David Reckhow CEE 680 #32 1 - UMass AmherstCEE 680 Lecture #32 3/25/2020 5 Cu‐NTA III Specific...

9
CEE 680 Lecture #32 3/25/2020 1 Lecture #32 Coordination Chemistry: Case Studies: NTA (cont.) (Stumm & Morgan, Chapt.6: pg.317319) Benjamin; Chapter 8.18.6 David Reckhow CEE 680 #32 1 Updated: 25 March 2020 Print version Biotic ligand model of the acute toxicity of metals. 2. Application to acute copper toxicity in freshwater fish and Daphnia Environmental Toxicology and Chemistry, Volume: 20, Issue: 10, Pages: 2397-2402

Transcript of David Reckhow CEE 680 #32 1 - UMass AmherstCEE 680 Lecture #32 3/25/2020 5 Cu‐NTA III Specific...

Page 1: David Reckhow CEE 680 #32 1 - UMass AmherstCEE 680 Lecture #32 3/25/2020 5 Cu‐NTA III Specific problem CuT = 10‐4 M6.35 mg/L NTAT = 10‐4 M19.1 mg/L Notes: this is a much higher

CEE 680 Lecture #32 3/25/2020

1

Lecture #32Coordination Chemistry: Case Studies: NTA 

(cont.)

(Stumm & Morgan, Chapt.6: pg.317‐319)

Benjamin; Chapter 8.1‐8.6

David Reckhow CEE 680 #32 1

Updated: 25 March 2020 Print version

Biotic ligand model of the acute toxicity of metals. 2. Application to acute copper toxicity in freshwater fish and Daphnia

Environmental Toxicology and Chemistry, Volume: 20, Issue: 10, Pages: 2397-2402

Page 2: David Reckhow CEE 680 #32 1 - UMass AmherstCEE 680 Lecture #32 3/25/2020 5 Cu‐NTA III Specific problem CuT = 10‐4 M6.35 mg/L NTAT = 10‐4 M19.1 mg/L Notes: this is a much higher

CEE 680 Lecture #32 3/25/2020

2

Algae and Copper

McKnight et al., 1983; Environmental Management 7(4)311-320

%𝜇

𝜇

Fresh and salt water algae

Depends on Cu+2 ion: 10‐7M seems to work for most

Add CuSO4

Smith et al., 2015, Applied Geochemistry 57:55

Page 3: David Reckhow CEE 680 #32 1 - UMass AmherstCEE 680 Lecture #32 3/25/2020 5 Cu‐NTA III Specific problem CuT = 10‐4 M6.35 mg/L NTAT = 10‐4 M19.1 mg/L Notes: this is a much higher

CEE 680 Lecture #32 3/25/2020

3

Modeling the Fate of Metal Concentrates in Surface Water

Environmental Toxicology and Chemistry, Volume: 38, Issue: 6, Pages: 1256-1272, First published: 23 March 2019, DOI: (10.1002/etc.4417)

Copper – NTA problem

NTA: nitrilotriacetate Used as a substitute “builder” in place of phosphate

Good example of moderately strong ligand

Research interests: 70’s & 80’s General Review

Perry et al., 1984 [Wat. Res., 18(3)255] Other Aspects

Photochemistry: e.g., Langford et al., 1973 [ES&T 7(9)820] Biodegradation: e.g., Kuhn et al., 1987 [Wat. Res. 21(10)1237], Vanbriesen

et al., 2000 [ES&T 34(16)3346]

Bioavailability of bound metals: e.g.,  Bressan & Brunetti, 1988 [Wat. Res. 22(5)553]

David Reckhow CEE 680 #32 6

N

CH2COOH

CH2COOH

CH2COOH

See: Knud-Hansen Paper

Page 4: David Reckhow CEE 680 #32 1 - UMass AmherstCEE 680 Lecture #32 3/25/2020 5 Cu‐NTA III Specific problem CuT = 10‐4 M6.35 mg/L NTAT = 10‐4 M19.1 mg/L Notes: this is a much higher

CEE 680 Lecture #32 3/25/2020

4

Cu‐NTA II

Thermodynamics  (20ºC)

Acid/Base

H3NTA = H+ + H2NTA‐ pK1 =   1.6

H2NTA‐ = H+ + HNTA‐2 pK2 =   3.0

HNTA‐2 = H+ + NTA‐3 pK3 = 10.3

Cu complex

Cu+2 + NTA‐3 = CuNTA‐ p1 = ‐13.0

Others are rather weak

CuHNTA

David Reckhow CEE 680 #32 7

David Reckhow CEE 680 #32 8

From: Snoeyink & Jenkins, 1980

Page 5: David Reckhow CEE 680 #32 1 - UMass AmherstCEE 680 Lecture #32 3/25/2020 5 Cu‐NTA III Specific problem CuT = 10‐4 M6.35 mg/L NTAT = 10‐4 M19.1 mg/L Notes: this is a much higher

CEE 680 Lecture #32 3/25/2020

5

Cu‐NTA III Specific problem

CuT = 10‐4 M 6.35 mg/L

NTAT = 10‐4 M 19.1 mg/L

Notes: this is a much higher concentration of NTA than is generally found, but it can be used to represent background natural organic matter

Copper concentrations may sometimes be this high when used as an algicide

We are ignoring other complexes such as copper hydroxides or carbonates

David Reckhow CEE 680 #32 9

Cu‐NTA IV Mass Balance Equations

CuT = [Cu+2] + [ CuNTA‐]

NTAT = [CuNTA‐] + [H3NTA] + [H2NTA‐] + [HNTA‐2] + [NTA‐3]

Definition: total free concentration (TF) is that which is unbound to any metal except H+

NTAT = [CuNTA‐] +NTATF

David Reckhow CEE 680 #32 10

Page 6: David Reckhow CEE 680 #32 1 - UMass AmherstCEE 680 Lecture #32 3/25/2020 5 Cu‐NTA III Specific problem CuT = 10‐4 M6.35 mg/L NTAT = 10‐4 M19.1 mg/L Notes: this is a much higher

CEE 680 Lecture #32 3/25/2020

6

Cu‐NTA V Equilibria

Acid/base

Complexation

David Reckhow CEE 680 #32 11

1

321

3

32

2

3

3

3

][][][1

][

KKK

H

KK

H

K

H

NTA

NTA

TF

]][[

][321

NTACu

CuNTA

Cu‐NTA VI Substitute mass balance  and alpha equations into the beta equation

David Reckhow CEE 680 #32 12

])[(][

][

][][

][

][

][

]][[

][

23

2

2

32

2

32

2

321

CuCuNTACu

CuCu

CuNTANTACu

CuCu

NTACu

CuCu

NTACu

CuNTA

TT

T

T

T

TF

T

Page 7: David Reckhow CEE 680 #32 1 - UMass AmherstCEE 680 Lecture #32 3/25/2020 5 Cu‐NTA III Specific problem CuT = 10‐4 M6.35 mg/L NTAT = 10‐4 M19.1 mg/L Notes: this is a much higher

CEE 680 Lecture #32 3/25/2020

7

Cu‐NTA VII

Now solve, noting that CuT = NTAT

Which gives us a quadratic which can be solved for a given pH

David Reckhow CEE 680 #32 13

][][

][

])[(][

][

23

2

2

23

2

2

1

CuCu

CuCu

CuCuNTACu

CuCu

T

TT

T

0][][ 22213

TCuCuCu

Cu‐NTA VIIIThen determine other species from the free copper

Can use a spreadsheet to calculate 3 versus pH, and then calculate the other species

David Reckhow CEE 680 #32 14

][][ 2 CuCuCuNTA T

][ CuNTANTANTA TTF

TFNTANTA 33][

Page 8: David Reckhow CEE 680 #32 1 - UMass AmherstCEE 680 Lecture #32 3/25/2020 5 Cu‐NTA III Specific problem CuT = 10‐4 M6.35 mg/L NTAT = 10‐4 M19.1 mg/L Notes: this is a much higher

CEE 680 Lecture #32 3/25/2020

8

Cu‐NTA IX

Figure shows impact of ligand speciation on extent of complexation

Same thing happens with fulvic acid

David Reckhow CEE 680 #32 15pH

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Con

cent

ratio

n (m

oles

/L)

1e-19

1e-18

1e-17

1e-16

1e-15

1e-14

1e-13

1e-12

1e-11

1e-10

1e-9

1e-8

1e-7

1e-6

1e-5

1e-4

1e-3

Cu+2

NTA-3

CuNTA-

CuNTA X

David Reckhow CEE 680 #32 16pH

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Log

Con

cent

ratio

n (m

oles

/L)

-19

-18

-17

-16

-15

-14

-13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

Cu+2

NTA-3

CuNTA-

Page 9: David Reckhow CEE 680 #32 1 - UMass AmherstCEE 680 Lecture #32 3/25/2020 5 Cu‐NTA III Specific problem CuT = 10‐4 M6.35 mg/L NTAT = 10‐4 M19.1 mg/L Notes: this is a much higher

CEE 680 Lecture #32 3/25/2020

9

CuNTA XI

David Reckhow CEE 680 #32 17pH

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Log

Con

cent

ratio

n (m

oles

/L)

-19

-18

-17

-16

-15

-14

-13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3Cu+2

NTA-3

CuNTA-

CuOH+

Cu(OH)2(aq)

Cu(OH)3-

Cu(OH)4-2

CuOHNTA-2

Cu(NTA)2-4

Cu2(OH)2+2

To next lecture

David Reckhow CEE 680 #32 18