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 Taihu David Reckhow CEE 680 #25 3
algae David Reckhow CEE 680 #25 4
Algal cells David Reckhow CEE 680 #25 5
Photosynthesis Problem Principles of conservation of Alk and C T 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
Problem Statement Photosynthesis with nitrate assimilation - + HPO 4 -2 + 122 H 2 O + 18 H + 106 CO 2 + 16 NO 3 = C 106 H 263 O 110 N 16 P + 138 O 2 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
Simplified Solution: (const. alk.) Use standard alkalinity equation − − ( ) = α + α + + Alk 2 C [ OH ] [ H ] 1 2 T 1 1 α = ≈ ≈ 0 . 9523 1 + + − 10 . 3 [ H ] + + K 1 10 1 2 − So initially: 9 + K 10 [ H ] 1 0 − + − + Alk [ OH ] [ H ] = C T 1 1 α = ≈ ≈ α + α 0 . 0477 2 2 + + − + 1 2 2 9 + + [ H ] [ H ] 10 1 1 − 10 . 3 − − − + − K K K 10 4 5 9 8 . 5 x 10 10 10 1 2 2 = 0 + 0 . 9523 2 ( 0 . 0477 ) − = 4 8 . 017 x 10 M David Reckhow CEE 680 #25 8
Simplified Solution: (const. alk.) 1 1 α = ≈ ≈ 0 . 8632 And 3 hours later 1 + + − 10 . 3 + + K 1 [ H ] 10 1 2 − 9 . 5 K + 10 [ H ] 1 − + − + Alk [ OH ] [ H ] 0 = C T α + α 2 1 1 1 2 α = ≈ ≈ 0 . 1368 − − − − − 4 4 . 5 9 . 5 8 . 5 x 10 10 10 2 + + − 2 9 . 5 + + + 1 [ H ] [ H ] 10 1 = − 10 . 3 K K K 10 1 2 2 + 0 . 8632 2 ( 0 . 1368 ) 0 − = 4 7 . 199 x 10 M So the rate is: − − − 4 4 ∆ 8 . 017 x 10 M 7 . 199 x 10 M C T = ∆ t 3 hr = − 5 2 . 7 x 10 M / hr David Reckhow CEE 680 #25 9
Alternative Solution Allow alkalinity to vary in accordance with reaction ∆ = + Alk Alk C stoichiometry 18 f i T 106 - + HPO 4 -2 + 122 H 2 O + 18 H + 106 CO 2 + 16 NO 3 = C 106 H 263 O 110 N 16 P + 138 O 2 Now, incorporating this into the final alkalinity value: ( ) − + = α + α + − Alk 2 C [ OH ] [ H ] 1 2 T ) ( ) ( − ∆ − = α + α − Alk [ OH ] 2 C C f f 1 2 T T f i ( ) − + ∆ − − = − − ∆ 4 4 . 5 4 8 . 5 x 10 18 C 10 1 . 136 8 . 017 x 10 C T T 106 ∆ = − 5 1 . 3066 C 9 . 2995 x 10 T ∆ = − 5 C 7 . 1 x 10 T And the rate becomes: − 5 ∆ 7 . 1 x 10 M C T − = = 5 2 . 4 x 10 M / hr ∆ t 3 hr David Reckhow CEE 680 #25 10
Buffer Intensity & Closed System ( ) − + β ≈ + + α α + α α 2 . 303 [ OH ] [ H ] C C T 0 1 T 1 2 From: Butler, 1991; pg 67 David Reckhow CEE 680 #25 11
Buffer Intensity & Open System ( ) β = − + + + − + − 2 2 . 303 [ OH ] [ H ] [ HCO ] 4 [ CO ] 3 3 From: Butler, 1991; pg 68 David Reckhow CEE 680 #25 12
To next lecture DAR David Reckhow CEE 680 #25 13
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