1. Background in Biogeochemistry Some aspects of element composition and behavior are illustrated in Table 1. The major elements include Si, C, Al and Ca. Most of the major elements are largely found in the lithosphere and exhibit long residence times in this pool. T = pool/mass input or mass output For some elements, a significant fraction of the mass is found in the oceans (Na +, Cl -, S). For the most part, the atmosphere and biosphere are minor element pools.

2. Background in Biogeochemistry (cont.) Only for N is the atmosphere a significant pool. Some elements exhibit more than one oxidation state and therefore can participate in redox (oxidation-reduction) reactions (e.g. C, Fe, S, N, O).

3. Table 1. Oxidation State, Abundancy and Residence Times of Major Elements (units of 10 20 moles and 10 6 years) ElementOxidation StateOcean Pool T Lithosphere Pool T Atmosphere Pool T Biosphere Pool T Total Si+40.00140.016220450220 C+4 - -40.0330.088610.000590.190.00420.07861 Al+35750457 Ca+20.151.25035150 Mg+20.82152538126 Na+17.119312528822 Fe+3, +21748117 Cl8.23055.921814 K+10.158.21339913 S+6, +4, +2, 0, -20.42224.92425.3 N+5, +4, +3, +2, +1, 0, -30.0160.72.73.5 P+50.27 O2O2 -2, 00.3870.38

4. Acid-Base Chemistry Major ionic solutes Cations Basic - Ca 2+, Mg 2+, K +, Na + largely derived from cation supply in the lithosphere Acidic- H +, Al 3+, Fe 3+ occurs under acidic conditions Reduced- Fe 2+, Mn 2+ occurs under reducing conditions (sediments, wetlands)

5. Acid-Base Chemistry (cont.) Anions Strong acid anions- SO 4 2-. NO 3 -. Cl - Weak acid anions- HCO 3 -, CO 3 2-, A n- (organic anions) Basic- OH - C B - sum of basic cations = 2[Ca 2+ ] + 2[Mg 2+ ] + [Na + ] + [K + ] C A - sum of strong acid anions = 2[SO 4 2- ] + [NO 3 - ] + [Cl - ]

6. Acid-Base Chemistry (cont.) Dissolved inorganic carbon (DIC) = C T [H 2 CO 3 * ] + [HCO 3 - ] + [CO 3 2- ] The distribution of inorganic carbon species is a function of pH (see figure). H 2 CO 3 * =H + + HCO 3 - ;pK a1 = 6.3 HCO 3 - =H + + CO 3 2- ;pK a2 = 10.3

8. Acid-Base Chemistry (cont.) An important measurement of the acid-base status of waters is acid neutralizing capacity or alkalinity. ANC= [HCO 3 - ] + 2[CO 3 2- ] + n[A n- ] + [OH - ] - [H + ] = the ability of a system to neutralize inputs of strong acid. HCO 3 - + H + =H 2 CO 3 * CO 3 2- + 2H + =H 2 CO 3 OH - + H + =H 2 O ANC = C B - C A

9. Acid-Base Chemistry (cont.) An important concept is electroneutrality. All solutions (and systems) must be electrically neutral. c i - concentration of ionic solute z i - charge 2[Ca 2+ ] + 2[Mg 2+ ] + [Na + ] + [K + ] + [H + ] = 2[SO 4 2- ] + [NO 3 - ] + [Cl - ] + [HCO 3 - ] + 2[CO 3 2- ] + n[A n- ] + [OH - ] Rearranging ANC= [HCO 3 - ] + 2[CO 3 2- ] + n[A n- ] + [OH - ] - [H + ] = 2[Ca 2+ ] + 2[Mg 2+ ] + [Na + ] + [K + ] - 2[SO 4 2- ] - [NO 3 - ] - [Cl - ] = C B - C A

10. Acid-Base Chemistry (cont.) Increases in Ca 2+ by weathering of minerals increases ANC. CaCO 3 + H + = Ca 2+ + HCO 3 - Inputs of H 2 SO 4 from acid rain decreases ANC. H 2 SO 4 = 2H + + SO 4 2- Any process which affects the concentration of ionic solutes changes ANC. There is a non-linear relationship between ANC and pH (see figure). e.g. ANC production is an important indicator of the abiotic fixation or removal of CO 2.

12. Acid-Base Chemistry (cont.) Weathering is an important process by which CO 2 is fixed from the atmosphere. NaAlSi 3 O 8(s) + H 2 O + CO 2 = Na + + HCO 3 - + Al(OH) 3(s) + 3H 4 SiO 4 (albite) This HCO 3 - is transported by rivers to the oceans. A budget for bicarbonate of the ocean is shown in Table 2 (from Berner and Berner).

13. Present-Day Budget InputsOutputs Rivers1980CaCO 3 deposition: Biogenic pyrite formation145 Shallow water1580 Deep sea1340 Total2125Total2920 Table 2. The Oceanic Bicarbonate Budget (rates in Tg HCO 3 - /yr) Budget for Past 25 Million Years InputsOutputs Rivers1980CaCO 3 deposition: Biogenic pyrite formation73 Shallow water730 Deep sea1340 Total2053Total2070 Note: Tg = 10 12 g. Replacement time for HCO 3 (river input only) is 83,000 years.

14. Acid-Base Chemistry (cont.) There are two budgets presented. The first is the current budget. The second is the budget over the past 25 MY. Note that the major inputs of HCO 3 - are riverine inputs and biogenic pyrite formation (see figure). About 7% of the total HCO 3 - inputs is associated with SO 4 2- reduction. 2CH 2 O + SO 4 2- = H 2 S + 2HCO 3 -

16. Organic Matter Dissolved SO 4 2- Iron Minerals Pyrite FeS 2 Organic S Compounds H2SH2S Bacteria

17. Acid-Base Chemistry (cont.) The sink of HCO 3 - inputs to the oceans is precipitation of CaCO 3. Ocean water is not oversaturated with respect to the solubility of CaCO 3 over the entire depth. The upper waters are oversaturated while the lower waters are undersaturated. The reason for this pattern is that the solubility of CaCO 3 increases with increasing depth due to increases in pressure (see figure). At the average ocean depth, the pressure is 400 atm. Under these conditions, CaCO 3 is about twice as soluble as at the surface.

18. Acid-Base Chemistry (cont.) Also, the production of CO 2 from respiration of organic matter facilitates the dissolution of CaCO 3. About 83% of the precipitated CaCO 3 redissolves at depth. CO 2 + H 2 O + CaCO 3 = Ca 2+ + HCO 3 - Note that the current ocean is not at steady-state with respect to inputs of HCO 3 -. Over the short-term, HCO 3 - is being depleted, sediment deposition exceeds inputs. At the current rate of deposition, all of the HCO 3 - in the ocean would be removed in 200,000 yr. This condition would never exist, as the ocean would eventually become undersaturated with respect to the solubility of CaCO 3 and precipitation would stop.

19. Acid-Base Chemistry (cont.) The condition of elevated CaCO 3 deposition to sediments has only been occurring during the past 11,000 yr. This condition is due to the rapid post-glacial rise of sea level over the continent of shelves.

21. over saturation under saturation

23. Redox Chemistry Redox reactions involve the transfer of electrons. All redox reactions must be coupled and require an electron donor and electron acceptor. e.g. CH 2 O + H 2 O=CO 2 + 4e - + 4H + O 2 + 4e - + 4H + =2H 2 O CH 2 O + O 2 =CO 2 + H 2 O Redox reactions are often characterized in stoichiometric half reactions. e.g.Fe 3+ + e - = Fe 2+

24. Redox Chemistry (cont.) This reaction can be written as a mass law. where K is a thermodynamic equilibrium constant. If we take the logarithm of this expression. or pe = - log[e - ] and is the indicator of the redox status of the system. oxidizing conditions - high positive pe values reducing conditions - low or negative pe values

25. Redox Chemistry (cont.) Only a few elements dominate redox reactions in natural waters. Major redox elements - C, N, O, S, Fe, Mn The most important electron donor in natural waters is organic matter. The electron acceptor that is coupled with the electron donor (organic matter) is variable. It depends on the quantity and energetics of the electron acceptor. Redox reactions in the natural environment can be thought of as an electron titration. The source of electrons is the electron donor (organic matter). These electrons are released to electron acceptors in order of their electron affinity or energetics.

26. Log K pe 0 pe 0 w Aerobic respiration¼O 2 + H + + e - = ½H 2 O20.7513.75 DenitrificationNO 3 - + 6/5 H + + e - = 1/10 N 2 + 3/2 H 2 O21.0512.65 Mn reduction½MnO 2(s) + 2H + + e - = ½Mn 2+ + H 2 O20.86.8 Fe reductionFe(OH) 3(s) + 3H + + e - = Fe 2+ + 3H 2 O16.0-5.0 SO 4 2- reduction c SO 4 2- +9/8 H + + e - = c HS - + ½H 2 O 4.25-3.6 Methanogenesis c CO 2(g) + H + + e - = c CH 4(g) + ¼ H 2 O 2.9-4.1

27. Redox Chemistry (cont.) The energy yield of these electron acceptor reactions decreases. So, O 2 is the preferred electron acceptor. When O 2 is consumed, electrons are transferred to NO 3 - and so on. Note that when the electron acceptor O 2 is in excess, conditions are aerobic (i.e. Earth's surface). When the quantity of electron donor (organic matter) exceeds the quantity of electron acceptor (O 2 ), then anaerobic conditions result. These conditions occur in wetlands or in lake sediments.

28. Redox Chemistry (cont.) See figure of the natural electron cycle. If photosynthetic products were oxidized completely by respiration, the atmosphere would be devoid of O 2. There is a loss of reduced species CH 2 O and FeS, which represents a new loss of bound electrons. This coincides with a net yield of oxidant O 2 to the atmosphere. The electron cycle is responsible for the partitioning of an oxidizing atmosphere and reducing lithosphere.

31. Methods Water Column Monitoring – 1981, 1989, 1990, 1991, 2000 O 2 NO 3 - NH 4 + SO 4 2- (H 2 S) T Fe 2+ CH 4 Alkalinity pH DIC – calculated from alkalinity, pH, temperature Sediment Traps – 10m – 1989, 1990, 1991 POC

34. Constituent Rate (mmol m -2 d -1 ) Year 1981198919901991 DIC35324942 NH 4 + 5.311.810.910.0 O2O2 -62-63-41-48 NO 3 - --8.8-4.7-5.4 Fe 2+ 0.91.61.11.6 SO 4 2- --11.2-11.0-7.0 H2SH2S16.59.914.713.5 CH 4 * -15.7 (10.5)14.5 (9.7)18.1 (12.1) Rates of Solute Accumulation (+) or Loss (-) in the Hypolimnion of Onondaga Lake * Includes ebullitive loss, assumed to be 33% of total; soluble component in parentheses.

35. Effective Equilibrium Constants of Aquatic Redox Couples Reactionpe 0 pe 0 w (pH = 7) ¼O 2 + H + + e - = ½ H 2 O20.7513.75 1 / 5 NO 3 - + 6 / 5 H + + e - = 1 / 10 N 2(g) + 3 / 5 H 2 O21.0512.65 ½ MnO 2(s) + 2H + + e - = ½ Mn 2+ + H 2 O20.86.8 Fe(OH) 3(am) + 3H + + e - = Fe 2+ + 3H 2 O16-5.0 1 / 8 SO 4 2- + 5 / 4 H + + e - = 1 / 8 H 2 S (g) + ½ H 2 O 5.25-3.5 1 / 8 CO 2 (g) + H + + e - = 1 / 8 CH 4 + ¼ H 2 O 2.9-4.1

36. Oxygen ReductionDIC Equivalents per Mole O 2 : -1 (CH 2 O) 106 (NH 3 ) 16 H 3 PO 4 + 106 O 2 106 CO 2 + 16 NH 3 + H 3 PO 4 + 106 H 2 O DenitrificationDIC Equivalents per Mole NO 3 - : -1.25 (CH 2 O) 106 (NH 3 ) 16 H 3 PO 4 + 84.9 HNO 3 106 CO 2 + 16 NH 3 + H 3 PO 4 + 42.4 N 2 + 148.4 H 2 O Manganese ReductionDIC Equivalents per Mole MN: 0.5 (CH 2 O) 106 (NH 3 ) 16 H 3 PO 4 + 212 MnO 2 + 424 H + 106 CO 2 + 16 NH 3 + H 3 PO 4 + 212 Mn 2+ + 318 H 2 O Iron ReductionDIC Equivalents per Mole Fe 2+ : 0.25 (CH 2 O) 106 (NH 3 ) 16 H 3 PO 4 + 424 FeOOH + 848 H + 106 CO 2 + 16 NH 3 + H 3 PO 4 + 424 Fe 2+ + 742 H 2 O Sulfate ReductionDIC Equivalents per Mole H 2 S: -2 (CH 2 O) 106 (NH 3 ) 16 H 3 PO 4 + 53 SO 4 2- 106 CO 2 + 16 NH 3 + H 3 PO 4 + 53 S - + 106 H 2 O Iron & Sulfate ReductionDIC Equivalents per Mole SO 4 2- : -2.25 (CH 2 O) 106 (NH 3 ) 16 H 3 PO 4 + 47.1 FeOOH + 47.1 SO 4 2- + 94.2 H + 106 CO 2 + 16 NH 3 + H 3 PO 4 + 47.1 FeS + 176.6 H 2 O MethanogenesisDIC Equivalents per Mole CH 4 : 1 (CH 2 O) 106 (NH 3 ) 16 H 3 PO 4 53 CO 2 + 16 NH 3 + H 3 PO 4 + 53 CH 4 FermentationDIC Equivalents per Mole C: 0.5 (CH 2 O) 106 (NH 3 ) 16 H 3 PO 4 35.3 CO 2 + 16 NH 3 + H 3 PO 4 + 35.3 C 2 H 5 OH HumificationDIC Equivalents per Mole C: 0.5 (CH 2 O) 106 (NH 3 ) 16 H 3 PO 4 35.3 CO 2 + ( C 2 H 5 OH) 35.3 (NH 3 ) 16 H 3 PO 4

39. Organic Carbon Budget 1987-92 Rate (mmol C/m -2 d -1 ) % Organic C Inputs (Sediment Traps)62100 DIC Release (Water Column)4166 CH 4 (Water Column)1626 Total C Release5792 Net Sediment C Accumulation58 Organic C Burial (Sediments)26

40. Hypolimnion Electron Budget 1987-92 Rate (meeq/m 2 -day) % Electron Donor (Sediment Traps)248100 Electron Acceptors (Water Column) O 2 6928 NO 3 - 187 Fe 3+ 0.79<1 SO 4 2- 4920 CH 4 4116 Total Electron Acceptors Measured17872 Net Electron Transfer to Sediments7028 Net Electron Burial in Sediments (Sediments) 105

41. Comparison of Hypolimnetic Electron Budgets During Summer Stratification Lake Fluxes Electron Donor (meeq m 2 d -1 ) Fluxes Electron Acceptors (meeq m -2 d -1 ) O2O2 NO 3 - Fe 2+ SO 4 2- CH 4 Total Bleltham Tarm15418 (49)7.3 (20)ND0.85 (2)10.7 (29)36.8 Dart’s18.815.2 (81)2.8 (14)0.4 (2)0.7 (4)-18.8 Mirror409.2 (69)0 (0)0.43 (3)1.5 (20)2.3 (17)13.4 L226ND15.2 (30)4.2 (8)0.97 (2)8.1 (16)23 (45)51.8 L227841.6 (5)0.3 (1)0.6 (2)8.8 (28)20.1 (64)31.3 L223ND2.7 (10)0.3 (1)1.8 (7)9.1 (34)12.8 (48)26.8 Onondaga Lake 1989-91 24069 (39)18 (10)0.79 (<1)49 (27)41 (23)178