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CEE 680 Lecture #29 3/11/2020
1
Lecture #29
Complexation: Speciation in Fresh Waters
(Stumm & Morgan, Chapt.6: pg.289‐305)
Benjamin; Chapter 8.1‐8.6
David Reckhow CEE 680 #29 1
Updated: 11 March 2020 Print version
Cadmium
David Reckhow CEE 680 #29 2
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
2
David Reckhow CEE 680 #29 3
Flux in 107
M/yr
David Reckhow CEE 680 #29 4
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
3
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
David Reckhow CEE 680 #29 5
-value21
166204
71.5
Cadmium Chloride
David Reckhow CEE 680 #29 6
From: Butler, 1964; pg.268Reproduced in Langmuir, 1997; pg.96
Similar to: Butler, 1998; pg.241
CEE 680 Lecture #29 3/11/2020
4
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]?
David Reckhow CEE 680 #29 7
12210 ][][][1
][ n
nM
LLLC
M
nn
M
nn L
C
ML][
][0
Cadmium Chloride
David Reckhow CEE 680 #29 8
From: Butler, 1964; pg.268Reproduced in Langmuir, 1997; pg.96
Similar to: Butler, 1998; pg.241
CEE 680 Lecture #29 3/11/2020
5
Cadmium Chloride (cont.)
David Reckhow CEE 680 #29 9
Butler, 1964; pg.269
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Pankow, 1991; pg.375
Cadmium Chloride (cont.)
CEE 680 Lecture #29 3/11/2020
6
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
David Reckhow CEE 680 #29 11
Determination of Ligand #
Equilibrium constant approach
And substituting in for the apha’s
David Reckhow CEE 680 #29 12
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
7
Determination of ligand #Mass balance approach
CM = [M]+[ML]+[ML2]+[ML3]+ +[MLn]
CL = [L]+[ML]+2[ML2]+3[ML3]+ +n[MLn]
David Reckhow CEE 680 #29 13
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
David Reckhow CEE 680 #29 14
CEE 680 Lecture #29 3/11/2020
8
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
David Reckhow CEE 680 #29 16
𝑛 𝐶 𝐿𝐶
CEE 680 Lecture #29 3/11/2020
9
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
10
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
11
Speciation in Natural Waters Mostly Cl complexes in sea water
David Reckhow CEE 680 #29 21
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
David Reckhow CEE 680 #29 22
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
CEE 680 Lecture #29 3/11/2020
12
K1=106,
K2=102
K1=104.5,
K2=103.5
David Reckhow CEE 680 #29 23
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
David Reckhow CEE 680 #29 24
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
13
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:
David Reckhow CEE 680 #29 25
]][[
][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:
David Reckhow CEE 680 #29 26
]][[
][
)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
CEE 680 Lecture #29 3/11/2020
14
Height of alpha maximum Dependent on difference between K1 and K2
David Reckhow CEE 680 #29 27
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
David Reckhow CEE 680 #29 28