curve here is higher than the one in Figure 20 at pH’sabove 6.5. The upshot is that having calcium in excess of phosphate, as italmost always is in a soil, results in a dramatic depression of the amount of

FIGURE 22. Iron(III) hydroxide “crystalline” (Fe(OH)3), apatite (Ca10(PO4)6(OH)2), and dis-solved calcium with the molality at 0.342. (pH + pe) = 8.763 as in Figures 17 and 20.

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FIGURE 23. Silica (SiO2), bobberite (Mg3(PO4)2·8H2O), apatite (Ca10(PO4)6(OH)2), and cal-cite (CaCO3): elemental activities vs. pH. This system is not thermodynamically stable asdescribed in the text.

dissolved phosphate available to change lead minerals such as galena. Eitherthe amount will be so low that no reaction will occur at all, or when it ishigh enough to allow reaction, the reaction itself will deplete the phosphateto the point that the change will stop. Furthermore, the chemical potentialsfor the formations of iron-phosphate minerals are positive, indicating thatphosphate must bind with calcium rather than iron.

The magnesium content of the “average” soil is a bit less than that ofcalcium at 200 µmol/g soil. This element forms a number of phosphateminerals as well as the hydroxide (brucite), two carbonates, and dolomite,MgCa(CO3)2. Values for pK0

sp for each of these were calculated, and modelswere made. The least soluble of the phosphates, as shown by the sizes of pK0

spis boberrite, Mg3(PO4)2·8H2O, and Figure 23 shows the results when this min-eral is added to the calcite-apatite system. Silica was also put into the modelbecause magnesium is known to form a number of silicate minerals. The silicawhich Lindsay22 labels “soil” is also used here. It is the most active and mostlikely to model a real soil environment. The line for pP(t) reaches a maximumof 5.4, sufficiently low to convert galena to chloropyromorphite, but calcula-tions of chemical potentials show that a number of other minerals may form.These other minerals include several magnesium silicates, two carbonates,and dolomite. Surely this system will not be thermodynamically stable.

Two other models were run by substituting carbonates for the boberrite.Magnestite, MgCO3, gave a pP(t) curve with a maximum of 7.34 at pH 6.5,and only dolomite’s formation had a negative chemical potential. Figure 24shows the results when dolomite, MgCa(CO3)2, is substituted for magnestite,

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FIGURE 24. The system of Figure 23 with dolomite (MgCa(CO3)2) in place of bobberite. Thissystem is stable, but the phosphate level is controlled by the apatite-calcite combination.

and here pP(t) peaks at 8.36 at pH 7.0. This curve is very similar to that seenin Figure 20, and in fact the level of phosphate is being controlled by theapatite-calcite combination in both cases.

The upshot of this analysis is that phosphate in soils will not be con-trolled by aluminum, iron, or magnesium as long as calcium is present in itsusual abundance. This leaves only manganese to consider, and this elementmay be important in spite of the fact that its typical abundance is only abouthalf that of phosphate. Figure 25 shows the system apatite-calcite-Mn3(PO4)2.The phosphate curve has not shifted at all, that is the addition of this com-pound of manganese has no effect on the phosphate level. On the other hand,three other compounds of manganese, Mn(OH)2, MnCO3, and MnHPO4 allhave negative chemical potentials for their precipitation reactions.

Figure 26 shows what happens when MnHPO4 is substituted forMn3(PO4)2 in the system of Figure 25. The phosphate and calcium curves arethe same, but the curves for carbonate and manganese are shifted upward.MnHPO4 is less soluble than Mn3(PO4)2, and the chemical potentials showthat no other compound of this element could precipitate. It would seemfrom this that the presence of manganese has no effect on the phosphate,but this is not quite so. Consider reaction [4].

1/5 H2O + CaCO3 + MnHPO4 = MnCO3 + 1/10 Ca10(PO4)6(OH)2

+ 2/5 H3PO04 [4]

pK0eq = 19.679 + 26.253 − 16.007 − 23.809 = 6.116

at equilibrium, pH3PO04 = 15.290.

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FIGURE 25. Manganese phosphate (Mn3(PO4)2), apatite (Ca10(PO4)6(OH)2), and calcite(CaCO3): elemental activities vs. pH.

This result represents a very low activity of phosphate, very similar to thatseen in Figure 26. As a consequence very low concentrations of solublephosphate will be sufficient to shift phosphate from apatite to MnHPO4, andany added phosphate will surely do so. This salt is, therefore, the ultimatephosphate sink in these systems.

FIGURE 26. Manganese hydrogen phosphate (MnHPO4), apatite (Ca10(PO4)6(OH)2), and cal-cite (CaCO3): elemental activities as functions of pH.

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TABLE 2. Equilibria Between Various Metal Oxidesand Carbonates with Phosphates, Ordered by theEquilibrium Values of pH3PO0

4

Reactant Product pH3PO04

Al2Si2O5(OH)4 AlPO4·2H2O + SiO2 6.87Fe(OH)3 FePO4·2H2O 7.40Al(OH)3 AlPO4·2H2O 8.69MnCO3 Mn3(PO4)2 9.45MgCO3 Mg3(PO4)2·8H2O 12.20CaCO3 Ca10(PO4)6(OH)2 15.22MnCO3 MnHPO4 15.25

Furthermore, the equilibrated system of [4] is short one degree offreedom, which is to say that one crystalline phase must disappear. Fourmodels were constructed with one each of the four solid phases missing,and the stable system, as shown by the chemical potentials for precipitat-ing the missing phase, was that shown by Figure 26. Therefore, MnCO3,rhodochrosite, will disappear when this system becomes thermodynamicallystable.

Table 2 summarizes the equilibria of the phosphates of the commonmetallic elements in soils with their oxides and carbonates. Each line showsthe result from either a carbonate or a hydroxide (depending on which ismore stable when exposed to the atmosphere) in terms of pH3PO0

4 as in thisreaction:

H2O + 1/3 M3(PO4)2 + CO02 = MCO3 + 2/3 H3PO0

4 [5]

Here pCO02 = 5.002 in equilibrium with the atmosphere. The table is ordered

so that the most stable phosphates are at the bottom while the most stableoxides and carbonates are at the top.

When dissolved phosphate is added to soil, its concentration at first willprobably be sufficient to react with all the reactants in the first column of thetable. As the system moves toward chemical equilibrium, at whatever rate,the phosphate will move down the table. This analysis describes, therefore,the environment in which the immobilization of lead, or any other element,by phosphate must occur based on thermodynamic principles.

3. Immobilized Lead in the Soil Environment

Without the interference of the elements discussed above, the conversion oflead minerals and salts to chloropyromorphite would be easy as the ratiosshown in Figure 14 would not be a problem. For example pS(t) over galenamight be about 9.0 (Figure 8), and a pP(t) of 1 or 2 would be sufficient forthe conversion. Section 4 will be a discussion of some of the many possible

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treatment strategies, and an examination of whether or not the chloropyro-morphite will be likely to endure in the soil environment for long periods.Thus we have worked backwards by first assuming that we have succeededin changing the lead in soil to this mineral.

This was done by thermodynamic analyses of systems containing chloro-pyromorphite, cerussite, and one of three pairs of minerals. The first wascalcite-apatite, the second MnCO3−MnHPO4, and the third Ca(OH)2-apatite.The carbonates of calcium and manganese were studied because they are themost active reactants in Table 2 (for a natural system), and calcium hydroxidewas studied because it might be added to a soil in a liming treatment.

The first of these systems (calcite, etc.) has six phases and seven com-ponents: PbO, CaO, P2O5, CO2, HCl, H2O, and Na2O. Thus there are threedegrees of freedom, and only one test needs to be made on the stoichiometryand that is on sodium, which here is forced to be equal to the total chloride.To understand this system, consider the equation below with Figure 27.

1/5 H2O + CaCO3 + 1/5 Pb5(PO4)3Cl = 1/10 Ca10(PO4)6(OH)2

+ PbCO3 + 1/5 HCl0 [6]

The only solute in this equation is the HCl. So long as the actual total chlorideis greater than that needed to maintain the equilibrium, that is the pCl(t) linein the Figure, the reaction will not go to the right, and chloropyromorphitewill be stable. In Figure 27 we see that this line represents a very smallconcentration of chloride, especially at moderate pH, and pyromorphite will

FIGURE 27. Chloropyromorphite (Pb5(PO4)3Cl), cerrusite (PbCO3), apatite (Ca10(PO4)6-(OH)2), and calcite (CaCO3): elemental activities vs. pH.

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be very unlikely to change. In fact modest amounts of chloride in solutionwill drive this system left, a desirable result for immobilization.

The second system comes from this equation:

MnHPO4 + PbCO3 + 1/5 HCl0 = MnCO3 + 1/5 Pb5(PO4)3Cl

+ 2/5 H3PO04 + H2O [7]

The equilibrium constant for this equation may be found from the pK0sp’s of

the minerals as

pK0eq = 26.253 + 14.533 − 16.007 − 22.070 = 2.709

And

pH3PO04 = 1/2 pHCl0 + 6.773

This system is pictured in Figure 28. In a soil environment this system isunlikely to be at equilibrium, and it will shift one way or the other dependingon the concentrations of phosphate and chloride relative to their lines onthe figure. Nonetheless, we may say that during any phosphate treatmentof a soil the total phosphate activity in the water will almost certainly bebelow the phosphate line. Even when the phosphate activity is controlled bycalcite as in Figure 27, the phosphate activity is sufficient to move phosphatefrom chloropyromorphite to MnHPO4. This reaction would also give leadcarbonate, an undesirable result. The only way out of this dilemma would

FIGURE 28. Chloropyromorphite (Pb5(PO4)3Cl) cerrusite (PbCO3), manganese hydrogenphosphate (MnHPO4), and rhodochrosite (MnCO3): elemental activities vs. pH.

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seem to be adding sufficient phosphate to the system to insure that all themanganese is changed to MnHPO4. Thus the equilibrium system of Figure 28and Equation [7] would be destroyed, and chloropyromorphite could remain.

Finally we consider the stability of Pb5(PO4)3Cl in the presence of lime,Ca(OH)2. This is shown by Equation [8].

CO02 + Ca(OH)2 + 1/5 Pb5(PO4)3Cl = 1/10 Ca10(PO4)6(OH)2

+ PbCO3 + 1/5 HCl0 [8]

pK0eq = 5.190 + 22.069 − 23.809 − 14.533 = −11.083

If pCO02 = 5.002, then pHCl0 = −30.5.

Such an activity of chloride is of course absurd, and this system will have togo to the right-hand side. Chloropyromorphite cannot be fully stable if it ismixed with hydrated quick lime (Ca(OH)2). We would have to insure thata soil containing Pb5(PO4)3Cl would never be treated with hydrated quicklime.

4. Possible Phosphate Treatments

The number of ways that phosphate might be added to a soil is quite large,and a few possibilities that seem to be or that have proved to be promisingwill be discussed. The simplest and most straight-forward technique is sim-ply to add soluble phosphate.21,31,32 This could be phosphoric acid in someconcentration, or it might be a salt of a cation such as sodium or ammoniumion. There are obvious cautions with regard to the pH of the addition, al-though the natural buffering tendency of any soil will mitigate the effects tosome extent. Adding large amounts of sodium could be the cause of laterproblems, but this could be avoided by using ammonium salts, perhaps as abuffered mixture of NH4H2PO4 and (NH4)2HPO4.

Surely the dissolved phosphate would react with all the reactants inTable 2 as well as some forms of lead. The ratios plotted in Figure 14 wouldhave to be exceeded when the particular lead mineral came in contact withthe solution. Such a situation is difficult to model, very complex, and close tochaotic. It does seem very likely that any application of dissolved phosphateto a soil contaminated with lead will change some of it to chloropyromor-phite. This is to say that treatment will probably provide an improvement inthe sense that less lead would be taken up by biological systems. Whethersuch a treatment could ever provide a permanent reduction to acceptablelevels in lead’s availability to organisms is still an open question.

In spite of the complexity we can gain an understanding of what ispossible by remembering that calcium is almost always more abundant thanphosphate in soils and that the stable phosphate sinks in a soil are apatiteand MnHPO4. The latter will form in an unwanted side reaction, from our

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point of view, but the former might even be part of a treatment. Surely addedsoluble phosphate will make apatite in most soils, and other minerals of itscomponents might also form. According to thermodynamic calculations, theonly way to avoid making only apatite and manganese hydrogen phosphateis to add more than enough phosphate to change all the calcium in the soil toapatite. To do this we would have to increase the amount of phosphorus in asoil up to 400 µmol/g, which is many times the normal 20 µmol/g. Whetherthis high P application would be either desirable or acceptable is beyond thescope of this discussion, but suspicion is advised. On a mass basis 400 µmol/gof phosphorus is approximately 12 parts per thousand phosphorus or closeto four percent as phosphoric acid. If we do a bench-scale experiment withone kilogram of dry soil, we would need 100 mL of 4.0 M H3PO4 to achievethe change. Furthermore, a soil containing large amounts of apatite couldcause unacceptable changes in soil structure.

5. Minerals of the Components CaO, P2O5, H2O, and CO2

To begin we will study the system calcite-apatite which is a stable part of anumber of the systems considered earlier. This is shown in Figure 29, whichincludes the phosphate curve from apatite alone from Figure 18. The mostinteresting feature here is the extremely low level to which phosphate issuppressed by the presence of calcite at all pH’s below 9.5. A recipe for asuccessful treatment will be one which increases this level of phosphate inspite of abundant calcium. The discussion which follows will discuss possible

FIGURE 29. Apatite (Ca10(PO4)6(OH)2) with calcite (CaCO3): elemental activities vs. pH. ThepP(*) curve from Figure 18, apatite alone, is included for sake of comparison.

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ways that this might be done, and we will take the calcite-apatite system tobe the base system.

The pK0sp equations show that pCa(OH)02 = 14.677 so long as the ground

and/or pore water is exposed to the atmosphere and the carbon dioxide in itand that pH3PO0

4 = 15.220. These activities are constant over the pH range.The pCa(t) curve goes below zero at pH 6.7, and this simply means thatcalcite dissolves. When it does, the calcium activity will rise to whatever levelis allowed by its abundance, and the phosphate activity will be suppressedbelow that shown by the pP(t) curve in Figure 29. Surely we must add solublephosphate to this system and do it in such a way that it will not all precipitateas apatite.

Apatite itself is, nonetheless, a possible treatment,33–38 and when it isadded to a soil, the level of phosphate in solution will reach that describedin the previous paragraph. It is certainly possible that the parts of the veryheterogeneous soil mixture will have solutions in which the phosphate levelwill be closer to that from apatite alone, especially soon after it is added.Since this is so, we would expect that a treatment of apatite alone wouldchange some of the PbCO3, Pb(OH)2, PbCl2, and PbSO4 to Pb5(PO4)3Cl, andsuch reactions would surely be beneficial. Such a treatment would of coursecause these changes to lead minerals in soils which did not contain apatitein the first place, and soils that had not been fertilized in some time (if ever)might very well fit this description. It is also true that the level of phosphatein the apatite-calcite system is great enough to change MnCO3 to MnHPO4

(see Equation [4]).Theoretical thought experiments with modeling programs give some in-

teresting results with pairs of minerals from the CaO-P2O5-H2O system. Oneof these is apatite-monetite (CaHPO4), and the total element concentrationsare shown as a function of pH in Figure 30. This figure also shows the pP(t)curve from the apatite-calcite system, and there clearly has been a dramaticchange. To understand this system a bit more we note that the two pK0

sp equa-tions have two unknowns, and pCa(OH)02 is constant at 21.312 while pH3PO0

4is fixed at 4.162. This compares to 15.220 for this quantity in the apatite-calcitesystem. Figure 31 shows the many important phosphate species.

Such a system certainly appears to describe a possible treatment as thelevel of phosphate is certainly high enough to effect the desired changes.Unfortunately, if we were to put this system into a soil, the amount of cal-cium in the soil environment would surely be several times the amount ofphosphate. While a small amount of calcite would dissolve in the apatite-monetite system, a large amount of calcite, or dissolved calcium, will haveanother effect altogether as shown by

CaHPO4 + 2/3 Ca(OH)02 = 1/6 Ca10(PO4)6(OH)2 + H2O [9]

pK0eq = 25.474 − 5 ∗ 23.809/3 = −14.208

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FIGURE 30. Apatite (Ca10(PO4)6(OH)2) with monetite (CaHPO4): elemental activities vs. pH.The pP(*) curve from Figure 29, apatite with calcite, is included.

At equilibrium, pCa(OH)02 = 21.31, which is a negligible amount of calciumwhen compared to that supplied by calcite. The reaction [9] surely goes tothe right, and we are back to the apatite-calcite system once again. The onlyescape is to add so much monetite that all the calcite reacts. Only if we getthe total amount of phosphate in soil high enough to make the phosphate-calcium ratio greater than three-fifths, will we succeed. If the soil to be treated

FIGURE 31. The activities vs. pH of the several phosphate species in the system of Figure30.

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has 350 µmol/g of calcium and 20 of phosphorus, we would need to add475 µmol/g of CaHPO4 to make the P:Ca ratio equal to three-fifths. This is65 g of monetite per kilogram of dry soil, and we would have to wait for thesolid monetite to react with the solid calcite.

6. “Super Phosphate,” Ca(H2PO4)2·H2O

The salt calcium dihydrogen phosphate, which has a P:Ca ratio of two, canbe shown to be quite soluble in water by thermodynamic analysis, but theroute it takes on the way to dissolving is an interesting and possibly usefulone.26,32,37 This salt dissolves incongruently forming CaHPO4·2H2O and thenCaHPO4 as it dissolves. Furthermore, the process takes days, and the systemwith two solids and the solution persists for quite a long time. The followingequations are valid during that time:

{Ca(H2PO4)2·H2O}, pK0sp = 24.846 = pCa(OH)0

2 + 2pH3PO04

{CaHPO4}, pK0sp = 25.474 = pCa(OH)0

2 + pH3PO04

These equations give pCa(OH)02 = 26.102, and pH3PO04 = −0.628. In this sys-

tem calcium is severely depressed while phosphate’s level in solution is ex-traordinarily high. It is the phosphate fertilizer without peer, and thus itspopular name. Figure 32 gives the activities of the principal solute species as

FIGURE 32. The principal solute species above the metastable combination ofCa(H2PO4)2·H2O and CaHPO4 as the former, known as super phosphate, dissolves incon-gruently. The range is 0.00 < pH < 2.40, and the natural pH of the system is approximately1.6.

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functions of pH up to 2.4. The natural pH of this system is a little above 1.6,and the acidic conditions could be quite useful in the making of chloropyro-morphite. Even metallic lead will dissolve under these conditions and changeto Pb5(PO4)3Cl. Surely the soil itself will buffer the action of this acid, butwe could expect an acidic pH in any case. Applications of super phosphateare often followed by lime to restore a neutral pH. As noted above calciumhydroxide must not be used. Calcium carbonate could be or perhaps somecombination of ammonium and potassium phosphates.

Of course when super phosphate is put into a soil, much of its phosphatewill go to apatite. This result suits plants very well indeed, but it does notserve the purposes being explored in this paper. It may be true that we will asbefore have to raise the over-all P:Ca ratio to above three-fifths. To treat thesoil described in section 2, we would need 136 mmol of Ca(H2PO4)2·H2O perkilogram of soil, that is 34 g. Most of this would react eventually accordingto

3 Ca(H2PO4)2·H2O + 7 CaCO3 = Ca10(PO4)6(OH)2 + 7 CO2 + 8H2O [10]

We would need super phosphate beyond this amount to react with the leadin the system and with the manganese as well. It should be noted that thesituation described by Figure 32 is not likely to ever be realized throughoutthe ground and pore water in a soil system. If we have 100 mL of watermixed with one kilogram of dry soil, the solution would have to be 6.91 Min phosphoric acid. Such would require 174 g of super phosphate solid.

Since soil systems are quite heterogeneous, there will be pockets ofphosphoric acid solution of high concentration, and we certainly do not needto have the concentration as high as 6.91 M to effect the desired changes.The amount that would be effective is probably between 34 and 174 g, andexperiment is needed. Thorough mixing or tilling will certainly be importantas well.

Super phosphate certainly could be an effective agent in changing otherlead minerals to chloropyromorphite. A model was made with super phos-phate and galena under various conditions as shown by (pH + pe), and thechemical potential for changing PbS to chloropyromorphite was calculated.The results are shown in Figure 33, and they show that super phosphate mayeffect the desired change without the oxidation of galena to lead sulfate. Whatis required is that (pQ0

sp − pK0sp) < 0.

Even so it is surely easier to convert lead sulfate to pyromorphite thanit is galena, so the oxidation of galena to PbSO4 (anglesite) was considered.Figure 34 shows the results. Under this condition (pH + pe) = 5.055, the twominerals may co-exist at all pH’s, and it is interesting to note that virtuallyall the sulfur in solution is in the form of sulfate. We also note in passingthat under these mildly oxidizing conditions that gypsum, CaSO4·2H2O, mayform as well. Simple metathesis reactions between calcium phosphates and

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FIGURE 33. Chemical potentials vs. pH at different (pH + pe) for the formation of chloropy-romorphite (Pb5(PO4)3Cl) from galena (PbS) in the solution formed by super phosphate,Figure 32.

lead sulfates or sulfides to give gypsum and chloropyromorphite ought to beboth possible and beneficial.

A model like that used to make Figure 33 was made by substitutingwulfenite, PbMoO4, for galena, and it is shown in Figure 35. The chemical

FIGURE 34. Galena (PbS) with anglesite (PbSO4) at equilibrium when (pH + pe) = 5.055:elemental activities as functions of pH and the activities of the important solute species.

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FIGURE 35. Wulfenite (PbMoO4) in the presence of super phosphate (Ca(H2PO4)2·H2O andCaHPO4) elemental activities as functions of pH up to 2.4.

potential for the formation of chloropyromorphite is shown in Figure 36, andwe see that this mineral may form at pH > 1.3. In principle at least superphosphate will convert any form of lead to Pb5(PO4)3Cl, but it will probablybe necessary always to add sufficient treatment to make the phosphate-to-calcium ratio greater than three-fifths.

FIGURE 36. The chemical potential for the formation of chloropyromorphite (Pb5(PO4)3Cl)in the system of Figure 35 as a function of pH.

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There could be a distinct advantage to super phosphate as compared tosoluble phosphate. The fact that it is a solid would mean that once mixedwith a soil it would remain in place. By contrast a liquid application to a fieldmight simply preferentially flow in the path of least resistance. Furthermore,soluble phosphate will simply react with the several soil constituents that aremore basic than itself, but the solution in equilibrium with the two mineralsfrom super phosphate will stay acidic for as long as it is in contact with thoseminerals. Since super phosphate dissolves very slowly, this could be quite along time.

7. Conclusions Concerning Phosphate Treatments of Lead

Of the several lead phosphate minerals and salts only chloropyromorphiteis stable in soil environments and then only if the soil is never treated withquick lime, Ca(OH)2. In order to make all other forms of lead into chloropy-romorphite, it is necessary to change all of the calcium in the soil, whetherit is solid calcite, some other mineral, or in solution, to hydroxyapatite,Ca10(PO4)6(OH)2. Since the typical abundance of calcium is ten or twentytimes that of phosphate, very large amounts of phosphate must be addedto bring the phosphate-calcium ratio up to three-fifths. This may be done inprinciple with soluble phosphate or with solid calcium phosphate mineralswhose phosphate-calcium ratio is above three-fifths. The possibilities includemonetite, CaHPO4, and super phosphate, Ca(H2PO4)2·H2O.

An added complication is the high stability and very low solubility ofmanganese hydrogen phosphate, and enough phosphate must be added dur-ing treatment to convert all of the manganese in the soil to this salt.

The treated soil may have a very low level of lead available to bio-logical systems, but it will also have very high levels of both calcium andphosphate. Furthermore, virtually all of the calcium will be precipitated asapatite. Whether such is desirable and acceptable is beyond the scope ofthis paper, but since it is well known that apatite and other minerals of itscomponents are found in bones and teeth, the friability of any soil so treatedmay very well disappear.

8. Wulfenite, PbMoO4, as a Possible Remediation

Wulfenite is one of the three least soluble minerals of lead as shown inFigure 8, and this section will examine first whether it would remain unre-acted if left in a soil and second what actions might be required to form itfrom other compounds of lead.

We will first consider the acid-base chemistry of lead molybdate withoutthe possibility of reducing Mo(VI) to Mo(IV). The solubility curve of Figure 8shows that acids will react with wulfenite only if they are quite strong, and we

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would expect that it would dissolve only in strong mineral acids. Bases couldbe a different story, and paralleling the earlier discussion of pyromorphites,we will examine the possible reactions of wulfenite with calcite and lime.Either of these bases might change wulfenite to powellite, CaMoO4, and wecan calculate the pK0

eq of each reaction from the pK0sp’s for the minerals.

PbMoO4 + CaCO3 = CaMoO4 + PbCO3 [11]

pK0eq = 25.549 + 19.679 − 27.688 − 14.533 = 3.007

PbMoO4 + Ca(OH)2 = CaMoO4 + Pb(OH)2 [12]

pK0eq = 25.549 + 5.190 − 27.688 − 9.595 = −6.544

Wulfenite will not be changed by calcite, but calcium hydroxide willeasily change it to lead hydroxide. These results parallel those found in thestudy of chloropyromorphite done earlier.

If dissolved molybdate is added to a soil, there are several metal, Me2+,salts and minerals that may form by reactions of the type shown by equation[13].

H2MoO4 + MeCO3 = MeMoO4 + CO2 + H2O [13]

Table 3 is constructed with the same logic as Table 2, which was concernedwith the formation of phosphates. The third column gives the activity ofH2MoO0

4 when the reactants and products are at equilibrium.This table is ordered with the most stable products at the bottom of the

second column and the most stable reactants at the top of the first column,and we see that wulfenite, PbMoO4, is indeed in the most advantageousposition. Nonetheless, an added solution of molybdate will make several ofthe products in the table at least for a time. Since calcium is quite abundant,

TABLE 3. Equilibria Between Various Metal Oxidesand Carbonates with Molybdates, Ordered by Equi-librium Values of pH2MoO0

4

Reactant Product pH2MoO04

MgCa(CO3)2 CaCO3, MgMoO4 4.648CuO CuMoO4 5.898MgCO3 MgMoO4 6.629ZnO ZnMoO4 7.408MnCO3 MnMoO4 7.531MgCa(CO3)2 MgMoO4, CaMoO4 8.820FeCO3 FeMoO4 10.925MgCa(CO3)2 MgCO3, CaMoO4 11.010CaCO3 CaMoO4 12.991Ag2CO3 Ag2MoO4 13.922Pb(OH)2 PbMoO4 15.953

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we will consider what happens when some of the added molybdate makespowellite, CaMoO4.

As shown by equation [11] (reversed) and its equilibrium constant, pow-ellite will react with cerrusite to form wulfenite. Table 3 shows that the sameproduct will be produced by Pb(OH)2. This is surely a desirable result, but itdoes not tell us whether galena might be changed to wulfenite. The issue iscomplicated by the fact that while molybdate will make CaMoO4, it will notchange all the calcium in the system to this mineral. The mole ratio of calciumto molybdenum will be a large number in any conceivable scenario, and theexcess calcium will severely reduce the activity of molybdate in solution.

To study the possible reactions, a system with five minerals (sevenphases) and nine components was modeled. The components were PbO,MoO3, CaO, H2S, P2O5, CO2, Na2O, HCl, and H2O. The five minerals on thefirst try were calcite, apatite, powellite, galena, and wulfenite. It is assumedthat the calcite or other more active forms of calcium will control the activitiesof both phosphate and molybdate because the calcium will be much moreabundant than either phosphorus or molybdenum. The two lead mineralswere put into the systems to see how easily one might be converted to theother.

Along with the several possible precipitation reactions it was necessaryto consider the redox chemistry of both molybdenum and sulfur. The Mo(VI)in molybdate may very well be reduced to Mo(IV), and both MoS2 and MoO2

are possible. Sulfur has several oxidation states, and as was shown in thediscussion with Figure 34 the formation of sulfate from sulfide can be doneunder mildly oxidizing conditions. The consideration of such redox chemistryactually increases the number of components and the degrees of freedom byone, and models were made at a variety of constant (pH + pe)’s.

In this system there are three values of (pH + pe) at which two mineralsof the components are in equilibrium. If (pH + pe) = 4.489, then molybden-ite (MoS2) and gypsum (CaSO4·2H2O) are in equilibrium. At (pH + pe) =4.381 there is equilibrium between powellite (CaMoO4) and MoO2. Finallyand most importantly there is equilibrium between wulfenite (PbMoO4) andgalena (PbS) at (pH + pe) = 4.122. This is also of course the region of the(pH + pe) scale where we see the change from S(-II) to S(VI). Galena andanglesite (PbSO4) are in equilibrium at 5.055, and the hydrogen sulfate ionis in equilibrium with the hydrogen sulfide ion with equal activities at 3.963.

The upshot of all this is that molybdate will make wulfenite from galenaif at least some of the sulfur in the system has been oxidized to sulfate, butunder reducing conditions galena will not be changed. The relative positionsof the curves on Figure 8 also indicate this.

Several models were made of the system, and two are included. Fig-ure 37, done under reducing conditions has PbS, MoS2, MoO2, CaCO3, andCa10(PO4)6(OH)2. Figure 38, done under somewhat more oxidizing condi-tions has PbMoO4, CaSO4·2H2O, CaMoO4, CaCO3, and Ca10(PO4)6(OH)2.

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FIGURE 37. Galena (PbS), molybdenite (MoS2), molybdenum(IV) oxide (MoO2), calcite(CaCO3), and apatite (Ca10(PO4)6(OH)2): elemental activities as functions of pH at (pH +pe) = 3.50.

The curve for pPb(t) has shifted very little, and both models show a very lowlevel of dissolved lead at neutral pH’s. In one sense the wulfenite is superior,in that it could never be oxidized to a more soluble form as galena can beand is. On the other hand it is possible that the Mo(VI) in wulfenite could be

FIGURE 38. Wulfenite (PbMoO4), gypsum (CaSO4·2H2O), powellite (CaMoO4), calcite(CaCO3), and apatite (Ca10(PO4)6(OH)2): elemental activities as functions of pH at (pH +pe) = 5.00.

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reduced to Mo(IV), and PbMoO4 would be destroyed. Apart from all of thisany soil treatment that uses molybdenum in any form would not be accept-able. The reasons for this are beyond the scope of this paper, but there arequite compelling. We may conclude that existing wulfenite can probably beleft alone but that we should never attempt a treatment that would make it.

9. Galena, PbS, as a Possible Remediation

Since galena is very difficult to change to more thermodynamically stableminerals, we can elect to leave it alone while changing other more solublelead minerals such as PbCO3. PbS will oxidize in the air to PbSO4, but theprocess is usually very slow, and PbS itself has a very low solubility. How-ever, it is important to remember that a mixture of these two solids, of anyproportions, will have the lead solubility of the more soluble sulfate. Wecould even consider trying to make the more soluble minerals of lead intogalena in order to reduce the level in solution.

To make galena it would be necessary to add H2S or some other sul-fide to the soil. At low concentrations this would make the soil smell likethe seashore at low tide; at higher concentrations the gas would be highlypoisonous. It might be possible to use a solid compound like thioacetamidewhich releases hydrogen sulfide as it dissolves in water, but we would alwayshave to worry about the effect of this gas on biological systems. Furthermore,the sulfide would make several compounds in addition to PbS. A numberof dipositive metal cations (Me2+) in soil make insoluble sulfides as Me2+S(i.e., FeS, ZnS, MnS, and SnS). To be sure, the addition of hydrogen sulfidewould make most or all the lead, except for the metal, into galena, but it isdoubtful that the side reactions or the effect of high sulfide concentrationswould be acceptable.

10. Changing Some Lead (but not all) to Pb5(PO4)3Cl

If we elect to try to change the more soluble forms of lead, the carbonate,the sulfate, the chloride, etc., into chloropyromorphite without touching thegalena, the important question becomes: Can we effect these changes witha phosphate-to-calcium ratio of less than three-fifths? This would be highlydesirable by making the treatment easier, and the result could certainly bean acceptable remediation. We could treat with phosphoric acid or with anyof the minerals of the CaO-P2O5-H2O system, including apatite itself, and wemight be able to use modest amounts. Apatite is highly basic, so it could payto use a more acidic mineral, perhaps even super phosphate.

As was discussed in section 2, any MnO, Mn(OH)2, or MnCO3 that ispresent will likely react with any added phosphate to make MnHPO4 beforechloropyromorphite can form. The resulting system is shown by Figure 26,

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FIGURE 39. Chloropyromorphite (Pb5(PO4)3Cl), manganese hydrogen phosphate (MnHPO4),lead hydroxide (Pb(OH)2), apatite (Ca10(PO4)6(OH)2), and calcite (CaCO3): elemental activitiesvs. pH.

the level of phosphate is the key, and we may now ask whether this levelof phosphate will be sufficient to change the relatively more soluble miner-als and salts of lead to chloropyromorphite. The system shown in Figure 39provides the probable answer. It has lead hydroxide (slightly more stablethan the carbonate), chloropyromorphite, apatite, calcite, manganese hydro-gen phosphate, a solution with sodium chloride, and carbon dioxide in thesolution and the atmosphere. It has seven phases, eight components, andthree degrees of freedom. Since two of these are temperature and pressure,only one test on the stoichiometry is necessary. This was done by requiringthat pCl(t) equal pNa(t).

Since the hydroxide of lead exists in this system, we need to ask whetherits remediation is possible. This can be done by considering the equilibriumof equation [14].

6/5 H2O + CaCO3 + 1/5 Pb5(PO4)3Cl = Pb(OH)2 + 1/10 Ca10(PO4)6(OH)2

+ 1/5 HCl0 + CO02 [14]

pKeq = 19.679 + 22.069 − 9.595 − 23.809 = 8.344

If pCO02 = 5.002, then pHCl0 = 16.710.

The system as described is quite stable, and lead hydroxide is not reacting.The level of chloride as shown by the pCl(t) curve of Figure 39 (whichshows this equilibrium system plus MnHPO4) is, however, quite low, and it

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is not unreasonable to imagine that the naturally occurring chloride in anyreal soil system will be orders of magnitude higher. This chloride will drivethe chemical reaction above to the left until either lead hydroxide or apatiteis gone. Thus the addition of apatite or other calcium phosphate mineralswill change lead hydroxide and other lead minerals, which are more solublethan the hydroxide, to chloropyromorphite without raising the phosphate-to-calcium ratio to three-fifths. Hydroxypyromorphite, if it is present, willbe changed at the same time as shown by Figure 9, and so will any otherlead phosphate. We conclude, therefore, that if a soil system is treated byany calcium phosphate so that there is sufficient phosphate to change allthe manganese present to manganese hydrogen phosphate and to changeall the lead to chloropyromorphite, assuming thorough mixing, that the onlyforms of lead which would remain would be galena, chloropyromorphite,wulfenite, and the element itself.

11. Conclusions Concerning Lead Remediation

The three least soluble minerals of lead are galena, chloropyromorphite, andwulfenite, and any of the three would be an acceptable result for renderinglead inert to biolgical systems, i.e., immobilization, except that galena slowlyoxidizes to the much more soluble anglesite. Furthermore, if a pair of leadminerals is in contact with a solution, the solubility of lead will be determinedby one with the higher solubility, regardless of their relative amounts.

It is possible in principle to change any form of lead, including galena,to chloropyromorphite by bringing the phosphate-to-calcium ratio up fromwhat is typically found in soils, one to ten, to three to five or higher. Suchcan be done with phosphoric acid or an acidic calcium phosphate. Superphosphate is especially efficient as it produces a solution of phosphoric acidwhich is 6.9 M and with a pH of 1.6. This solution will dissolve and reactwith lead metal as well. The drawback to this idea is that the amounts ofphosphate which must be added to and left in the soil are quite large.

A much less drastic treatment can be done by adding sufficient phos-phate to react all the manganese carbonate and hydroxide and all the moresoluble lead salts and minerals, the carbonate, the hydroxide, the chloride,and the sulfate. This will leave only the three least soluble minerals and themetal, and such could be an acceptable result, or at least an improvement.

IV. CADMIUM IN SOILS

Cadmium is similar to lead in that it is almost always found as a divalentcation, and the two metals form a number of minerals with similar formulas.39

Figure 40 shows the solubilities of several cadmium salts as functions of pH.

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FIGURE 40. Solubilities as pCd(t) vs. pH for several minerals and salts of cadmium.

We may say in general that, except for the sulfide, the compounds of cad-mium are more soluble than those of lead. Cadmium sulfate is quite solubleat all pH’s, and cadmium phosphate is more soluble across the range thanany of the several lead phosphates. The silicate, carbonate, and hydroxide areall insoluble at neutral pH, but their solubilities increase sharply in acid. Theonly two possibilities for immobilization are, therefore, CdS and Cd3(PO4)2,and we shall examine how stable each of these might be in a soil environ-ment and whether making them in those environments is feasible.

Except for the fact that cadmium sulfide could and does oxidize to cad-mium sulfate, it is surely the mineral of choice for our purposes. It is quiteinsoluble at all pH’s above one, and it could probably be ingested withoutharm. Unfortunately, CdS is quite easily oxidized, and it is in equilibrium withthe sulfate at (pe + pH) = 5.751. The question then becomes whether theoxidation process in a soil system will be slow enough to make remediationfeasible. This is certainly possible, but our predictive thermodynamic calcu-lations can not tell us one way or the other. If the oxidation is slow enough,and such depends on the condition of the CdS itself as well as the envi-ronment, the mineral might simply be left alone. However, trying to changeother forms of cadmium into the sulfide would probably not be desirable forthe same reasons that making lead into PbS would not be.

We may examine the possibility of making cadmium into its phosphateby the same methods that were used above to test the feasibility of makingchloropyromorphite out of lead. A glance at Figure 40 tells us that this may bedifficult because it shows that at pH’s above 7 or 8 Cd3(PO4)2 is not even the

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second least soluble compound of cadmium. Nonetheless, we first examinethe following:

H2O + pCO02 + 1/3 Cd3(PO4)2 = CdCO3 + 2/3 H3PO0

4

pK0eq = 18.531 − 15.568 = 2.963

if pCO02 = 5.002, pH3PO0

4 = 11.948.

Reference to Table 2 shows that a cadmium row would be above that of Mn,Ca, and Mg, which is to say that there would have to be sufficient phosphateto change all the carbonates and oxides of these three elements before thechange from cadmium carbonate to cadmium phosphate would work. As wasdiscussed in Section 7 of Part III for Pb, this would require high amounts ofphosphate.

In the case of lead we came to the conclusion that it might not be nec-essary to add massive amounts of phosphates if we would be satisfied withchanging only the carbonate, sulfate, and hydroxide. By analogy with thediscussion of Section 10 of Part III for Pb, we look at the system representedby the following equilibrium:

CaCO3 + 1/3 Cd3(PO4)2 = 1/10 Ca10(PO4)6(OH)2 + CdCO3 + 1/15 H3PO04

pK0eq = 19.679 + 18.531 − 23.809 − 15.568 = −1.167

at equilibrium pH3PO04 = −17.505

This is impossible, and this system will have to go the right-hand side untileither the calcite or the cadmium phosphate disappears. Very large amountsof phosphate would be required. If this were achieved, we would havethe system of Figure 41 which shows the four minerals, less calcite, in theequation above. Of course the cadmium goes into solution at moderatelyacidic pH’s, and the cadmium curve is essentially the cadmium carbonatecurve of Figure 40. This system also has the active form of silica found insoils in order to see if cadmium silicate might form, but the chemical potentialof that reaction is positive. The system also has ammonium chloride with anactivity of 10−5 molal, and both the chloro- and ammine-complexes haveimportant activities. Finally we have Figure 42, which comes from the samesystem, showing the important cadmium species when the activity of NH4Cl is0.10 M. Surely the ammine-complexes and the chloro-complexes of cadmiumincrease the element’s solubility.

The conclusion that one must come to is that the chemical immobiliza-tion of cadmium in soils is probably not feasible, at least with the commonanions that are likely to be acceptable treatments to soils.

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FIGURE 41. Cadmium phosphate (Cd3(PO4)2), cadmium carbonate (CdCO3), apatite(Ca10(PO4)6(OH)2), and silica (SiO2): elemental activities as functions of pH. Also presentare sodium chloride, pNa(t) = 3.0 and ammonium chloride, pN(t) = 5.0.

V. ARSENIC

Arsenic is a widespread pollutant which can and has caused cancer and deathin thousands.40–46 It can be found in wells, most famously and tragicallyin Bangladesh in recent years,41,47–49 mine wastes,50–54 pesticides,55,56 and

FIGURE 42. The same system as Figure 41 with pN(t) = 1.0: Activities of the important speciesof cadmium as functions of pH.

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wood treatment compounds.57 Its disposal or immobilization presents oneof the most urgent problems of modern society, and it has been extensivelystudied. A recent review58 discussed the chemistry of the technology in useat the present time. Most of the techniques presently employed use someform of iron with the hope of making insoluble iron arsenates or havingarsenate sorbed onto the surface of an iron oxide or hydroxide. This sectionwill discuss the former as well as other possible forms of arsenic which areonly slightly soluble, but the surface phenomenon of sorption is beyond thescope of the present work.

1. The Chemistry of Arsenic

Unlike the other three elements discussed in this paper, arsenic is not truly ametal. It is in the middle of Group 15 of the periodic table, and the elementsabove it are non-metals which form acidic oxides, while the elements be-low it are metals. Arsenic itself lies somewhere in between, and its chemicalbehavior is mixed.59 Its oxides and sulfides are acidic like those of phos-phorus, but it is much easier to reduce As(V) to As(III) than it is to reduceP(V) to P(III). The latter reaction requires severe conditions which are mostunlikely to occur in the systems under discussion here, while conditions forthe former are quite common.

The range of insoluble compounds of arsenic is quite impressive. Wemay be able to use arsenic cations with either oxide or sulfide to immobilizearsenic, and we may be able to use other metals in combination with theoxyanions of either As(III) or As(V). There are many more possible precipi-tates with arsenic than with any of the other metals discussed in this paper.The text which follows will speak to several of the reasonable possibilities.

The situation is complicated by the fact that oxidation-reduction chem-istry (redox) is very important to the understanding of what is stable underwhat conditions. At least three of the elements which are present in possibleimmobilizing precipitates undergo fairly easy redox reactions: arsenic itself,iron, which can be found as Fe(II), Fe(III), and the metal, and sulfur, whichis found in solution as S(-II) [sulfide], S(VI) [sulfate], and a menagerie of ionsand molecules with intermediate oxidation states.

The approach taken here uses pe, the negative log of the virtual activityof free electrons. As has been done by many other authors, the sum (pH +pe) was used. If (pH + pe) is zero, the conditions are strongly reducing, andwater will be in equilibrium with hydrogen gas. At the other extreme used,(pH + pe) is equal to twenty, nearly oxidizing enough to oxidize water tooxygen. The conditions in soil systems are almost always much more mod-erate than either of these extremes, but to cover all the possible situations,calculations of solubilities have been done over these ranges of pH and ofpe: 0 < (pH + pe) <20, and 2 < pH <12. The method of calculation was like

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that used earlier to make spreadsheets and graphs, but here each spreadsheetwas made assuming a constant value of (pH + pe). Several spreadsheetswere then made on each system at intervals of 0.2 for (pH + pe) with a likeinterval for pH, and the results were arrays of numbers, pE(t)’s, that wereused to make contour maps of solubilities over the entire area.

The systems discussed below are defined first by their components.Analysts have long calculated which solid phases and solute species arestable under what conditions of pH and pe. That was done here as well,and combined with the contours of solubility to give a complete pictureof the equilibrium states over the range of conditions. The boundaries ofstability for each solid phase were calculated by using chemical potentialsfor precipitation, as before, and all the diagrams show systems that are stablewith one exception, that of the iron oxides.

The situation with iron is especially interesting and important becauseiron oxides have been used as arsenic treatments. This element forms severaloxides, some from Fe(III), some from Fe(II), and some mixed. There is alsoa carbonate known as siderite, FeCO3. (There is no carbonate of Fe(III).)The solubilities of all these phases vary widely, and to cover the possibilitiestwo different kinds of iron systems were used, and thus the exception tostability. Using free energies from Bard,24 the most stable iron oxides underdifferent conditions were found to form one kind of system. This workedout as follows:

Fe2O3·H2O: stable if (pH + pe) >2.397

Fe3O4: stable if 2.397 > (pH + pe) > 0.722

FeO1.062: stable if 0.722 > (pH + pe)

These numbers were calculated using the pK0sp numbers tabulated in

Appendix A. Figure 43 is a map of pFe(t) contours over the pH-pe space. Anamount of iron was assumed sufficient to make a solution of 1.0 M when allis in solution as it is at low pH’s and low pe’s. This map shows the situationwith stable iron oxides, and it will be applied below.

Figure 44 parallels this using reactive compounds of iron. Under oxi-dizing conditions the hydroxide described by Bard24 as Fe(OH)3(c) is used,and siderite, FeCO3, is used under reducing conditions. The two are in equi-librium at (pH + pe) = 3.638. While this system is not stable thermody-namically, it could approximate the situation soon after an iron treatment ofarsenic-bearing waste, and it might persist for a long time. This system willalso be applied to a number of models in what follows. The minimum solu-bilities of iron as shown by the heights of the peaks in the middle of Figures43 and 44 differ by about four orders of magnitude, and such a change willsurely have an affect on the solubility of arsenic in minerals of iron.

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FIGURE 43. Contour map of pFe(t) in pe-pH space over stable oxides of iron.

2. The System Orpiment, As2S3

The first system of arsenic to be considered will be the simplest, with nocomponents from metallic elements other than arsenic itself. The componentsof the system seen in Figure 45 are As2O5, H2S, HCl, Na2O, CO2, H2, andO2. In the region where a precipitate exists, at low pH and low pe, there arethree phases and six degrees of freedom. Two are satisfied by using standardtemperature and pressure, and two by stoichiometric tests on Na and Cl. Oneis satisfied by the ratio of As to S over As2S3 or by a test on the amount ofS over As4O6. The last degree of freedom is done by calculating at constantvalues of (pH + pe). Orpiment itself is thermodynamically stable only upto approximately (pH + pe) = 6 and then only at low pH’s. Just above thisarea is a region where As4O6 can precipitate if the amount of arsenic in thesystem is quite large. In this diagram it is 0.82 M. Above about 10 in (pH + pe)even this amount of As4O6 completely dissolves. Under even more stronglyoxidizing conditions the stable compound is As4O10, and this is even moresoluble, forming a solution of arsenic acid.

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FIGURE 44. Contour map of pFe(t) in pe-pH space over reactive iron oxide, Fe(OH)3(c), andsiderite, FeCO3.

Since the conditions in the area where orpiment is stable are difficult torealize, we can say that this system will give soluble arsenic, sooner or later.It is probably true, however, that orpiment is like most metal sulfides in theenvironment, metastable but long lived nonetheless.

Other sulfides of arsenic are possible. One of these does appear, and thatis realgar, As4S4, which is stable in a small region at about pH 6 and pe = −6.Such strongly reducing conditions are unlikely in the environment. The otheris As2S5. Numerous chemistry books speak of this “compound”, but its struc-ture seems to be a mystery, and thermodynamic data were not found. If it ex-ists, it would likely show up at low pH and intermediate pe’s. It would surelybe quite acidic, and it is hard to imagine that it could exist above pH three orso. Furthermore, one has to wonder about the coexistence in one moleculeof As(V) and S(−II). The reduction potentials would seem to say that suchis impossible. Even so, many chemists seem to believe that the compoundexists. We take the existence to be an open question but believe that even ifit does, it would contribute little to the stability of arsenic in the environment.

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FIGURE 45. Contour map of pAs(t) in pe-pH space over orpiment, As2S3, and oxides ofarsenic. Except for the area at low pH and low pe, the arsenic is in solution, and no contoursappear.

3. Orpiment plus Reactive Iron Oxides

The next system shown is a combination of what we find in Figure 45 withwhat we have in Figure 44. Arsenates, arsenites, and arsenides of iron becomepossible, and four of them do appear as stable precipitates. Under stronglyoxidizing conditions and at low pH we find scorodite, FeAsO4·2H2O. At lowpH and low pe we find orpiment and arsenolite as before, and the iron hasno effect. At neutral pH’s and moderate pe we find iron(II) arsenate, andunder strongly reducing conditions both arsenopyrite, FeAsS, and loellingite,FeAs2 are stable. Figure 46a shows contours of pAs(t), and Figure 46b showsthe regions of stability for each precipitate of arsenic. The hole in the contourmap between scorodite and orpiment is arsenolite, here going into solutionentirely since we have assumed only enough arsenic to make pAs(t) = 1.00.

In places on this diagram the solubility of arsenic is low, but rarely does itfall below the recommended level for drinking water of 10 ppb, which comes

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(a)

(b)FIGURE 46. (a.) Contour map in pe-pH space of pAs(t) for orpiment, As2S3, with reactive ironoxides as in Figure 44. (b.) Phase diagram for the same system showing the stable crystallinephases of arsenic.

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out to pAs(t) > 6.87. The area with loellingite, which is stable under severelyreducing conditions, does so. Such conditions are possible only in the vicinityof a cathode with an applied reducing potential sufficient to reduce water tohydrogen. The center of the diagram, the area where soil systems are almostalways found, does have iron(II) arsenate, but the minimum solubility is suchthat pAs(t) is close to 6. Furthermore, this minimum solubility covers quite asmall area, and the gradients in the diagram are strong. This means that smallchanges in a system’s conditions will dramatically change the solubility ofarsenic. These strong gradients are a common theme throughout this analysisas will be seen. Since this model was made using reactive iron oxide, it canbe assumed that the solubility of arsenic in a model using the stable ironoxides would be a few orders of magnitude higher.

4. Adding Calcium to form Arsenates

A number of calcium arsenates have been reported,60–62 and a model wasmade using orpiment, the stable iron oxides as in Figure 43 and either calciumcarbonate as in Figure 18 or dissolved calcium at low pH. The pAs(t) contoursare shown in Figure 47a, while Figure 47b is a phase diagram for the system.The difference between the stable and reactive iron oxides shows up herein that the solubility of arsenic over scorodite (low pH, high pe) is muchhigher if the stable oxides are in the model, and iron(II) arsenate does notappear at all. The presence of calcium does make an impact under highlyoxidizing conditions and moderate pH: Both calcium hydrogen arsenate andcalcium arsenate form. pAs(t) is never above 4 at all in these areas, and forthe most part it is 1 to 3. Across the middle of this diagram there is a bandcovering the entire pH range in which the stable mineral of arsenic is As4O6,a compound which is highly soluble in water. If this model were done withreactive iron oxides, the area for Fe3(AsO4)2 would reappear as in Figure 46.

5. Oxides of Manganese

The last metallic element examined for this study is manganese. Two modelswere made, and both had the same components: As2O5, H2S, MnO, Fe2O3,CaO, HCl, Na2O, CO2, H2, and O2. One system had the reactive iron oxidesand the other the stable. It is assumed here that the stable oxides and car-bonates of manganese coexist with the solution or are dissolved in it. It isalso assumed that sufficient Mn is in the system to make pMn(t) = 0.50 whenit is fully dissolved. Three minerals are stable as follows:

pyrolusite, MnO2: (pH + pe) > 16.618

manganite, MnOOH: 16.618 > (pH + pe) > 13.614

rhodochrosite, MnCO3: 13.614 > (pH + pe)

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(a)

(b)

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FIGURE 48. Contours of pMn(t) over pe-pH space to show the solubility of manganese overthe stable oxides and carbonates of the element.

There is a large area of acidic pH’s where manganese is soluble, and evenat 7 or 8 the amount of manganese in solution is fairly substantial, especiallyat intermediate pe’s. The solubility of manganese as pMn(t) is shown inFigure 48. The area of soluble manganese is large as noted, and there isalso a large area contered around pH 10 and pe zero, where pMn(t) is on aplateau with a value of 4.4. In this region the neutral solute species MnSO0

4 isprevalent, accounting for virtually all the manganese in solution. This resultis just one instance of several in which soluble complexes of the metallicelements are quite important.

←−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−FIGURE 47. (a.) Contour map in pe-pH space of pAs(t) for orpiment, As2S3, with stable ironoxides as in Figure 43 and either calcium carbonate as in Figure 18 or dissolved calcium atlow pH. (b.) Phase diagram of the same system.

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The formation of manganese arsenates ought to be a possibility. Theamount of thermodynamic data on such salts is not large at all, but one ref-erence was found to Mn3(AsO4)2·8H2O, and this information proved to bequite useful. In these models nine different minerals or salts of arsenic includ-ing this one proved to be stable in different parts of the pe-pH space. Figure49a is a contour map of pAs(t) for the system with the stable iron oxides(Figure 43), and Figure 49b is a phase diagram in pe-pH space. Figure 50ais a contour map of pAs(t) with the reactive iron oxides (Figure 44), and50b is the phase diagram. In Figure 49b we see areas of possible precip-itation for eight salts of arsenic, viz.: FeAsO4·2H2O (scorodite), CaHAsO4,Ca3(AsO4)2, Mn3(AsO4)2·8H2O, As4O6 (arsenolite), As2S3 (orpiment), FeAsS(arsenopyrite), and FeAs2 (loellingite). The list for 50b is the same exceptthat CaHAsO4 does not appear, and there is an area of Fe3(AsO4)2.

The contour maps bear some resemblance to a map of Switzerlandwith a number of hills and valleys and several sharp gradients. The mostpromising area for possible remediation is as always near the center of thegraph where the pH’s are close to neutral, and the redox conditions are whatone would expect in the environment. On both maps there is an area withMn3(AsO4)2·8H2O which could be quite useful. If pAs(t) = 7, then we haveless than ten parts per billion. Unfortunately, the area with this condition isnot large, and the solubility of arsenic increases strongly in all directions.

Under strongly oxidizing conditions (up) the nature of the manganese-oxygen phase changes, and manganese becomes less soluble (Figure 48). Theresult is that Mn3(AsO4)2·8H2O is no longer the stable phase of arsenic, andone or both of the calcium arsenates precipitate. Here solubilities are muchhigher than those of Mn3(AsO4)2·8H2O. If the pH is reduced as well (up andleft), the stable mineral becomes FeAsO4·2H2O (scorodite). With the reactiveiron oxides, the pH must be below 5, and with the stable iron oxides the pHmust be below 3.5. Scorodite is truly stable only in quite acidic conditionsthat are strongly oxidizing, certainly extreme for most environmental systems.

Below scorodite and to the left of Mn3(AsO4)2·8H2O is a wedge-shapedarea where the stable mineral is arsenolite, and this oxide of arsenic(III)is highly soluble. Below this and stable at the extreme conditions of lowpH and low pe is orpiment. This sulfide, like many other metal sulfides,probably persists for long periods in conditions where it is truly not stable,but the thermodynamic stability is confined to extreme conditions. To theright of orpiment and below Mn3(AsO4)2·8H2O are found two minerals whichare under the right conditions extremely insoluble. The first is arsenopyrite,which occurs in a small area, and loellingite, which occurs over a large areaalong the lower limit of (pH + pe). The numbers on the contours get intothe teens, very impressive indeed, but the conditions required are so stronglyreducing that water will go to hydrogen.

At high pH’s, except for the very bottom of the map, arsenic goes intosolution. Under strongly oxidizing conditions the arsenates become more

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(a)

(b)

FIGURE 49. (a.) Contour map in pe-pH space of pAs(t) in the system with the stable ironoxides (Figure 43) plus manganese and calcium. (b.) Phase diagram for the same systemshowing eight minerals and salts of arsenic.

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(a)

(b)

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soluble as the solubilities of their metallic elements fall with increasing pH.Under reducing conditions the solubility of arsenic increases with increasingpH because orpiment is a Lewis acid. This means that one or more complexions of the type AsnS3n−2m

m will form. Thermodynamic information on the ionwith n = 3 and m = 6 was found, and its stability accounts for the finger ofhigh arsenic solubility between Mn3(AsO4)2·8H2O and FeAs2 in Figure 49a.

Finally there is an important area of stability for iron(II) arsenate onthe diagram for reactive iron oxides (50b), and it is just below the areafor Mn3(AsO4)2·8H2O. The Fe3(AsO4)2 area covers most of the area of higharsenic solubility discussed in the previous paragraph. This yields pAs(t)peaks at about 5 in this area, and the conditions under which this compoundis stable are certainly more common than those for scorodite. It might evenform the basis of a useful remediation strategy.

6. Sulfur and Phosphorus in the Manganese System

Contour maps and phase diagrams were also made for the minerals of sulfurand phosphorus in each of these systems. Figure 51a gives the contours forphosphorus in the system with the reactive iron oxides, and 51b shows thestable phases. Figure 52 does the same for sulfur.

The phosphorus diagram shows four phases: FePO4·2H2O (strengite),MnHPO4, Ca10(PO4)6(OH)2 (hydroxyapatite), and Fe3(PO4)2·2H2O (vivian-ite). The manganese salt occupies the center of the diagram and has a lowsolubility as was noted above in the section on lead. Strengite appears onlyat low pH and high pe, and its area disappears when the stable iron oxidesare in the system. Vivianite appears only at high pH and low pe, and its areaalso disappears when the stable iron oxide is used. Apatite is stable underhighly oxidizing conditions, when the solubility of manganese itself is quitelow, and with moderate to high pH. The boundaries of the phases are shownin Figure 51b.

The sulfur diagram shows six solid phases: CaSO4·2H2O (gypsum), As2S3

(orpiment), FeS2 (pyrite), FeAsS (arsenopyrite), Fe2S3, and FeS1.053 (iron-pyrrhotite). Two of these phases also occur on the arsenic diagram. Thestable mineral when (pH + pe) > 6 is gypsum. The others appear undermore reducing conditions as shown in Figure 52b.

Most of the gradients on these contour maps are much less severe thanthose on the arsenic maps, and keeping either sulfur or phosphorus in acrystalline state is much simpler than keeping arsenic out off solution.

←−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−FIGURE 50. (a.) Contour map in pe-pH space of pAs(t) in the system with reactive iron oxides(Figure 44) plus manganese and calcium. (b.) Phase diagram for the same system showingeight minerals and salts of arsenic.

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(a)

(b)FIGURE 51. (a.) Contour map in pe-pH space of pP(t) in the system with reactive iron oxides(Figure 44) and oxides and carbonates of manganese (Figure 48). (b.) Phases diagram of thissystem showing the areas of stability for four phosphate minerals.

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(a)

(b)

FIGURE 52. (a.) Contour map in pe-pH space of pS(t) in the system with reactive iron oxides(Figure 44) and oxides and carbonates of manganese (Figure 48). (b.) Phases diagram of thissystem showing the areas of stability for four phosphate minerals.

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7. Conclusions Regarding the Immobilization of Arsenic

It is possible to use precipitation reactions to make arsenic highly insoluble,especially by the formation of Mn3(AsO4)2·8H2O. This can be done undermoderate conditions, and the region of lowest solubility is centered aroundpH = 6.2 and pe = 6.0. pAs(t) becomes as high as 7.0. Under more reducingconditions, but not severely so, it also possible to make Fe3(AsO4)2, andits solubility is also low with pAs(t) as high as 6.0. These two solids arewithout doubt the best choices. The conditions under which the other arsenicminerals are stable are simply too severe as shown on Figures 49b and 50b.Furthermore, their solubilities are with one exception not as low as those overMn3(AsO4)2·8H2O or Fe3(AsO4)2. That exception is FeAs2, and its formationwould require a strong reducing potential.

Many of the gradients in the contour map are steep, and this means thatchanging conditions can significantly change the equilibrium solubility of ar-senic. For example if pe is fixed at 5.0, and the pH increases from 6.2 to 7.2,the solubility of arsenic increases by a factor of 29. Any scheme for immobi-lizing arsenic and leaving it in the environment must take this into account.

8. Possibilities for Further Work

Two projects can be suggested. The first is calculating the contours with afiner resolution. The maps done here are done at intervals of 0.2 in both peand pH. Maps with intervals of 0.1 would give smoother curves, and thereare places in our maps that are difficult to follow.

The second concerns the thermodynamic data. Every effort has beenmade to be as complete and accurate as possible in searching the literature,but no thermodynamic measurements were made. The models cannot bebetter than the data used. Two classes are of special concern. The first isthe manganese arsenates. As noted earlier, there seems to be a paucity ofdata, and knowing if the free energy of formation used for the octa-hydrateis accurate is important. Knowing if this is the only arsenate of the element isalso important, and the authors do not. MnHPO4 is an important solid. DoesMnHAsO4 exist? If it does, how insoluble is it?

The second class is the thioarsenate ions of the form AsnS3n−2mm and

their conjugate acids. Thermodynamic information was found on three only:HAsS0

2, AsS−2 (but not AsS+), and As3S

3−6 . These ions seem to have a pro-

found effect on the solubility of orpiment at neutral and high pH’s, and morecomplete and accurate information would be quite useful.

VI. MERCURY

The final element to be modeled will be mercury. The possible precipitates in-clude sulfides, chlorides, sulfates, phosphates, oxides, and carbonates. There

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is also the metal itself, and this may be insoluble enough for purposes of im-mobilization.

1. Modeling Systems with Mercury

The modeling techniques used for mercury were very similar to those used forarsenic. Calculations were made over wide ranges of pH’s and pe’s, and bothcontour maps and phase diagrams were made. Two model systems were run,and the first was exactly like the system with stable oxides of iron discussed inthe arsenic section, but with mercury substituted for arsenic. Sulfur solubilityis controlled by iron sulfides under reducing conditions and gypsum underoxidizing conditions. Phosphate is controlled by either hydroxyapatite ormanganese hydrogen phosphate. The carbon dioxide activity is controlledby the atmosphere. The second was like the first except that an activity ofiodine was added to make pI(t) = 3.00. This element is usually soluble,commonly present in the environment at least in small concentrations, andit can be precipitated as either HgI2 or Hg2I2.

The technique of using pK0sp was adapted for possible precipitates of

mercury. No information on Hg(OH)02 was found, but the neutral mercuryatom itself, Hg0, seems to be an important solute species. Furthermore, thethermodynamics of its formation have been studied with some care.63 Con-sequently, pK0

sp’s were calculated using the free energy of formation of thisspecies. Two examples follow:

Hg(liq.), pK0sp = pHg0 = 6.517

HgS, pK0sp = pHg0 + pH2S

0 − 2(pH + pe) = 9.764

The others are tabulated in the Appendix. There is a significant area in thepH-pe space over which the liquid metal is the stable condensed phase, andthe stable species in solution is the neutral atom with pHg(t) = 6.517.

2. The Mercury System without Iodine

The results for this system are shown in the several parts of Figure 53.Figure 53a is a contour map showing pHg(t) over the ranges of pH and pe,and Figure 53b shows which condensed phases are stable under which con-ditions. At no point does mercury become highly soluble, and pHg(t) >3.20everywhere. There are two regions where the gradients are steep, but in thelarge region near the center the map is close to level. The phase diagram hasfive areas but only four condensed phases with the element itself appearingtwice. The stable sulfide is mercury(I) sulfide rather than HgS (cinnabar), andboth Hg2Cl2 (calomel) and HgCO3 show up under oxidizing conditions. Asnoted in the previous paragraph there is an area at neutral pH and moderately

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reducing conditions where the stable mercury species in solution is the neu-tral Hg atom.

Mercury has three common oxidation states, and Figures 53c, 53d, and53e are contour maps for the total activities for each of these states over thepH-pe range. Figure 53c shows the activity of Hg0 only as this is the singlespecies of the zero oxidation state. Comparing Figures 53c to 53a shows thatthis species is dominant under reducing conditions. Figure 53d shows theactivity of Hg(I), and this state is always negligible in solution in spite of thefact that two precipitates of it are stable. Hg(II) is important under oxidizingconditions, and there are several species of this state including a number ofcomplex ions. The ion Hg2+ or its conjugate base is never the most importantHg(II) species in solution. The most common species of Hg(II) are bothneutral: HgCl02 and HgClOH0. Figure 53f shows the boundary between thedominance of Hg0 and Hg(II) in this system. The position of this boundaryis in an area where the conditions are at least moderately oxidizing with(pH + pe) in the neighborhood of ten.

3. Mercury Vapor

The thermodynamics for the formation of mercury vapor was also studied byGlew and Hames,63 and the conclusions they came to need to be considered.In particular the �S0 of vaporization is extraordinarily large, and accordingto these authors shows the largest known positive deviation from the valueexpected by Trouton’s Rule for a liquid that does not contain ions. At 298 K,�S0 = 210.9 J/K, and this quantity rises with falling temperature. (The �S0

from the “rule” is usually quoted as 88 J/K.) The free energy of formation forthe vapor is also known, and it is possible to write

pHg(gas) = pHg0 − 0.937

The reference state for gas activities is 100 kPa, and if pHg0 = 6.517, thenpHg = 0.263 Pa. Perhaps by itself this is not a large pressure, but certainlyany system with such an aqueous phase exposed to the atmosphere will losea lot of mercury over time. Oxidizing the mercury, difficult as this may be,to lower the activity of pHg0 could be very useful.

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→FIGURE 53. A model with mercury rather than arsenic added to the stable iron oxides, theoxides and carbonate of manganese, and the carbonate of calcium. (a.) The total solubility ofmercury as pHg(t) as a function of pH and pe. (b.) The stable phases of mercury over the samerange of pH and pe. (There are no areas where this element goes into solution completely.)(c.) The solubility of the neutral mercury atom as pHg0 as a function of pH and pe. (d.) Thesolubility of Hg(I) over the same space as pHg(I). (e.) The solubility of Hg(II) as pHg(II). (f.)A diagram showing the areas in pH-pe space in which Hg0 and Hg(II) are dominant. (Hg(I)never is in spite of the fact that both Hg2S and Hg2Cl2 have areas of stability.)

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(a)

(b)

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(c)

(d)

FIGURE 53. (Continued).

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(e)

(f)


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4. The Mercury System with Iodine Added

Iodine is a common element, usually found in the environment in solution.The most abundant elements in pore water do not precipitate it, but veryheavy elements such as mercury will. The several parts of Figure 54 showwhat happens when pI(t) is fixed at 3.00, and the parts of the figure parallelFigure 53. Figure 54a is the contour map, and the solubility of mercury islowered by the addition of iodine over much of the pH and pe ranges. Figure54b is the phase diagram, and the iodides of both Hg(I) and Hg(II) do formunder conditions that are not strongly reducing.

Figures 54c is a map of pHg0, and the activities of it and, therefore, thevapor pressure are lowered over much but not all of the area by the additionof iodine. Iodine will not, however, eliminate the problem discussed in theprevious section.

Figure 54d is a map of pHg(I), and its activity is still quite insignificant.In the upper left portion of the diagram, the activity of this state does increaseby several orders of magnitude, but it is still quite small.

Figure 54e and 54f parallel those in Figure 53. Under oxidizing conditionsthe activity of mercury falls significantly with the addition of iodine, and theboundary between the dominance of Hg(II) and the dominance of Hg(0) islowered to more strongly reducing conditions.

The abundances of species of Hg(II) in solution under oxidizing con-ditions are dominated by a series of complexes of the form HgI2-n

n . Thesecomplexes make mercury more soluble than it would be otherwise at thesame time iodide is precipitating the element.

5. The Solubility Product Constant

The solubility product, Ksp and its negative log pKsp are often used to cal-culate solubilities of slightly-soluble salts and minerals such as Hg2S. Thefree energies of formation used in this work can be used to calculate thesenumbers, so it is of interest to compare the results to what we have found.The pKsp for Hg2S describes the reaction

Hg2S = Hg2+2 + S2−

and pKsp = pHg2+2 + pS2−

Free energies of formation give us pKsp = 54.77, and it is often thought thatthe solubility of this salt can be found by assuming that pHg2

2+ = pS2− =pKsp/2 = 27.38. If this were true, the activity of mercury in a solution over

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→FIGURE 54. The same model and the same diagrams except that iodine is added and fixedso that pI(t) = 3.00. This includes the several species and oxidation states of iodine listed inAppendix A.

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(a)

(b)

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(c)

(d)


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(e)

(f)


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mercury(I) sulfide would be very low indeed. (It also assumed that the onlymercury species in solution is the mercury(I) ion.) These things are not true,and they would not be worth mentioning except for the fact that millions ofcollege students have learned that they are. The pKsp equation itself is, bythe way, true, but it is also irrelevant. If we look at the model in section 4 forHg and the results at pH = 7.2 and pe = −3.0, we find the following: Hg2Sis the stable mineral phase containing mercury; pHg(t) = 9.241, meaningthat the calculation using pKsp is off by 18 orders of magnitude; and pS(t) =3.467 (almost all sulfate), pHg2+

2 = 38.359, and pS2− = 16.413. (The last twonumbers do add to give the pKsp.)

6. Conclusions Concerning the Possible Immobilization of Hg

It is beyond the scope of this work to wonder what levels of mercury aresafe, but the authors are suspicious of anything above zero. The results ofthis section can be used to predict what might be possible or not possibleif some immobilization strategy or another is tried. It would seem to be truethat immobilization is much harder than it appears, and that it is probablynot a good technique for risk reduction under any circumstances. If thesystem with Hg in it is exposed to the atmosphere under any but the mostsevere oxidizing conditions, the metal will get into the atmosphere sooneror later. Such is surely the case with any strategy using sulfide, the formationof which requires highly reducing conditions. Even a strategy using HgI2runs into the difficulty that the iodide ion itself is a fairly active reducingagent. When (pH + pe) is 18.345, iodide is in equilibrium with iodate. Thisis a strong condition but certainly not beyond the realm of the possible.The metal itself should not be left in an environment open to the air orworse, the water at the bottom of a lake or river. Closed containers probablyneed to be the rule as the old name quicksilver fits even better than mostimagine.

VII. CONCLUSIONS REGARDING IMMOBILIZATION

This paper has used equilibrium calculations to test the stabilities and prob-able efficacies of reasonable immobilization strategies for lead, cadmium,arsenic, and mercury. The need for such strategies is urgent as the damageto the environment has already been done. On the basis of the thermody-namic calculations made, it would seem that virtually all such treatments aredoomed to failure. Making chloropyromorphite the only Pb form in soil mayrequire huge amounts of phosphate, amounts large enough to potentiallyturn the soil into a very hard and intractable substance. No compound ofcadmium is both thermodynamically stable and insoluble enough to work.

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Arsenic could be made into manganese arsenate and be left alone so long asstable conditions of pH and pe could be guaranteed for very long periods oftime. Mercury is perhaps the most difficult case of the four metals examined,slowly but quite surely living up to its ancient name.

The objection could be raised that systems in the environment do notcome to equilibrium and that, therefore, such calculations are useless. It cer-tainly is true that systems of interest are not at equilibrium, but the fact isthat interesting chemical systems are never at equilibrium. The point is thatthey always go that way unless they are driven in the opposite directionby stronger potentials. Systems left in any ecosystem will move toward thesystems described by these models if they change at all. Some immobiliza-tion strategies, such as leaving liquid mercury or cinnabar exposed to air, aresimply slow but deadly. In parting, the endeavor of this work shows that ther-modynamic calculations, as presented here, can tell us what is theoreticallypossible or impossible.

ACKNOWLEDGMENTS

The US EPA has not subjected this manuscript to internal policy review, thusit does not necessarily reflect Agency policy. Mention of trade names ofcommercial products does not constitute endorsement or recommendationfor use. The use of existing literature-derived data not generated by US EPAwas not subjected to US EPA quality assurance procedures, therefore noattempt to was made to verify the quality of the data. The authors wish tothank P. Burke for his careful review of the manuscript.

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58. Welham, N.J., Malatt, K.A., and Vukcevic, S. The Stability of Iron Phases PresentlyUsed for Disposal from Metallurgical Systems—A Review, Miner. Eng. 13, 911–931, 2000.

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59. Smedley, P.L., and Kinniburgh, D.G. A review of the source, behaviour anddistribution of arsenic in natural waters, Appl. Geochem. 17, 517–568, 2002.

60. Sterling, R.O., and Helble, J.J. Reaction of arsenic vapor species with fly ash com-pounds: kinetics and speciation of the reaction with calcium silicates, Chemo-sphere, 51, 1111–1119, 2003.

61. Mollah, M.Y.A., Lu, F., and Cocke, D.L. An X-ray diffraction (XRD) andFourier transform infrared spectroscopic (FT-IR) characterization of the speci-ation of arsenic (V) in Portland cement type-V, Sci. Total Environ. 224, 57–68,1998.

62. Bothe Jr., J.V., and Brown, P.W. The stabilities of calcium arsenates at23+/−1[deg]C, Journal of Hazardous Materials 69, 197–207, 1999.

63. Glew, D.N., and Hames, D.A. Can. J. Chem. 49, 3114–3118, 1971.64. Hem, J.D. Metal ions at surfaces of hydrous iron oxide, Geochim. Cosmochim.

Acta, 41, 527–538, 1977.65. Itagaki, K., and Nishimura, T. Thermodynamic properties of compounds and

aqueous species of VA elements, Metall. Rev. of MMIJ, 3, 29–48, 1986.66. USGS, Thermodynamic Properties of Materials, Bulletin 2131. 1995, U. S. Gov-

ernment Printing Office: Washington DC.67. Naumov, G.B., Ryzhenko, B.N., and Khodakovsky, I.L. Handbook of thermody-

namic data. in Report Available Only through NTIS. 1974.68. Smith, L.A. Summary of data analyses and experimental design for investiga-

tion of iron chemistry in lead-contaminated materials. 1994, US EPA: Cincinnati.63.

69. Barton, P.D. Geochim. Cosmochim. Acta 33, 841–857, 1969.

APPENDIX OF THERMOCHEMICAL DATA AND EQUILIBRIUMCONSTANT EQUATIONS

This appendix is divided into three parts. Part I is a table of free energies offormation for the neutral species, one for each element, that are used in theequilibrium-constant equations, including those for pK0

sp. These species areordered according to position on the periodic table, going from left to right,group 1 to group 18. The other two parts of this appendix are ordered thesame way.

Part II of the appendix is information on condensed phases, organizedas follows:

FORMULA, mineral name, �G0f in kJ/mol as the formula is written, reference,

pK0sp: numeric value, formula using the conventions described on page four

Part III of the appendix gives the information for the solute species not listedin part one. The data are given as follows:

chemical formula, �G0f in kJ/mol as the formula is written, reference, al-

gebraic formula for the negative log of the activity as the formula iswritten

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Part I. Reference Species

Formula ∆G0f in kJ/mol Reference

e− (the electron) 0 standard stateH+ (the proton) 0 standard stateH2O −237.178 Bard24

NaOH0 −419.17 BardKOH0 −436.89 Lindsay22

Mg(OH)02 −769.1 Bard

Ca(OH)2 −869.06 LindsayH2MoO0

4 −883.16 LindsayMn(OH)02 −610.45 Weast12

Fe(OH)03 −659.4 BardCd(OH)02 −442.6 BardHg0 +37.2 Bard, Glew63

Al(OH)03 −1094.6 Lindsay

CO02 −386.225 Bard

H4SiO04 −1308.17 Lindsay

Pb(OH)02 −397.73 LindsayNH0

3 −26.6 LindsayH3PO0

4 −1149.68 LindsayH3AsO0

4 −766.1 Welham,58 Hem64

H2S0 −27.87 Lindsay, BardHCl0 −114.14 Bard and est. pKa = −3.0

Part II. Minerals and Other Condensed Phases

KEY:FORMULA mineral name �G0

f in kJ/mol as the formula is written refer-ence pK0

sp: numeric value formula using the conventions described on pagefour

SODIUMNaAlSiO4 nepheline −1996.77 Lindsay21.644 pNaOH0 + pAl(OH)0

3 + pH4SiO04

NaAlSi3O8 Na glass −366.52 Lindsay22.016 pNaOH0 + pAl(OH)0

3 + 3pH4SiO04

NaAlSi3O8 high albite −3707.65 Lindsay29.222 pNaOH0 + pAl(OH)0

3 + 3pH4SiO04

NaAlSi3O8 low albite −3712.92 Lindsay30.145 pNaOH0 + pAl(OH)0

3 + 3pH4SiO04

NaAlSi2O6 jadeite −2854.20 Lindsay25.781 pNaOH0 + pAl(OH)0

3 + 2pH4SiO04

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NaAlSi2O6·H2O analcime −3091.7 Bard25.838 pNaOH0 + pAl(OH)0

3 + 2pH4SiO04

NaAl3Si3O10(OH)2 paragonite −5559.82 Lindsay53.281 pNaOH0 + 3pAl(OH)03 + 3pH4SiO0

4

NaAl7Si11O30(OH)6 beidellite −16,081.75 Lindsay127.162 pNaOH0 + 7pAl(OH)03 + 11pH4SiO0

4

NaAlSi3O8 anabite −3706.5 Bard29.020 pNaOH0 + pAl(OH)0

3 + 3pH4SiO04

Na3As −187.44 Itagaki65

−10.071 pNaOH0 + pH3AsO04/3 + 8(pH + pe)/3

NaAs −89.12 Itagaki15.925 pNaOH0 + pH3AsO0

4 + 6(pH + pe)

NaAs2 −103.76 Itagaki50.485 pNaOH0 + 2pH3AsO0

4 + 11(pH + pe)

MAGNESIUM

MgO periclase −569.2 Lindsay,Bard

6.243 pMg(OH)02

Mg(OH)2 brucite −834.3 Lindsay11.142 pMg(OH)0

2

MgOHCl −732.2 Weast14.951 pMg(OH)0

2 + pHCl0

MgCO3 magnestite −1026.6 Lindsay18.719 pMg(OH)02 + pCO0

2

MgCO3·3H2O nesquehodite −1722.2 Lindsay15.926 pMg(OH)0

2 + pCO02

MgCa(CO3)2 dolomite −2168.4 Lindsay40.379 pMg(OH)02 + pCa(OH)0

2 + 2pCO02

MgHPO4·3H2O newberryite −2297.2 Lindsay24.456 pMg(OH)02 + pH3PO0

4

Mg3(PO4)2 −3503.3 Lindsay18.382 pMg(OH)02 + 2pH3PO0

4/3

Mg3(PO4)2·8H2O boberrite −5460.1 Lindsay21.850 pMg(OH)02 + 2pH3PO0

4/3

Mg3(PO4)2·22H3O −8769.7 Lindsay21.216 pMg(OH)02 + 2pH3PO0

4/3

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MgCO3·5H2O lansfordite −2195.85 Lindsay15.797 pMg(OH)02 + pCO0

2

MgSiO3 clinoenstatite −1461.9 Lindsay16.567 pMg(OH)02 + pH4SiO0

4

Mg2SiO4 forsterite −2055.6 Lindsay13.547 pMg(OH)02 + pH4SiO0

4/2

Mg2SiO6(OH)4 sepolite −4271.7 Lindsay20.038 pMg(OH)02 + 3pH4SiO0

4/2

Mg3Si2O5(OH)4 chrystolite −4034.2 Lindsay17.027 pMg(OH)02 + 2pH4SiO0

4/3

Mg3Si4O10(OH)2 talc −5525.2 Lindsay20.564 pMg(OH)02 + 4pH4SiO0

4/3

Mg3Si4O10(OH)2·2H2O vermiculite −5953.2 Lindsay17.855 pMg(OH)0

2 + 4pH4SiO04/3

Mg6Si4O10(OH)8 serpentine −8091.2 Lindsay17.693 pMg(OH)02 + 2pH4SiO0

4/3

CALCIUM

CaS oldhamite −469.5 USGS66

8.214 pCa(OH)02 + pH2S0

CaSO4·2H2O gypsum −1799.83 Lindsay−8.033 pCa(OH)02 + pH2S0 − 8(pH + pe)

CaSO4 −1320.3 Lindsay−8.265 pCa(OH)0

2 + pH2S0 − 8(pH + pe)

Ca10(PO4)6(OH)2 hydroxyapatite −12,678.5 Lindsay23.809 pCa(OH)02 + 3pH3PO0

4/5

Ca3(PO4)2 α −3860.6 Lindsay,Naumov67

22.022 pCa(OH)02 + 2pH3PO04/3

Ca3(PO4)2 whitelockite −3880.1 Lindsay23.163 pCa(OH)02 + 2pH3PO0

4/3

Ca8H2(PO4)6·5H2O octa-calcium phos. −12,311.9 Lindsay23.439 pCa(OH)02 + 3pH3PO0

4/4

CaHPO42H2O brushite −2162.7 Lindsay25.215 pCa(OH)02 + pH3PO0

4

CaHPO4 monetite −1690.17 Lindsay25.540 pCa(OH)02 + pH3PO0

4

Ca2P2O7 β −3105.91 Lindsay22.277 pCa(OH)02 + pH3PO0

4

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Ca(H2PO4)2 “super phosphate” −3073.1 Lindsay24.846 pCa(OH)02 + 2pH3PO0

4

CaCO3 calcite −1130.4 Lindsay19.679 pCa(OH)02 + pCO0

2

CaCO3·6H2O ikaite −2541.9 Lindsay17.806 pCa(OH)0

2 + pCO02

CaO lime −603.58 Lindsay−4.957 pCa(OH)02Ca(OH)2 portlandite −898.68 Lindsay5.190 pCa(OH)02CaO·Fe2O3 −1412.81 Bard30.586 pCa(OH)02 + pFe(OH)032CaO·Fe2O3 −2001.8 Bard11.615 pCa(OH)02 + pFe(OH)03/2

CaSiO3 wollastanite −1549.71 Bard, Lindsay14.719 pCa(OH)02 + pH4SiO0

4

CaSiO3 pseudo-wollastanite −1497.04 Weast13.766 pCa(OH)02 + pH4SiO0

4

Ca2SiO4 β-laruite −2192.8 Bard8.183 pCa(OH)02 + pH4SiO0

4/2

Ca2SiO4 γ -olivine −2201.2 Bard9.085 pCa(OH)02 + pH4SiO0

4/2

CaMoO4 powellite −1435.9 Lindsay27.688 pCa(OH)02 + pH2MoO0

4

Ca3(AsO4)2 −3063.1 Bard20.252 pCa(OH)02 + 2pH3AsO0

4/3

Ca3(AsO4)2·4H2O −4018.73 Naumov20.656 pCa(OH)02 + 2pH3AsO0

4/3

Ca(H2AsO4)2 −2053.93 Itagaki, Naumov22.257 pCa(OH)02 + 2pH3AsO0

4

CaHAsO4 −1287.42 Itagaki, Naumov22.184 pCa(OH)02 + pH3AsO0

4

Ca(AsO2)2 −1292.02 Itagaki, Naumov54.985 pCa(OH)02 + 2pH3AsO0

4 + 4(pH + pe)

CaAsO2OH −1112.94 Itagaki, Naumov33.169 pCA(OH)02 + pH3AsO0

4 + 2(pH + pe)

Ca5H2(AsO4)4 −5636.7 Itagaki, Naumov20.983 pCa(OH)02 + 4pH3AsO0

4/5

Ca2AsO4OH −1987.82 Itagaki, Naumov17.094 pCa(OH)02 + pH3AsO0

4/2

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MOLYBDENUM

Mo metal 0 standard state11.485 pH2MoO0

4 + 6(pH + pe)

MoO2 −533.08 Lindsay21.773 pH2MoO0

4 + 2(pH + pe)

H2MoO4 −912.45 Lindsay5.131 pH2MoO0

4

MoO3 molybdite −668.0 Lindsay3.861 pH2MoO0

4

MoS2 molybdenite −266.48 Lindsay48.391 pH2MoO0

4 + 2pH2S0 + 2(pH + pe)

MANGANESE

Mn metal 0 standard state−17.736 pMn(OH)02 + 2(pH + pe)

MnS “green” −211.38 Lindsay14.414 pMn(OH)0

2 + pH2S0

MnS alabandite −218.07 Lindsay15.586 pMn(OH)0

2 + pH2S0

MnS2 hauerite −232.34 Lindsay13.203 pMn(OH)0

2 + pH2S0 − 2(pH + pe)

MnSO4 −955.54 Lindsay−21.422 pMn(OH)0

2 + pH2S0 − 8(pH + pe)

MnSO4H2O −1209.64 Lindsay−18.458 pMn(OH)0

2 + pH2S0 − 8(pH + pe)

Mn2(SO4)3 −2469.15 Lindsay−58.083 pMn(OH)0

2 + 3pH2S0/2 − 13(pH + pe)

Mn3(PO4)2 −2899.3 Lindsay17.302 pMn(OH)0

2 + 2pH3PO04/3

MnHPO4 −1400.8 Lindsay26.253 pMn(OH)0

2 + pH3PO04

MnCl2 scacchite −440.50 Bard19.444 pMn(OH)0

2 + pHCl0

MnCl2H2O −696.1 Bard22.672 pMn(OH)0

2 + pHCl0

MnCO3 rhodochrosite −816.01 Lindsay16.007 pMn(OH)0

2 + pCO02

MnO manganosite −362.80 Lindsay4.272 pMn(OH)0

2

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Mn(OH)2 pyrochorite −618.23 Lindsay7.470 pMn(OH)02MnO2 pyrolusite −465.85 Lindsay−19.227 pMn(OH)02 − 2(pH + pe)

MnO1.8 birnessite −455.60 Lindsay−12.712 pMn(OH)02 − 8(pH + pe)/5

MnO1.9 nsutite −459.28 Lindsay−16.222 pMn(OH)02 − 9(pH + pe)/5

Mn2O3 bixbyite −879.02 Lindsay−3.065 pMn(OH)02 − (pH + pe)

Mn3O4 hausmannite −1280.76 Lindsay1.655 pMn(OH)02 − 2(pH + pe)/3

MnOOH manganite −560.70 Lindsay−2.609 pMn(OH)02 − (pH + pe)

MnAs kaneite −57.32 Itagaki, Naumov24.301 pMn(OH)02 + pH3AsO0

4 + 7(pH + pe)

Mn3(AsO4)2 −2145.14 Itagaki, Naumov18.060 pMn(OH)02 + 2pH3AsO0

4/3

Mn3(AsO4)2·8H2O −4055.13 Naumov18.793 pMn(OH)02 + 2pH3AsO0

4/3

IRON

Fe metal 0 standard state9.134 pFe(OH)03 + 3(pH + pe)

FeO −251.454 Welham11.635 pFe(OH)03 + (pH + pe)

Fe(OH)2 −486.6 Bard11.279 pFe(OH)03 + (pH + pe)

Fe3O4 magnetite −1015.359 Bard13.026 pFe(OH)0

3 + (pH + pe)/3

FeO1.062 wustite −276.336 Bard13.418 pFe(OH)0

3 + 0.876(pH + pe)

Fe3(OH)8 −1921.4 Lindsay10.533 pFe(OH)03 + (pH + pe)/3

FeOOH goethite −490 Welham12.190 pFe(OH)03Fe2O3·H2O −984.03 Bard12.277 pFe(OH)03

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Fe2O3 maghemite −726.97 Lindsay10.486 pFe(OH)0

3

Fe2O3 hematite −742.2 Bard11.820 pFe(OH)0

3

FeOOH lepidocrocite −483.25 Lindsay10.587 pFe(OH)0

3

Fe(OH)3 (c) −705.535 Bard8.083 pFe(OH)0

3

Fe(OH)3 “soil” −712.95 Lindsay9.382 pFe(OH)0

3

FeCO3 siderite −677.60 Lindsay16.723 pFe(OH)0

3 + pCO02 + (pH + pe)

FeCl2 lawrencite −302.38 Lindsay22.82 pFe(OH)0

3 + 2pHCl0 + (pH + pe)

FeCl3 molysite −334.97 Lindsay7.667 pFe(OH)0

3 + 3pHCl0

FeOCl −359.234 Bard10.521 pFe(OH)0

3 + pHCl0

FeS2 pyrite −162.26 Lindsay28.594 pFe(OH)0

3 + 2pH2S0 − (pH + pe)

FeS2 markasite −158.28 Lindsay27.084 pFe(OH)0

3 + 2pH2S0 − (pH + pe)

Fe7S8 S-rich pyrrhotite −748.5 Bard22.279 pFe(OH)0

3 + 8pH2S0/7 + 5(pH + pe)/7

FeS1.053 Fe-rich pyrrhotite −105.61 Lindsay22.487 pFe(OH)0

3 + 1.053pH2S0 + 0.894(pH + pe)

FeS troilite −97.91 Lindsay21.397 pFe(OH)0

3 + pH2S0 + (pH + pe)

Fe2S3 −278.40 Lindsay26.187 pFe(OH)03 + 3pH2S0/2

FeSO4 −820.61 Lindsay−18.199 pFe(OH)03 + pH2S0 − 7(pH + pe)

FeSO4·7H2O tauriscite −2510.3 Lindsay−13.046 pFe(OH)0

3 + pH2S0 − 7(pH + pe)

KFe3(SO4)2(OH)6 jarosite −3316.5 Lindsay−18.069 pKOH0 + 3pFe(OH)0

3 + 2pH2S0 − 16(pH + pe)

FePO4 −1184.9 Lindsay,Naumov

15.298 pFe(OH)03 + pH3PO04

P1: GIM



FePO4·2H2O strengite −1667.7 Lindsay16.784 pFe(OH)0

3 + pH3PO04

Fe3(PO4)2·8H2O vivianite −4428.18 Lindsay22.647 pFe(OH)0

3 + 2pH3PO04/3 + (pH + pe)

Fe2P2O7 −2195.93 Lindsay20.850 pFe(OH)0

3 + pH3PO04 + (pH + pe)

FeAs −28.0 Barton46.032 pFe(OH)0

3 + pH3AsO04 + 8(pH + pe)

Fe2As −20.9 Barton26.961 pFe(OH)0

3 + pH3AsO04/2 + 11(pH + pe)/2

FeAs2 loellingite −52.3 USGS82.282 pFe(OH)0

3 + 2pH3AsO04 + 13(pH + pe)

FeAsS arsenopyrite −109.6 Barton55.438 pFe(OH)0

3 + pH3AsO04 + pH2S0 + 6(pH + pe)

Fe3(AsO4)2 −1765.3 Itagaki23.832 pFe(OH)0

3 + 2pH3AsO04/3 + (pH + pe)

FeAsO4 −774.58 Barton11.539 pFe(OH)0

3 + pH3AsO04

FeAsO4·2H2O scorodite −1280.0 Bard16.062 pFe(OH)0

3 + pH3AsO04

CADMIUMCd metal 0 standard state5.564 pCd(OH)02 + 2(pH + pe)

CdO monteponite −228.66 Bard4.070 pCd(OH)02Cd(OH)2 β −474.34 Bard, Lindsay5.561 pCd(OH)02CdCO3otavite −674.29 Lindsay, Bard14.484 pCd(OH)02 + pCO0

2

CdCl2 −343.93 Bard25.825 pCd(OH)02 + 2pHCl0

Cd3(PO4)2 −2502.70 Lindsay17.438 pCd(OH)02 + 2pH3PO0

4/3

CdSiO3 −1105.33 Lindsay11.579 pCd(OH)02 + pH4SiO0

4

CdS greenokite −146.57 Lindsay26.358 pCd(OH)02 + pH2S0

CdSO4 −822.66 Bard, Lindsay−21.403 pCd(OH)02 + pH2S0 − 8(pH + pe)

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CdSO4·H2O −1068.68 Bard, Lindsay−19.854 pCd(OH)02 + pH2S0 − 8(pH + pe)

CdSO4·8/3H2O −1465.141 Bard−19.650 pCd(OH)02 + pH2S0 − 8(pH + pe)

CdSO4·2Cd(OH)2 −1797.57 Lindsay−1.895 pCd(OH)02 + pH2S0/3 − 8(pH + pe)/3

2CdSO4·Cd(OH)2 −2158.65 Lindsay−10.138 pCd(OH)02 + 2pH2S0/3 − 16(pH + pe)/3

MERCURYHg liquid metal 0 standard state6.517 pHg0

HgO red, orthorhombic −58.555 Bard−24.777 pHg0 − 2(pH + pe)

Hg(OH)2 −294.85 Lindsay−24.932 pHg0 − 2(pH + pe)

Hg2(OH)2 −290.75 Lindsay−9.567 pHg0 − (pH + pe)

HgCO3 −492.122 Lindsay−16.483 pHg0 + pCO0

2 − 2(pH + pe)

Hg2CO3 −468.2 Bard−7.078 pHg0 + pCO0

2/2 − (pH + pe)

HgCl2 −180.3 Bard−1.889 pHg0 + 2pHCl0 − 2(pH + pe)

Hg2Cl2 calomel −210.374 Bard4.949 pHg0 + pHCl0 − (pH + pe)

Hg2HPO4 −966.57 Bard−9.523 pHg0 + pH3PO0

4/2 − (pH + pe)

HgS cinnabar −46.4 Bard, Lindsay9.764 pHg0 + pH2S0 − 2(pH + pe)

HgSO4 −594 Bard−60.509 pHg0 + pH2S0 − 10(pH + pe)

Hg2SO4 −626.34 Bard, Lindsay−24.163 pHg0 + pH2S0/2 − 5(pH + pe)

Hg2S −73.18 Lindsay10.486 pHg0 + pH2S0/2 − (pH + pe)

HgI2 “red” −101.7 Bard25.435 pHg0 + 2pHI0 − 2(pH + pe)

Hg2I2 −111.002 Bard

P1: GIM



16.791 pHg0 + pHI0 − (pH + pe)

Hg(IO3)2 −167.2 Bard−212.955 pHg0 + 2pHI0 − 13(pH + pe)

Hg2(IO3)2 −179.9 Bard−101.832 pHg0 + pHI0 − 7(pH + pe)

ALUMINUM

Al(OH)3 gibbsite −1156.58 Lindsay10.564 pAl(OH)0

3

Al(OH)3 “amorphous” −1147.30 Lindsay9.233 pAl(OH)0

3

Al(OH)3 bayerite −1153.86 Lindsay10.382 pAl(OH)0

3

Al(OH)3 nordstrandite −1156.04 Lindsay10.764 pAl(OH)0

3

Al(OH)3·H2O −1376.42 Bard7.821 pAl(OH)03Al(OH)3·3H2O −1850.4 Bard14.763 pAl(OH)0

3

AlOOH α, diaspore −920.06 Lindsay11.051 pAl(OH)0

3

AlOOH γ , boehmite −918.85 Lindsay9.684 pAl(OH)0

3

Al2O3 corundum −158.26 Lindsay9.162 pAl(OH)0

3

Al2O3 γ −1562.22 Lindsay9.407 pAl(OH)0

3

AlPO4 berlinite −1625.48 Lindsay16.247 pAl(OH)0

3 + pH3PO04

AlPO4·2H2O variscite −2116.98 Lindsay19.250 pAl(OH)0

3 + pH3PO04

Al2SiO2O5(OH)4 kaolinite −3804.22 Lindsay16.168 pAl(OH)0

3 + pH4SiO04

Al2SiO5 andalusite −2444.50 Lindsay11.652 pAl(OH)0

3 + pH4SiO04/2

Al2SiO5 kyanite −2440.86 Lindsay11.333 pAl(OH)0

3 + pH4SiO04/2

Al2SiO5 sillimanite −2438.48 Lindsay11.125 pAl(OH)0

3 + pH4SiO04/2

P1: GIM



Al2Si2O5(OH)4 dickite −3796.3 Bard15.474 pAl(OH)03 + pH4SiO0

4

Al2Si2O5(OH)4 halloysite −3785.6 Lindsay14.533 pAl(OH)03 + pH4SiO0

4

Al2Si4O10(OH)2 pyrophillite −5276.74 Lindsay19.853 pAl(OH)03 + 2pH4SiO0

4

SILICON

SiO2 “soil” −851.49 Lindsay3.096 pH4SiO0

4

SiO2 quartz −856.67 Lindsay4.005 pH4SiO0

4

LEAD

Pb metal 0 standard state13.424 pPb(OH)0

2 + 2(pH + pe)

PbO “yellow” −188.28 Lindsay4.858 pPb(OH)0

2

PbO “red” −189.28 Lindsay5.034 pPb(OH)0

2

PbO “white” −183.72 Bard4.059 pPb(OH)0

2

Pb(OH)2 “aged” −452.5 Lindsay9.595 pPb(OH)0

2

Pb(OH)2 “fresh” −420.91 Smith(68)

4.542 pPb(OH)02

PbO2 −215.52 Bard−31.923 pPb(OH)02 − 2(pH + pe)

Pb3O4 −601.659 Bard−6.843 pPb(OH)02 − 2(pH + pe)/3

Pb2O3 −411.78 Bard−12.833 pPb(OH)02 − (pH + pe)

PbO1.57 −211.21 Bard−14.810 pPb(OH)02 − 1.14(pH + pe)

PbCO3 cerrusite −629.73 Lindsay14.533 pPb(OH)0

2 + pCO02

Pb2CO3Cl2 phosgenite −953.70 Lindsay−39.968 pPb(OH)0

2 + pCO02/2 + pHCl0

P1: GIM



Pb3(CO3)2(OH)2 −1711.6 Lindsay12.865 pPb(OH)0

2 + 2pCO02/3

PbO·PbCO3 −818.9 Lindsay9.772 pPb(OH)0

2 + pCO02/2

Pb(OH)2·(PbCO3)2 Smith12.178 pPb(OH)02 + 2pCO0

2/3

2PbO·PbCO3 −1012 Smith8.416 pPb(OH)0

2 + pCO02/3

PbCl2 −314.0 Wall(25)

28.436 pPb(OH)02 + 2pHCl0

PbS galena −95.86 Lindsay25.329 pPb(OH)0

2 + 2pH2S0

PbS2O3 −560.6 Smith−22.784 pPb(OH)0

2 + 2pH2S0 − 8(pH + pe)

PbS3O6 −894.5 Smith−93.827 pPb(OH)0

2 + 3pH2S0 − 16(pH + pe)

PbSO4 anglesite −813.70 Lindsay, Smith−15.111 pPb(OH)0

2 + pH2S0 − 8(pH + pe)

PbSO4·PbO −1032.2 Lindsay−2.408 pPb(OH)0

2 + pH2S0/2 − 4(pH + pe)

PbSO4·2PbO −1230.1 Lindsay0.527 pPb(OH)0

2 + pH2S0/3 − 8(pH + pe)/3

PbSO4·3PbO −1427.6 Lindsay2.013 pPb(OH)0

2 + pH2S0/4 −2(pH + pe)

PbHPO4 −1186.4 Lindsay19.853 pPb(OH)0

2 + pH3PO04

Pb(H2PO4)2 −2355.8 Lindsay23.306 pPb(OH)0

2 + 2pH3PO04

Pb3(PO4)2 −2378.9 Lindsay18.069 pPb(OH)0

2 + 2pH3PO04/3

Pb5(PO4)3OH hydroxypyromorphite −3796.5 Lindsay17.289 pPb(OH)0

2 + 3pH3PO04/5

Pb5(PO4)3Cl chloropyromorphite −3809.91 Lindsay22.069 pPb(OH)0

2 + 3pH3PO04/5 + pHCl0/5

Pb4O(PO4)2 −2598.0 Lindsay16.117 pPb(OH)0

2 + pH3PO04/2

PbMoO4 wulfenite −952.36 Lindsay25.548 pPb(OH)0

2 + pH2MoO04

P1: GIM



ARSENICAs2O5 −781.4 Bard−3.440 pH3AsO0

4

As4O6 arsenolite −1153.0 Bard20.164 pH3AsO0

4 + 2(pH + pe)

As4S4 realgar −142.26 Barton(69)

33.334 pH3AsO04 + pH2S0 + 3(pH + pe)

As2S3 orpiment −90.4 USGS38.417 pH3AsO0

4 + 2pH2S0/3 + 2(pH + pe)

III. Solute Species

chemical formula �G0f in kJ/mol as the formula is written reference

algebraic formula for the negative log of the activity as the formula is written

SODIUM

Na+ −261.87 BardpNa+ = pNaOH0 + pH −13.994

NaCl0 −393.04 BardpNaCl0 = pNaOH0 + pHCl0 − 16.978

NaCO−3 −797.05 Lindsay

pNaCO−3 = pNaOH0 + pCO0

2 − pH + 1.261

Na2CO03 −1051.77 Lindsay

pNa2CO03 = 2pNaOH0 + pCO0

2 − 11.682

NaHCO03 −850.19 Lindsay

pNaHCO03 = pNaOH0 + pCO0

2 − 8.048

NaHPO−4 Smith

pNaHPO−4 = pNaOH0 + pH3PO0

4 − pH − 5.841

NaSO−4 −1010.39 Lindsay

pNa2SO−4 = pNaOH0 + pH2S0 − 8(pH + pe) − pH + 25.767

Na2SO04 −1265.7 Bard

pNa2SO04 = pNaOH0 + pH2S0 − 8(pH + pe) + 13.116

MAGNESIUM

Mg2+ −456.10 LindsaypMg2+ = pMg(OH)02 + 2pH − 27.981

MgOH+ −627.93 LindsaypMgOH+ = pMg(OH)0

2 + pH − 16.534

MgCO03 −1002.53 Lindsay

pMgCO03 = pMg(OH)02 + pCO0

2 − 14.496

P1: GIM



MgHCO+3 −1049.10 Lindsay

pMgHCO+3 = pMg(OH)02 + pCO0

2 + pH − 22.655

MgCl+ −587.18 WeastpMgCl+ = pMg(OH)02 + pHCl0 + pH −29.94

MgCl02 −718.43 LindsaypMgCl02 = pMg(OH)0

2 + 2pHCl0 − 33.948

MgHPO04 −1569.04 Lindsay

pMgHPO04 = pMg(OH)0

2 + pH3PO04 − 21.546

MgP2O2−7 −2413.8 Bard

pMgP2O2−7 = pMg(OH)0

2 + 2pH3PO04 − 2pH − 9.964

MgSO04 −1211.81 Naumov

pMgSO04 = pMg(OH)0

2 + pH2S0 − 8(pH + pe) + 12.689

CALCIUMCa2+ −554.46 LindsaypCa2+ = pCa(OH)0

2 + 2pH − 27.989

CaOH+ −719.19 LindsaypCaOH+ = pCa(OH)0

2 + pH − 15.295

CaPO−4 −1617.16 Lindsay

pCaPO−4 = pCa(OH)0

2 + pH3PO04 − pH 12.750

CaHPO04 −1666.45 Lindsay

pCaHPO04 = pCa(OH)0

2 + pH3PO04 − 21.384

CaH2PO+4 −1699.88 Lindsay

pCaH2PO+4 = pCa(OH)0

2 + pH3PO04 + pH − 27.241

CaP2O2−7 −2506.38 Lindsay

pCaP2O2−7 = pCa(OH)0

2 + 2pH3PO04 − 2pH − 8.671

CaHP2O−7 −2541.82 Lindsay

pCaHP2O−7 = pCa(OH)0

2 + 2pH3PO04 − pH − 14.880

CaOHP2O3−7 −2675.63 Lindsay

pCaOHP2O3−7 = pCa(OH)0

2 + 2pH3PO04 − 3pH + 3.252

CaCO03 −1100.39 Lindsay

pCaCO03 = pCa(OH)0

2 + pCO02 − 14.416

CaHCO+3 −1147.80 Lindsay

pCaHCO+3 = pCa(OH)0

2 + pHCl0 + pH − 22.721

CaCl+ −680.03 LindsaypCaCl+ = pCa(OH)0

2 + pHCl0 + pH − 29.990

CaCl02 −816.97 LindsaypCaCl02 = pCa(OH)0

2 + 2pHCl0 − 33.984

CaSO04 −1312.19 Lindsay

pCaSO04 = pCa(OH)0

2 + pH2S0 − 8(pH + pe) + 12.648

P1: GIM



MOLYBDENUM

MoO2+2 −411.16 Lindsay

pMoO2+2 = pH2MoO0

4 + 2pH − 0.413

MoO2OH+ −645.76 LindsaypMoO2OH+ = pH2MoO0

4 + pH + 0.039

HMoO−4 −860.31 Lindsay

pHMoO−4 = pH2MoO0

4 − pH + 4.002

MoO2−4 −836.13 Lindsay

pMoO2−4 = pH2MoO0

4 − 2pH + 8.239

Mo7O6−24 −5251 Bard

pMo7O6−24 = 7pH2MoO0

4 − 6pH − 3.524

MANGANESEMn2+ −230.58 Lindsay,

BardpMn2+ = pMn(OH)0

2 + 2pH −22.661

MnOH+ −407.31 LindsaypMnOH+ = pMn(OH)0

2 + pH − 12.071

Mn(OH)−3 −748.10 LindsaypMn(OH)−3 = pMn(OH)0

2 − pH + 11.330

Mn(OH)2−4 −903.70 Lindsay

pMn(OH)2−4 = pMn(OH)0

2 − 2pH + 25.621

Mn2OH3+ −637.85 LindsaypMn2OH3+ = 2pMn(OH)0

2 + 3pH − 34.725

Mn2(OH)+3 −1036.34 LindsaypMn2(OH)+3 = 2pMn(OH)0

2 + pH − 21.432

HMnO−2 −507 Bard

pHMnO−2 = pMn(OH)0

2 − pH + 12.017

Mn3+ −84.77 Lindsay,Bard

pMn3+ = pMn(OH)02 − (pH + pe) + 3pH + 2.885

MnOH2+ −324.22 LindsaypMnOH2+ = pMn(OH)0

2 − (pH + pe) + 2pH + 2.487

Mn4+ +60.79 LindsaypMn4+ = pMn(OH)0

2 − 2(pH + pe) + 4pH + 28.387

MnO3−4 −527 Bard

pMnO3−4 = pMn(OH)0

2 − 3(pH + pe) − 3pH + 91.618

MnO2−4 −504.09 Lindsay,

BardpMnO2−

4 = pMn(OH)02 − 4(pH + pe) − 2pH + 95.632

P1: GIM



MnO−4 −477.2 Bard

pMnO−4 = pMn(OH)0

2 − 5(pH + pe) − pH + 100.344

MnCl+ −493.34 LindsaypMnCl+ = pMn(OH)0

2 + pHCl0 + pH −26.824

MnCl02 −754.46 LindsaypMnCl02 = pMn(OH)0

2 + 2pHCl0 − 29.258

MnCO03 −754.46 Lindsay

pMnCO03 = pMn(OH)0

2 + pCO02 − 5.225

MnHCO+3 −827.76 Lindsay

pMnHCO+3 = pMn(OH)0

2 + pCO02 + pH − 18.067

MnSO04 −988.01 Lindsay

pMnSO04 = pMn(OH)0

2 + pH2S0 − 8(pH + pe) + 15.735

IRON(III)

Fe3+ −4.6 BardpFe3+ = pFe(OH)0

3 + 3pH − 9.940

FeOH2+ −229.41 BardpFeOH+

2 = pFe(OH)03 + 2pH − 7.668

Fe(OH)+2 −438.1 BardpFe(OH)+2 = pFe(OH)0

3 + pH − 2.677

Fe(OH)−4 −842.2 BardpFe(OH)−4 = pFe(OH)0

3 − pH + 9.632

Fe2(OH)4+2 −491.45 Lindsay

pFe2(OH)4+2 = 2pFe(OH)0

3 + 4pH − 23.286

Fe3(OH)5+4 −963.28 Lindsay

pFe3(OH)5+4 = 3pFe(OH)0

3 + 5pH − 32.988

FeCl2+ −143.9 BardpFeCl2+ = pFe(OH)0

3 + pHCl0 + 2pH − 14.242

FeCl+2 −291.5 LindsaypFeCl+2 = pFe(OH)0

3 + 2pHCl0 + pH − 17.932

FeCl03 −415.0 LindsaypFeCl03 = pFe(OH)0

3 + 3pHCl0 − 19.222

FeHPO+4 −1175.4 Lindsay

pFeHPO+4 = pFe(OH)0

3 + pH3PO04 + pH − 15.020

FeH2PO2+4 −1244.0 Lindsay

pFeH2PO2+4 = pFe(OH)0

3 + pH3PO04 + 2pH − 16.742

FeSO+4 −772.8 Bard

pFeSO+4 = pFe(OH)0

3 + + pH2S0 − 8(pH + pe) + pH + 26.568

P1: GIM



Fe(SO4)−2 −1524.6 Bard

pFe(SO4)−2 = pFe(OH)0

3 + + 2pH2S0 − 16(pH + pe) − pH + 65.949

FeH2AsO2+4 −793.7 Welham

pFeH2AsO2+4 = pFe(OH)03 + pH3AsO0

4 + 2pH − 13.864

FeHAsO+4 −788.2 Welham

pFeHAsO+4 = pFe(OH)0

3 + pH3AsO04 + pH − 12.900

FeAsO04 −773.6 Welham

pFeAsO04 = pFe(OH)03 + pH3AsO0

4 − 10.343

IRON(II)Fe2+ −78.87 BardpFe2+ = pFe(OH)0

3 + (pH + pe) + 2pH − 22.951

FeOH+ −277.4 BardpFeOH+ = pFe(OH)0

3 + (pH + pe) + pH − 16.180

Fe(OH)02 −441.0 BardpFe(OH)0

2 = pFe(OH)03 + (pH + pe) − 3.290

Fe(OH)−3 −615.0 BardpFe(OH)−3 = pFe(OH)0

3 + (pH + pe) − pH + 7.779

Fe(OH)2−4 −769.9 Bard

pFe(OH)2−4 = pFe(OH)03 + (pH + pe) − 2pH + 22.193

Fe2(OH)2+2 −467.27 Bard

pFe2(OH)2+2 = pFe(OH)0

3 + 2(pH + pe) + 2pH − 17.026

HFeO−2 −377.8 Bard

pHFeO−2 = pFe(OH)0

3 + (pH + pe) − pH + 7.783

FeO2−2 −455.2 Bard

pFeO2−2 = pFe(OH)03 + (pH + pe) − 2pH − 5.778

FeCl+ −361.08 WeastpFeCl+ = pFe(OH)0

3 + pHCl0 + (pH + pe) + pH − 25.949

FeCl02 −341.37 BardpFeCl02 = pFe(OH)0

3 + 2pHCl0 + (pH + pe) − 28.947

FeHPO04 −1208.09 Lindsay

pFeHPO04 = pFe(OH)0

3 + pH3PO04 + (pH + pe) − 19.367

FeSO04 −823.49 Bard

pFeSO04 = pFe(OH)03 + pH2S0 − 7(pH + pe) + 17.687

FeSH+ −120 WeastpFeSH+ = pFe(OH)0

3 + pH2S0 + (pH + pe) + pH − 25.213

Fe(SH)02 −119.24 NaumovpFe(SH)0

2 = pFe(OH)03 + 2pH2S0 + (pH + pe) − 20.259

Fe(SH)−3 −118.83 NaumovpFe(SH)−3 = pFe(OH)0

3 + 3pH2S0 + (pH + pe) − pH − 15.304

P1: GIM



CADMIUM

Cd2+ −77.86 LindsaypCd2+ = pCd(OH)0

2 + 2pH − 19.205

CdOH+ −257.4 LindsaypCdOH+ = pCd(OH)0

2 + pH − 9.106

Cd(OH)−3 −601.0 LindsaypCd(OH)−3 = pCd(OH)0

2 − pH + 13.803

Cd(OH)2−4 −756.68 Lindsay

pCd(OH)2−4 = pCd(OH)0

2 − 2pH + 28.080



2 − 3pH + 42.716



2 − 4pH + 57.594

Cd2OH3+ −356.39 LindsaypCd2OH3+ = 2pCd(OH)02 + 3pH − 32.013

Cd4(OH)4+4 −1100.81 Lindsay

pCd4(OH)4+4 = 4pCd(OH)02 + 4pH − 48.901

CdCO03 −629.19 Lindsay

pCdCO03 = pCd(OH)0

2 + pCO02 − 6.582

CdHCO+3 −676.72 Lindsay

pCdHCO+3 = pCd(OH)0

2 + pCO02 + pH − 14.909

CdNH2+3 −118.91 Lindsay

pCdNH2+3 = pCd(OH)0

2 + pNH03 + 2pH − 22.844

Cd(NH3)2+2 −156.86 Lindsay

pCd(NH3)2+2 = pCd(OH)0

2 + 2pNH03 + 2pH − 24.852



2 + 3pNH03 + 2pH − 26.194



2 + 4pNH03 + 2pH − 27.029

CdNO+3 −191.08 Lindsay

pCdNO+3 = pCd(OH)0

2 + pNH03 − 8(pH + pe) + pH + 89.169

Cd(NO3)02 −300.79 Lindsay

pCd(NO3)02 = pCd(OH)0

2 + 2pNH03 − 16(pH + pe) + 199.247

CdHPO04 −1192.4 Lindsay

pCdHPO04 = pCd(OH)0

2 + pH3PO04 − 14.147

CdP2O2−7 −2040.6 Lindsay

pCdP2O2−7 = pCd(OH)0

2 + 2pH3PO04 − 2pH − 2.880

P1: GIM



CdSO04 −836.38 Lindsay

pCdSO04 = pCd(OH)0

2 + pH2S0 − 8(pH + pe) + 17.911

CdCl+ −220.41 LindsaypCdCl+ = pCd(OH)0

2 + pHCl0 + pH − 25.270

CdCl02 −355.22 LindsaypCdCl02 = pCd(OH)0

2 + 2pHCl0 − 28.891

CdCl−3 −485.34 LindsaypCdCl−3 = pCd(OH)0

2 + 3pHCl0 − pH − 31.691

CdCl2−4 −617.14 Lindsay

pCdCl2−4 = pCd(OH)0

2 + 4pHCl0 − 2pH − 34.785

MERCURYHg2+ +164.703 Bard,

LindsaypHg2+ = pHg0 − 2pe + 22.338

Hg2+2 +153.607 Bard,

LindsaypHg2+

2 = 2pHg0 − 2pe + 13.877

HgOH+ −52.01 BardpHgOH+ = pHg0 − 2(pH + pe) + pH + 25.923

HHgO−2 −190.0 Bard

pHHgO−2 = pHg0 − 2(pH + pe) − pH + 43.301

Hg(OH)−3 −426.43 LindsaypHg(OH)−3 = pHHgO−

2

Hg(OH)02 −274.5 Bard

pHg(OH)02 = pHg0 − 2(pH + pe) + 65.616

HgCl+ −5.0 BardpHgCl+ = pHg0 + pHCl0 − 2(pH + pe) + pH + 12.604

HgCl02 −172.8 BardpHgCl02 = pHg0 + 2pHCl0 − 2(pH + pe) + 3.203

HgCl−3 −308.8 BardpHgCl−3 = pHg0 + 3pHCl0 − 2(pH + pe) − pH − 0.627

HgCl2−4 −446.4 Bard

pHgCl2−4 = pHg0 + 4pHCl0 − 2(pH + pe) − 2pH − 4.737

HgClOH0 −222.17 LindsaypHgClOH0 = pHg0 + pHCl0 − 2(pH + pe) + 16.109

HgI+ +40.2 BardpHgI+ = pHg0 + pHI0 − 2(pH + pe) + pH − 0.025

HgI02 −74.9 Bard

P1: GIM



pHgI02 = pHg0 + 2pHI0 − 2(pH + pe) − 20.740

HgI−3 −148.1 BardpHgI−3 = pHg0 + 3pHI0 − 2(pH + pe) − pH − 34.114

HgI2−4 −211.3 Bard

pHgI2−4 = pHg0 + 4pHI0 − 2(pH + pe) − 2pH − 45.737

HgIOH0 −173.22 LindsaypHgIOH0 = pHg0 + pHI0 − 2(pH + pe) + 4.138

Hg2P2O2−7 −1820 Bard

pHg2P2O2−7 = 2pHg0 + 2pH3PO0

4 − 2(pH + pe) − 2pH + 29.395

Hg2OH(P2O7)3− −2012 BardpHg2OH(P2O7)3− = 2pHg0 + 2pH3PO0

4 − 2(pH + pe) − 3pH + 37.310

Hg2(OH)2P2O4−7 −2197 Bard

pHg2(OH)2P2O4−7 = 2pHg0 + 2pH3PO0

4 − 2(pH + pe) − 4pH + 46.451

Hg2(P2O7)6−2 −3694 Bard

pHg2(P2O7)6−2 = 2pHg0 + 4pH3PO0

4 − 2(pH + pe) − 6pH + 62.293

Hg(SH)02 −26.53 Lindsay

pHg(SH)02 = pHg0 + 2pH2S0 − 2(pH + pe) − 1.399

HgS2−2 +45.27 Lindsay

pHgS2−2 = pHg0 + 2pH2S0 − 2(pH + pe) − 2pH + 11.179

HgSO04 −587.9 Bard

pHgSO04 = pHg0 + pH2S0 − 10(pH + pe) + 61.579

ALUMINUM

Al3+ −458 BardpAl3+ = pAl(OH)03 + 3pH − 17.858

AlOH2+ −694.1 BardpAlOH2+ = pAl(OH)03 + 2pH − 12.939

AlO+ −654.2 BardpAlO+ = pAl(OH)03 + pH − 5.949

Al(OH)+2 −699.44 LindsaypAl(OH)+2 = pAl(OH)03 + pH − 9.597

AlO−2 −823.4 Bard

pAlO−2 = pAl(OH)03 − pH + 5.961

Al(OH)−4 −1297.8 BardpAl(OH)−4 = pAl(OH)03 − pH + 5.953

Al(OH)2−5 −1481.39 Lindsay

pAl(OH)2−5 = pAl(OH)03 − 2pH + 15.341

Al2(OH)4+2 −1412.27 Lindsay

pAl2(OH)4+2 = 2pAl(OH)0

3 + 4pH − 30.095

P1: GIM



AlSO+4 −1253.69 Lindsay

PAlSO+4 = pAl(OH)0

3 + pH2S0 − 8(pH + pe) + pH + 18.570

Al(SO4)−2 −1990.83 Lindsay

pAl(SO4)−2 = pAl(OH)03 + 2pH2S0 − 16(pH + pe) − pH + 60.526

Al2(SO4)03 −3204.62 Lindsay

pAl2(SO4)03 = 2pAl(OH)0

3 + 3pH2S0 − 24(pH + pe) − 86.082

CARBON

H2CO03 −623.42 Bard

pH2CO03 = pCO0

2 − 0.007

HCO−3 −587.06 Bard

pHCO−3 = pCO0

2 − pH + 6.363

CO2−3 −527.90 Bard

pCO2−3 = pCO0

2 − 2pH + 16.727

SILICON

H3SiO−4 −1223.4 Bard

pH3SiO−4 = pH4SiO0

4 − pH + 9.215

H2SiO2−4 −1152.7 Bard

pH2SiO2−4 = pH4SiO0

4 − 2pH + 21.602

HSiO3−4 −1120.7 Lindsay

pHSiO3−4 = pH4SiO0

4 − 3pH + 32.838

SiO4−4 −1046.0 Lindsay

pSiO4−4 = pH4SiO0

4 − 4pH + 45.938

HSiO−3 −955.46 Bard

pHSiO−3 = pH4SiO0

4 − pH + 20.241

SiO2−3 −887 Bard

pSiO2−3 = pH4SiO0

4 − 2pH + 32.234

Si2O3(OH)2−4 −2211.2 Bard

pSi2O3(OH)2−4 = 2pH4SiO0

4 − 2pH + 18.154

Si4O6(OH)2−6 −4079.8 Bard


4 − 2pH + 13.225

Si4O8(OH)4−4 −3969.8 Bard


4 − 4pH + 32.497

LEAD

Pb2+ −24.69 LindsaypPb2+ = pPb(OH)0

2 + 2pH − 17.749

P1: GIM



PbOH+ −217.94 LindsaypPbOH+ = pPb(OH)0

2 + pH − 10.005

Pb(OH)−3 −575.89 LindsaypPb(OH)−3 = pPb(OH)0

2 − pH + 10.341

Pb(OH)2−4 −748.02 Lindsay

pPb(OH)2−4 = pPb(OH)0

2 − 2pH + 21.737

Pb2OH3+ −250.04 LindsaypPb2OH3+ = 2pPb(OH)0

2 + 3pH − 29.361

Pb3(OH)2+4 −886.42 Lindsay

pPb3(OH)2+4 = 3pPb(OH)0

2 + 2pH − 29.361



2 + 4pH − 50.108



2 + 4pH − 62.921

PbHPO04 −1138.72 Lindsay

pPbHPO04 = pPb(OH)0

2 + pH3PO04 − 11.504

PbH2PO+4 −1170.68 Lindsay

pPbH2PO+4 = pPb(OH)0

2 + pH3PO04 + pH − 17.104

PbP2O2−7 −2002.25 Lindsay

pPbP2O2−7 = pPb(OH)0

2 + 2pH3PO04 − 2pH − 2.925

PbPO−4 Smith

pPbPO−4 = pPb(OH)0

2 + pH3PO04 − pH − 5.152

Pb(PO4)4−2 Smith

pPb(PO4)4−2 = pPb(OH)02 + 2pH3PO0

4 − 4pH + 11.145

Pb(P2O7)6−2 Smith

pPb(P2O7)6−2 = pPb(OH)0

2 + 4pH3PO04 − 6pH + 26.119

Pb(HPO4)2−2 Smith

pPb(HPO4)2−2 = pPb(OH)0

2 + 2pH3PO04 − 2pH − 1.555

PbCl+ −165.06 LindsaypPbCl+ = pPb(OH)0

2 + pHCl0 + pH − 22.345

PbCl02 −297.36 LindsaypPbCl02 = pPb(OH)0

2 + 2pHCl0 − 25.527

PbCl−3 −428.07 LindsaypPbCl−3 = pPb(OH)0

2 + 3pHCl0 − pH − 28.429

PbCl2−4 −557.60 Lindsay

pPbCl2−4 = pPb(OH)0

2 + 4pHCl0 − 2pH − 31.127

PbCO03 −626.34 Weast

pPbCO03 = pPb(OH)02 + pCO0

2 − 8.124

P1: GIM



Pb(CO3)2−2 Smith

pPb(CO3)2−2 = pPb(OH)0

2 + 2pCO02 − 2pH + 5.357

PbSO04 −784.17 Lindsay

pPbSO04 = pPb(OH)0

2 + pH2S0 − 8(pH + pe) + 20.286

Pb(SO4)2−2 −1533.56 Lindsay

pPb(SO4)2−2 = pPb(OH)02 + 2pH2S0 − 16(pH + pe) − 2pH + 60.088

PbO2−3 −272.7 Bard

pPbO2−3 = pPb(OH)0

2 − 2(pH + pe) − 2pH + 63.458

PbO4−4 −282.1 Bard

pPbO4−4 = pPb(OH)0

2 − 2(pH + pe) − 4pH + 103.363

NITROGEN

NH+4 −79.45 Lindsay

pNH+4 = pNH0

3 + pH − 9.280

NH4OH0 −263.8 Lindsay,Bard

pNH4OH0 = pNH03 − 0.032

N2H04 +127.9 Lindsay

pN2H04 = 2pNH0

3 − 2(pH + pe) + 31.689

N2H+5 +82.4 Lindsay,

BardpN2H

+5 = 2pNH0

3 − 2(pH + pe) + pH + 23.721

N2H2+6 +94.14 Lindsay

pN2H2+6 = 2pNH0

3 − 2(pH + pe) + 2pH + 25.773

NH2OH0 −23.43 LindsaypNH2OH0 = pNH0

3 − 2(pH + pe) + 42.088

NH2OH+2 −56.65 Bard

pNH2OH2+ = pNH0

3 − 2(pH + pe) + pH + 36.268

N−3 +348.3 Lindsay

pN−3 = 3pNH0

3 − 8(pH + pe) − pH + 74.944

N2O0 +101.0 LindsaypN2O0 = pNH0

3 − 8(pH + pe) + 68.535

H2N2O02 +36 Bard

pH2N2O02 = 2pNH0

3 − 8(pH + pe) + 98.692

HN2O−2 Weast

pHN2O−2 = 2pNH0

3 − 8(pH + pe) − pH + 107.277

N2O2−2 +139 Bard

pN2O2−2 = 2pNH0

3 − 8(pH + pe) − 2pH + 116.692

NH2O−2 +76.1 Bard

pNH2O−2 = pNH0

3 − 4(pH + pe) − pH + 101.078

P1: GIM



NO0 +102.1 LindsaypNO0 = pNH0

3 − 5(pH + pe) + 64.086

HNO02 −55.7 Lindsay

pHNO02 = pNH0

3 − 6(pH + pe) + 77.981

NO−2 −37.7 Lindsay

pNO−2 = pNH0

3 − 6(pH + pe) − pH + 81.133

NO−3 −111.46 Lindsay

pNO−3 = pNH0

3 − 8(pH + pe) − pH + 109.770

PHOSPHORUS

H2PO−4 −1137.4 Lindsay

pH2PO−4 = pH3PO0

4 − pH + 2.148

HPO2−4 −1096.3 Lindsay

pHPO2−4 = pH3PO0

4 − 2pH + 9.346

PO3−4 −1025.8 Lindsay

pPO3−4 = pH3PO0

4 − 3pH + 21.697

H4P2O07 −2022.6 Lindsay

pH4P2O07 = 2pH3PO0

4 + 6.929

H3P2O−7 −2018.1 Lindsay

pH3P2O−7 = 2pH3PO0

4 − pH + 7.728

H2P2O2−7 −2005.1 Lindsay

pH2P2O2−7 = 2pH3PO0

4 − 2pH + 10.008

HP2O3−7 −1966.8 Lindsay

pHP2O3−7 = 2pH3PO0

4 − 3pH + 16.707

P2O4−7 −1913.1 Lindsay

pP2O4−7 = 2pH3PO0

4 − 4pH + 26.119

H3PO03 −846.8 Lindsay

pH3PO03 = pH3PO0

4 + 2(pH + pe) + 11.518

H2PO−3 −838.2 Lindsay

pH2PO−3 = pH3PO0

4 + 2(pH + pe) − pH + 13.021

HPO2−3 −799.4 Lindsay

pHPO2−3 = pH3PO0

4 + 2(pH + pe) − 2pH + 19.809

PH03 +9.0 Lindsay

pPH03 = pH3PO0

4 + 8(pH + pe) + 36.785

ARSENIC

H2AsO−4 −753.3 Welham

pH2AsO−4 = pH3AsO0

4 − pH + 2.243

P1: GIM



HAsO2−4 −714.7 Welham

pHAsO2−4 = pH3AsO0

4 − 2pH + 9.005

AsO3−4 −648.5 Welham

pAsO3−4 = pH3AsO0

4 − 3pH + 20.603

HAsO02 −402.7 Welham

pHAsO02 = pH3AsO0

4 + 2(pH + pe) − 19.439

AsO−2 −350.0 Welham

pAsO−2 = pH3AsO0

4 + 2(pH + pe) − pH − 10.206

H3AsO03 −639.9 Welham

pH3AsO03 = pH3AsO0

4 + 2(pH + pe) − 19.443

H2AsO−3 −587.2 Welham

pH2AsO−3 = pH3AsO0

4 + 2(pH + pe) − pH − 10.210

HAsO2−3 −524.3 Welham

pHAsO2−3 = pH3AsO0

4 + 2(pH + pe) − 2pH + 0.810

AsO+ −163.8 WelhampAsO+ = pH3AsO0

4 + 2(pH + pe) + pH − 19.137AsS+ −70.3 WelhampAsS+ = pH3AsO0

4 + pH2S0 + 2(pH + pe) + pH − 19.0

HAsS02 −48.58 Welham

pHAsS02 = pH3AsO0

4 + pH2S0 + 2(pH + pe) − 30.724

AsS−2 −27.4 Welham

pAsS−2 = pH3AsO0

4 + pH2S0 + 2(pH + pe) − pH − 27.014

As3S3−6 −252.38 Itagaki

pAs3S3−6 = 3pH3AsO0

4 + 6pH2S0 + 6(pH + pe) − 3pH − 110.861

SULFURHS− +12.05 BardpHS− = pH2S0 − pH + 6.994

S2− +86.31 BardpS2− = pH2S0 − 2pH + 20.004

S2−2 +79.5 Bard

pS2−2 = 2pH2S0 − 2(pH + pe) − 2pH + 23.693

S2−3 +73.6 Bard

pS2−3 = 3pH2S0 − 4(pH + pe) − 2pH + 27.543

S2−4 +69.0 Bard

pS2−4 = 4pH2S0 − 6(pH + pe) − 2pH + 31.619

S2−5 +65.7 Bard

pS2−5 = pH2S0 − 8(pH + pe) − 2pH + 35.924

H2S2O03 −529.1 Lindsay

pH2S2O03 = 2pH2S0 − 8(pH + pe) + 41.749

P1: GIM



HS2O−3 −525.6 Lindsay

pHS2O−3 = 2pH2S0 − 8(pH + pe) − pH + 42.350

S2O2−3 −518.8 Lindsay

pS2O2−3 = 2pH2S0 − 8(pH + pe) − 2pH + 43.532

S5O2−6 −956.0 Bard

pS5O2−6 = 5pH2S0 − 20(pH + pe) − 2pH + 105.908

S4O2−6 −1022.2 Bard

pS4O2−6 = 4pH2S0 − 18(pH + pe) − 2pH + 86.577

H2S2O04 −616.7 Bard

pH2S2O04 = 2pH2S0 − 10(pH + pe) + 67.944

HS2O−4 −614.6 Bard,

LindsaypHS2O

−4 = 2pH2S0 − 10(pH + pe) − pH + 68.310

S2O2−4 −600.4 Bard,

LindsaypS2O

2−4 = 2pH2S0 − 10(pH + pe) − 2pH + 70.803

S3O2−6 −958 Bard

pS3O2−6 = 3pH2S0 − 16(pH + pe) − 2pH + 96.125

SO02 −300.708 Bard

pSO02 = pH2S0 − 6(pH + pe) + 35.233

H2SO03 −537.90 Bard

pH2SO03 = pH2S0 − 6(pH + pe) + 35.304

HSO−3 −527.81 Bard

pHSO−3 = pH2S0 − 6(pH + pe) − pH + 37.092

SO32− −486.6 Bard

pSO32− = pH2S0 − 6(pH + pe) − 2pH + 44.298

S2O2−5 −791 Bard

pS2O2−5 = 2pH2S0 − 12(pH + pe) − 2pH + 79.003

S2O2−6 −966 Bard

pS2O2−6 = 2pH2S0 − 14(pH + pe) − 2pH + 89.768

S2O2−8 −1110.4 Bard

pS2O2−8 = 2pH2S0 − 18(pH + pe) − 2pH + 146.849

HSO−4 −756.01 Bard

pHSO−4 = pH2S0 − 8(pH + pe) − pH + 38.695

SO2−4 −744.63 Bard

pSO2−4 = pH2S0 − 8(pH + pe) − 2pH + 40.674

CHLORINE

Cl− −131.26 LindsaypCl− = pHCl0 − pH − 3.00

P1: GIM



IODINE

I02 +16.43 BardpI02 = 2pHI0 − 2(pH + pe) + 1.778

I− −51.67 BardpI− = pHI0 − pH − 9.602

I−3 −51.50 BardpI−3 = 3pHI0 − 2(pH + pe) − pH − 10.673

HIO0 −98.67 BardpHIO0 = pHI0 − 2(pH + pe) + 23.716

IO− −37.96 BardpIO− = pHI0 − 2(pH + pe) − pH + 34.352

I+·H2O −89.99 BardpI+·H2O = pHI0 − 2(pH + pe) + pH + 25.237

ICl0 −14.85 BardpICl0 = pHI0 − 2(pH + pe) + 16.845

pICl−2 −158.70 BardpICl−2 = pHI0 − 2(pH + pe) − pH + 11.640

HIO03 −139.94 Bard

pHIO03 = pHI0 − 6(pH + pe) + 99.696

IO−3 −134.94 Bard

pIO−3 = pHI0 − 6(pH + pe) − pH + 100.467

IO−4 −53.14 Bard

pIO−4 = pHI0 − 8(pH + pe) − pH + 156.350

H5IO06 −537.14 Bard

pH5IO06 = pHI0 − 8(pH + pe) + 154.661

H4IO−6 −518.35 Bard

pH4IO−6 = pHI0 − 8(pH + pe) − pH + 157.953

H3IO2−6 −480.11 Bard

pH3IO2−6 = pHI0 − 8(pH + pe) − 2pH + 164.652

H2IO3−6 −410.47 Bard

pH2IO3−6 = pHI0 − 8(pH + pe) − 3pH + 176.853

toxic metals in the environment: thermodynamic ... · kirk g. scheckel, christopher a....

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