CEE 680 Lecture #25 3/4/2020 Print version Updated: 4 March 2020 - PDF document

CEE 680 Lecture #25 3/4/2020 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 1

CEE 680 Lecture #25 3/4/2020  Lake Taihu David Reckhow CEE 680 #25 3  algae David Reckhow CEE 680 #25 4 2

CEE 680 Lecture #25 3/4/2020  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 3

CEE 680 Lecture #25 3/4/2020 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           Alk 2 C [ OH ] [ H ] 1 2 T 1 1     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        0 . 0477 2 2 1 2  2   9  [ H ]  [ H ]  10 1 1  10 . 3      K K K 10 8 . 5 x 10 4 10 5 10 9 1 2 2  0  0 . 9523 2 ( 0 . 0477 )   4 8 . 017 x 10 M David Reckhow CEE 680 #25 8 4

CEE 680 Lecture #25 3/4/2020 Simplified Solution: (const. alk.) 1 1     0 . 8632  And 3 hours later 1      K 1 10 . 3 [ H ] 10 1 2  9 . 5 K  10 [ H ] 1     Alk [ OH ] [ H ] 0  C T    2 1 1 1 2     0 . 1368      8 . 5 x 10 4 10 4 . 5 10 9 . 5 2  2   9 . 5    1 [ H ] [ H ] 1 10   10 . 3 K K K 10 1 2 2  0 . 8632 2 ( 0 . 1368 ) 0   7 . 199 x 10 4 M  So the rate is:     8 . 017 x 10 4 M 7 . 199 x 10 4 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    stoichiometry Alk Alk 18 C f i 106 T ‐ + 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:           Alk 2 C [ OH ] [ H ] 1 2 T             Alk [ OH ] 2 C C f f 1 2 f T T 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    C 7 . 1 x 10 5 T  And the rate becomes:   7 . 1 x 10 5 M C T    5 2 . 4 x 10 M / hr  t 3 hr David Reckhow CEE 680 #25 10 5

CEE 680 Lecture #25 3/4/2020 Buffer Intensity & Closed System              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            2 . 303 [ OH ] [ H ] [ HCO ] 4 [ CO 2 ] 3 3 From: Butler, 1991; pg 68 David Reckhow CEE 680 #25 12 6

CEE 680 Lecture #25 3/4/2020  To next lecture DAR David Reckhow CEE 680 #25 13 7