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 Updated: 4 March 2020 Lecture #25 Dissolved Carbon - - PowerPoint PPT Presentation
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
(Stumm & Morgan, Chapt.4 )
David Reckhow CEE 680 #25 1
Updated: 4 March 2020
David Reckhow CEE 680 #25 2
David Reckhow CEE 680 #25 3
David Reckhow CEE 680 #25 4
David Reckhow CEE 680 #25 5
From Stumm & Morgan (example 4.8, pg. 174)
General: assume that system is closed Simplified: treat alkalinity as constant Alternative: allow alkalinity to vary in accordance with
David Reckhow CEE 680 #25 6
106 CO2 + 16 NO3
Initial
Alk = 0.85 meq/L pH = 9.0
Final (3 hours later)
pH = 9.5
David Reckhow CEE 680 #25 7
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
And 3 hours later
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 − − − ∆ ∆
= − =
106 CO2 + 16 NO3
= C106H263O110N16P + 138 O2
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 − − ∆ ∆
= =
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 + + + ≈
+ −
David Reckhow CEE 680 #25 12
From: Butler, 1991; pg 68
] [ 4 ] [ ] [ ] [ 303 . 2
2 3 3 − − + −
+ + + = CO HCO H OH β
David Reckhow CEE 680 #25 13