CEE 680 Lecture #29 3/11/2020

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Lecture #29

Complexation: Speciation in Fresh Waters

(Stumm & Morgan, Chapt.6: pg.289‐305)

Benjamin; Chapter 8.1‐8.6

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Updated: 11 March 2020 Print version

Cadmium

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Cadmium is of no use to the human body and is toxic even at low levels. The negative effects of cadmium on the body are numerous and can impact nearly all systems in the body, including cardiovascular, reproductive, the kidneys, eyes, and even the brain.• Cadmium affects blood pressure.• Cadmium affects prostate function and testosterone levels.• Cadmium induces bone damage (Itai-ltai).• Exposure to cadmium can affect renal and dopaminergic systems in children.

Batteries & electroplating

Very high ratio of abundance to toxicity Like Pb, Hg, As

EPA Standards 0.005 g/L in drinking water

Concern over kidney damage

Click here for more on Cd

CEE 680 Lecture #29 3/11/2020

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Flux in 107

M/yr

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From: Morel & Malcolm, “The Biogeochemistry of Cadmium”, Chapt 8 in Metal Ions in Biological Systems, Vol 43 Sigel, Sigel & Sigel, eds.

CEE 680 Lecture #29 3/11/2020

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CdCl Example Consider the speciation of cadmium and chloride

First the four beta’s

Cd+2 + Cl‐ = CdCl+

Cd+2 + 2Cl‐ = CdCl2 Cd+2 + 3Cl‐ = CdCl3

‐

Cd+2 + 4Cl‐ = CdCl4‐2

Now plot the alpha curves

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-value21

166204

71.5

Cadmium Chloride

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From: Butler, 1964; pg.268Reproduced in Langmuir, 1997; pg.96

Similar to: Butler, 1998; pg.241

CEE 680 Lecture #29 3/11/2020

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CdCl Example Calculate the concentration of all species for the following solution 0.01 M Cd(NO3)2 + 1 M HCl

Use the equilibrium equations

But what do you use for [L]?

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12210 ][][][1

][ n

nM

LLLC

M

nn

M

nn L

C

ML][

][0

Cadmium Chloride

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From: Butler, 1964; pg.268Reproduced in Langmuir, 1997; pg.96

Similar to: Butler, 1998; pg.241

CEE 680 Lecture #29 3/11/2020

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Cadmium Chloride (cont.)

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Butler, 1964; pg.269

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Pankow, 1991; pg.375

Cadmium Chloride (cont.)

CEE 680 Lecture #29 3/11/2020

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Ligand Number Definition

The average number of bound ligands per atom of metal

Significance Defines Mixture Can be analytically determined

Used to evaluate ’s Can be calculated via 2 independent ways and used to solve problems, if you know the free ligand concentration Mass balance equations Equilibrium equations

Thus, we have 2 independent equations and two unknowns (free L, and n‐bar), so we can solve

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Determination of Ligand #

Equilibrium constant approach

And substituting in for the apha’s

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n

M

n

MMM

M

n

n

C

MLn

C

ML

C

ML

C

ML

C

MLnMLMLMLn

321

32

32

32

][][3

][2

][

][][3][2][

nn

nn

LnLLL

LnLLLn

][][3][2][

][][3][2][3

32

210

03

302

2010

CEE 680 Lecture #29 3/11/2020

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Determination of ligand #Mass balance approach

CM = [M]+[ML]+[ML2]+[ML3]+    +[MLn]

CL = [L]+[ML]+2[ML2]+3[ML3]+   +n[MLn]

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M

L

M

n

C

LC

C

MLnMLMLMLn

][

][][3][2][ 32

Take alpha diagram For CdClx Next add n‐bar curves

One based on equilibria

Other based on mass balances

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CEE 680 Lecture #29 3/11/2020

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David Reckhow CEE 680 #29 15Log [L]

-4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0

Alp

ha

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

n-b

ar

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

n-bar (equ)

Cadmium Chloride alpha diagram

In‐Class Problems Class problems with CdClx

Sea Water: CL=5 x 10‐1, CM=1 x 10

‐9

Contaminated Sea Water: CL=5 x 10‐1, CM=1 x 10

‐1

Desal Brine: CL=15 x 10‐1, CM=3 x 10

‐9

RO conc. of Cont. Sea Water: CL=15 x 10‐1, CM=3 x 10

‐1

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𝑛 𝐶 𝐿𝐶

CEE 680 Lecture #29 3/11/2020

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David Reckhow CEE 680 #29 17Log [L]

-4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0

Alp

ha

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

n-b

ar

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

n-bar (equ)

Cadmium Chloride problem 

𝑛 𝐶 𝐿𝐶

Sea Water: CL=5 x 10-1, CM=1 x 10-9

David Reckhow CEE 680 #29 18Log [L]

-4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0

Alp

ha

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

n-b

ar

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

n-bar (equ)

Cadmium Chloride problem 2

𝑛 𝐶 𝐿𝐶

Contaminated Sea Water: CL=5 x 10-1, CM=1 x 10-1

CEE 680 Lecture #29 3/11/2020

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David Reckhow CEE 680 #29 19Log [L]

-4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0

Alp

ha

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

n-b

ar

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

n-bar (equ)

Cadmium Chloride problem 3

𝑛 𝐶 𝐿𝐶

Desal Brine: CL=15 x 10-1, CM=3 x 10-9

David Reckhow CEE 680 #29 20Log [L]

-4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0

Alp

ha

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

n-b

ar

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

n-bar (equ)

Cadmium Chloride problem 4

𝑛 𝐶 𝐿𝐶

RO conc. of Cont. Sea Water: CL=15 x 10-1, CM=3 x 10-1

CEE 680 Lecture #29 3/11/2020

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Speciation in Natural Waters Mostly Cl complexes in sea water

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From: Morel & Malcolm, “The Biogeochemistry of Cadmium”, Chapt 8 in Metal Ions in Biological Systems, Vol 43 Sigel, Sigel & Sigel, eds.

Cellular metabolism of Cd gg

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From: Morel & Malcolm, “The Biogeochemistry of Cadmium”, Chapt 8 in Metal Ions in Biological Systems, Vol 43 Sigel, Sigel & Sigel, eds.

PCs are phytochelatins; metal binding proteins that help regulate Cd concentrations

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K1=106, 

K2=102

K1=104.5, 

K2=103.5

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Log [L]

-8 -6 -4 -2 0

Alp

ha

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

n-b

ar

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

n-bar (equ)

Log [L]

-8 -6 -4 -2 0

Alp

ha

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

n-b

ar

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

n-bar (equ)

Note: as K’s get closer, so do the intersections, and the middle alpha’s get compressed with diminished height

K1=104, K2=10

4

K1=103.5, 

K2=104.5

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Log [L]

-8 -6 -4 -2 0

Alp

ha

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2n

-ba

r

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

n-bar (equ)

Log [L]

-8 -6 -4 -2 0

Alp

ha

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

n-b

ar

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

n-bar (equ)

CEE 680 Lecture #29 3/11/2020

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Features of alpha diagrams

Intersection point for adjacent curves

Occur at: log [L] = pK’s

E.g., first intersection occurs where: 0 =1 which is where: [M]=[ML]

And in general, intersection of adjacent curves will occur where: [ML(i‐1)]=[ML(i)]

So at these intersection points, the metal terms in the equations for K will cancel each other out and:

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]][[

][1 LM

MLK

]][[

][

)1( LML

MLK

i

ii

1

1

]log[

1][

pKL

KL

i

i

pKL

KL

]log[

1][or

Features (cont.) An alpha curve will reach its maximum at the point where the preceding alpha and the following alpha intersect (i.e., i‐1 and i+1 ), 

Consider the intersection of i‐1 and i+1

The two K equations are:

And their product gives us:

Once again, we can cancelmetal concentrations to get:

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]][[

][

)1( LML

MLK

i

ii

]][[

][

)(

11 LML

MLK

i

ii

2)1(

11 ]][[

][

LML

MLKK

i

iii

)(5.0]log[

1][

1

1

ii

ii

pKpKL

KKL

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Height of alpha maximum Dependent on difference between K1 and K2

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LogKx-LogKx+1

-1 0 1 2

-m

ax v

alue

0.0

0.2

0.4

0.6

0.8

1.0

To next lecture

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