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Print version Updated: 4 March 2020 Lecture #25 Dissolved Carbon Dioxide: Open & Closed Systems VI (Stumm & Morgan, Chapt.4 ) Benjamin; Chapter 7 David Reckhow CEE 680 #25 1 Lake Erie David Reckhow CEE 680 #25 2 Lake


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SLIDE 1

Lecture #25 Dissolved Carbon Dioxide: Open & Closed Systems VI

(Stumm & Morgan, Chapt.4 )

Benjamin; Chapter 7

David Reckhow CEE 680 #25 1

Updated: 4 March 2020

Print version

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SLIDE 2

 Lake Erie

David Reckhow CEE 680 #25 2

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SLIDE 3

 Lake Taihu

David Reckhow CEE 680 #25 3

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SLIDE 4

 algae

David Reckhow CEE 680 #25 4

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SLIDE 5

 Algal cells

David Reckhow CEE 680 #25 5

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SLIDE 6

Photosynthesis Problem

 Principles of conservation of Alk and CT

 From Stumm & Morgan (example 4.8, pg. 174)

 Approach

 General: assume that system is closed  Simplified: treat alkalinity as constant  Alternative: allow alkalinity to vary in accordance with

reaction stoichiometry

David Reckhow CEE 680 #25 6

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SLIDE 7

Problem Statement

 Photosynthesis with nitrate assimilation

 106 CO2 + 16 NO3

  • + HPO4
  • 2 + 122 H2O + 18 H+

= C106H263O110N16P + 138 O2

 Conditions

 Initial

 Alk = 0.85 meq/L  pH = 9.0

 Final (3 hours later)

 pH = 9.5

 What is the net rate of carbon fixation?

David Reckhow CEE 680 #25 7

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SLIDE 8

Simplified Solution: (const. alk.)

 Use standard alkalinity equation

 So initially:

David Reckhow CEE 680 #25 8

( )

] [ ] [ 2

2 1 + − −

+ + = H OH C Alk

T

α α M x x H OH Alk CT

4 9 5 4 2 1

10 017 . 8 ) 0477 . ( 2 9523 . 10 10 10 5 . 8 2 ] [ ] [

− − − − + −

= + + − = + + − = α α

0477 . 1 1 1 1

3 . 10 9 2 2 1 2

10 10 ] [ ] [ 2

≈ + ≈ + + =

− − + +

K H K K H

α 9523 . 1 1 1 1

9 3 . 10 2 1

10 10 ] [ ] [ 1

≈ + ≈ + + =

− − + +

H K K H

α

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SLIDE 9

Simplified Solution: (const. alk.)

 And 3 hours later

 So the rate is:

David Reckhow CEE 680 #25 9

M x x H OH Alk CT

4 5 . 9 5 . 4 4 2 1

10 199 . 7 ) 1368 . ( 2 8632 . 10 10 10 5 . 8 2 ] [ ] [

− − − − + −

= + − − = + + − = α α

1368 . 1 1 1 1

3 . 10 5 . 9 2 2 1 2

10 10 ] [ ] [ 2

≈ + ≈ + + =

− − + +

K H K K H

α 8632 . 1 1 1 1

5 . 9 3 . 10 2 1

10 10 ] [ ] [ 1

≈ + ≈ + + =

− − + +

H K K H

α hr M x hr M x M x t CT / 10 7 . 2 3 10 199 . 7 10 017 . 8

5 4 4 − − − ∆ ∆

= − =

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SLIDE 10

Alternative Solution

 Allow alkalinity to vary in accordance with reaction

stoichiometry

 106 CO2 + 16 NO3

  • + HPO4
  • 2 + 122 H2O + 18 H+

= C106H263O110N16P + 138 O2

 Now, incorporating this into the final alkalinity value:  And the rate becomes:

David Reckhow CEE 680 #25 10

T i f

C Alk Alk

+ =

106 18

( ) ( ) (

) ( )

5 5 4 5 . 4 106 18 4 2 1 2 1

10 1 . 7 10 2995 . 9 3066 . 1 10 017 . 8 136 . 1 10 10 5 . 8 2 ] [ ] [ ] [ 2

− ∆ − ∆ ∆ − − ∆ − ∆ − + −

= = − = − + − + = − − + + = x C x C C x C x C C OH Alk H OH C Alk

T T T T T T f f f T

i

α α α α

hr M x hr M x t CT / 10 4 . 2 3 10 1 . 7

5 5 − − ∆ ∆

= =

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SLIDE 11

Buffer Intensity & Closed System

David Reckhow CEE 680 #25 11

From: Butler, 1991; pg 67

( )

2 1 1

] [ ] [ 303 . 2 α α α α β

T T

C C H OH + + + ≈

+ −

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SLIDE 12

Buffer Intensity & Open System

David Reckhow CEE 680 #25 12

From: Butler, 1991; pg 68

( )

] [ 4 ] [ ] [ ] [ 303 . 2

2 3 3 − − + −

+ + + = CO HCO H OH β

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SLIDE 13

To next lecture

David Reckhow CEE 680 #25 13

DAR