2 examples: Mg 2+ Rivers [Mg!! ]! (v r ) [Mg 2+ ] SW. **Answers with OR without vent effluent arrow are acceptable, since [Mg] in vent fluid is zero.

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1 OCN400 Winter 2015 PS2 - Key 1. You might think it is well known, but there is some debate about what controls the magnesium concentration in seawater. The main input is rivers (see Power Point Lecture Notes 4). The main removal is by hydrothermal processes- - the concentration of Mg 2+ in 350 C end member hydrothermal solutions is zero (you can see the table in the major ions lecture (Lecture Notes 2 or 4) for more details). Here we'll see how this balance between sources and sinks as a control on the composition of seawater works. a.) Draw a schematic diagram for a one- box model calculation for Mg 2+ in seawater which shows the source (rivers) and sink to a hypothetical mid- ocean ridge. (5) 2 examples: Mg 2+ Rivers [Mg ]! (v r ) [Mg 2+ ] SW Hydrothermal circulation ([Mg ]!" ) (v!" ) **Answers with OR without vent effluent arrow are acceptable, since [Mg] in vent fluid is zero. **NOTE: Answers that included flux and reservoir values from McDuff and Morel were 1

2 marked correct. In the future, use values given in the problem set first. If you must pull values from outside sources, please cite your sources (i.e. when using values not given in the problem set). - 1pt/arrow or missing label (like Mg 2+ ) b.) Calculate the residence time of water in the ocean, once relative to river input, and again relative to hydrothermal circulation. From notes use: Mass of ocean = 1.38 x kg; River discharge = 3.5 x kg y - 1 ; Hydrothermal circulation 1 x kg y - 1. (10) τ =!"##!"#$ Mass ocean = M ocean = kg River discharge = river flux = v R = kg/y Hydrothermal circulation =hydrothermal vent flux = v HV = kg/y τ!"#$! =!!"#$%!! =!.!"!"!"!"!.!!"!"!"/! = y τ!" =!!"#$%!!" =!.!"!"!"!"!.!!"!"!"/! = ! y 5 pts/each answer c.) Assume the ocean Mg cycle is at steady state, rivers are the only input and hydrothermal removal of Mg is the only sink. How long will it take for all the water in the ocean to pass through the mid- ocean ridge vent system? Is this an upper or lower limit? (5) NOTE: Due to the confusing wording of this problem, this problem was not graded on the homework [marked NG for Not Graded ]. There were a few ways to approach it: 1) Mass balance, where Mg in rivers and hydrothermal vents (out) are in steady state 2) Calculating residence time of Mg with respect to Hydrothermal Venting and River input Some others also restated answers to B, which because of the wording of the question, would technically be correct. If you want to discuss any of these answers, please stop by office hours or set up an appointment. **Note: Answers will vary depending on the values of [Mg] in rivers and seawater you used. 2

3 Approach 1: Assume Mg is in steady state (inputs = outputs) and that the end member has [Mg]=0 Inputs (sources): rivers, hydrothermal vent effluent coming back into the system. Output (sink): Hydrothermal circulation (seawater) Mg! v! + Mg!" (v!" ) = Mg!" (v!" ) Mg! v! = Mg!" v!" Mg!" (v!" ) Assume [Mg2+]HV, the concentration of the hydrothermal effluent is 0 (all the Mg precipitates out along the ridge crest). Mg! v! = v!" Mg!" Mg! v! v!" = Mg!" v!" = mmol kg (3.5 10!" kg y ) = !" kg y mmol kg Use this calculated hydrothermal vent flux and the mass of the ocean to determine the residence time of magnesium, with respect to the hydrothermal vents !" kg = ! y!" 8.5x10 kg y Approach 2 [Mg ]!"#$% (M!"#$% ) τ! = [Mg ]! (v! ) [52.82 mmol kg ] ( !" kg) τ! = = ! y [0.128 mmol kg ](3.5 10!" kg y ) **upper bound [Mg ]!"#$% (M!"#$% ) M!"#$% τ!" = = [Mg ]!"#$% (v!" ) v!" 3

4 !" kg = ! y !" kg y **lower bound τ!" = d. How would the Mg2+ concentration in seawater change and what would the steady state concentration be if the rate of removal due to hydrothermal circulation was to double? (5) Doubling the vhv τ!" [Mg ]!"#$% (M!"#$% ) = [Mg ]!"#$% (v!" ) simplifying, M!"#$% v!" If vhv doubles, τhv must decrease 50%, but how does this affect the [Mg2+]ocean? Assume at steady state, τriver= τvent [Mg ]!"#$% (M!"#$% ) M!"#$% = [Mg ]! (v! ) v!" Simplifying, and doubling the rate of hydrothermal venting: [Mg ]!"#$% 1 = [Mg ]! (v! ) v!" Solving for [Mg2+]Ocean [Mg ]! (v! ) [Mg ]!"#$% = v!" mmol kg !" kg y [Mg ]!"#$% = = 22.4 mmol kg !" kg y 4

5 [Mg 2+ ] ocn = 22.4 mmol/kg, which is about ½ the original [Mg 2+ ] ocn. This makes sense if the sink doubles, the concentration must decrease by ½ in accordance with the now shorter τ **2 points if you say it decreases, but with wrong reason/no calculation. 2. Composition of River Water Britta Voss ( has recently completed a comprehensive study of the dissolved inorganic composition of the Fraser River, Canada (Voss et al., 2014, GetCA, 124, ). Britta took this Chemical Oceanography course at UW (OCN 400) in Winter Here is a summary of the discharge- weighted average concentration of major ions (μmol L- 1). The global average river concentrations (from Lecture 4) are included for comparison. Ion Fraser River concentration (µmol L - 1 ) Cl Na Mg SO K Ca HCO SiO Global average (µmol L - 1 ) a. Assume that there are no evaporate rocks in the Fraser River drainage basin. Assume all the Cl- comes from marine aerosols. Calculate the seawater aerosol contributions to all the major ions in the Fraser River. (10) If we assume that there are no evaporate rocks in the Fraser River drainage basin and that all of the Cl - comes from marine aerosols. Aerosols have the same ion concentration as the parcel of seawater that they came from. If we know that the only concentration of Cl - is from aerosols (the ocean) and that elements are conservative in the ocean, we can use those two facts to determine the seawater aerosol contributions to the Fraser River Ion!"#$%$& = [Cl! ]!" [ion]!" [Cl! ]!" 5

6 Ion Cl- Na+ Fraser River concentration (µmol L- 1) 29 Global average (µmol L- 1) 220 Seawater concentration (mmol L- 1) Aerosol Concentration (µmol L- 1) Math calculation µμmol L HCO SiO Mg SO4 + K 2+ Ca mmol L mmol L mmol L 29µμmol L mmol L 54.2 mmol L 29µμmol L mmol L 27.6 mmol L 29µμmol L mmol L 10.0 mmol L 29µμmol L mmol L 9.7 mmol L 29µμmol L mmol L 2.3 mmol L 29µμmol L mmol L 0.1 mmol L 29µμmol L mmol L **Correct setup is worth 3 pts **Each ion is worth 1 pt **Minus 2 for not showing work (or sample calculation) **If you try the problem but the setup is incorrect 2pt b. Calculate the aerosol corrected concentrations for all ions and % of total that come from weathering. (10) To calculate the concentration due to weathering, [Ion]!"#$%" [Ion]!"#$%$& Then to calculate the percent due to weathering, divide by the total ion concentration [Ion]!"#$%" [Ion]!"#$%$& [Ion]!"#$%" Ion Concentrations % due to due to weathering weathering (µmol L- 1) Cl Na Mg SO K Ca HCO SiO

7 **Correct setup is worth 3 pts **Each ion is worth 1 pt **Minus 2 for not showing work (or sample calculation) **If you try the problem but the setup is totally wrong 2pt c. Calculate the % of how much of the HCO3- comes from silicate versus carbonate rocks. There are different ways you could do this calculation. Do the one you think best and state your assumption(s). (10) If we add up all of the concentrations of species due to weathering, we can determine the percent of that total that comes from carbonate rocks (CaCO 3 ) and the percent due to silicate. This assumes that other species contribute minimally. Also that aerosols and weathering are the only sources contributing to the overall total (ie no gas exchange with the atmosphere) Assume all calcium weathers as calcium carbonate, not as a silicate. Carbonate rock species: Method 1: CaCO 3 +H 2 O +CO 2 à Ca HCO 3 [HCO 3 - ] carbonate =2[Ca 2+ ] [HCO 3 - ] carbonate =2(364.5 µmol L - 1 ) = µmol L - 1 % carbonate = µmol L - 1 /850.9 µmol L - 1 (100) = 85.7% [HCO 3 - ] total = [HCO 3 - ] carbonate + [HCO 3 - ] silicate [HCO 3 - ] silicate = µmol L µmol L - 1 =121.9 µmol L - 1 % silicate = µmol L - 1 /850.9 µmol L - 1 (100) = 14.3% Method 2 [HCO 3 - ] silicate = H 4 SiO 4 /2 = 81/2 = 40.5 µmol L - 1 [HCO 3 - ] total = [HCO 3 - ] carbonate + [HCO 3 - ] silicate 40.5 µmol L - 1 /850.9 µmol L - 1 (100) =4.76% [HCO 3 - ] total - [HCO 3 - ] silicate = [HCO 3 - ] carbonate 7

8 [HCO 3 - ] carbonate = 810.4/850.9 µmol L - 1 (100) = 95.24% *State assumptions 2 pts No assumptions, automatically only 8 pts d. How does the % HCO 3 - from weathering silicate rocks for the Fraser River compare with that from the global average rivers? (5) If we do this problem with the global averages, we still need to determine the contribution of aerosols. Ion Global average (µmol L - 1 ) Seawater concentration (µmol L - 1 ) Aerosol Concentration (µmol/l) Concentrations due to weathering Cl Na Mg SO 4 2- % due to weathering K Ca HCO SiO Sample calculation to determine aerosol concentration, concentration due to weathering, and percent due to weathering for Ca 2+ (like 2a and 2b but consolidated ) Aerosol concentration 9.7 mmol 220µμmol L L = 3.99 µμmol L mmol L Concentration due to weathering: [Ca 2+ ] Total =[Ca 2+ ] aerosol + [Ca 2+ ] Weather 375 µmol L - 1 = 3.99 µmol L [Ca 2+ ] Weather [Ca 2+ ] Weather = µmol L - 1 % due to weathering: ([Ca 2+ ] Weather /[Ca 2+ ] Total )*100 = µmol L - 1 /375 µmol L - 1 =98.9% To determine the proportion of weathering by carbonate and silicate rocks, CaCO 3 +H 2 O +CO 2 à Ca HCO 3 [HCO 3 - ] carbonate =2[Ca 2+ ] 8

9 [HCO 3 - ] carbonate =2(371.01µmol L - 1 ) = µmol L - 1 % carbonate = µmol L - 1 /957.05µmol L - 1 (100) = 77.53% [HCO 3 - ] total = [HCO 3 - ] carbonate + [HCO 3 - ] silicate [HCO 3 - ] silicate = µmol L µmol L - 1 = µmol L - 1 % silicate = µmol L - 1 /957.05µmol L - 1 (100) = 22.47% OR [HCO 3 - ] silicate = H 4 SiO 4 /2 [HCO 3 - ] silicate = /2 = µmol L - % silicate = µmol L /957.05µmol L - 1 (100) = 11.39% [HCO 3 - ] total - [HCO 3 - ] silicate = [HCO 3 - ] carbonate [HCO 3 - ] carbonate = µmol L µmol L [HCO 3 - ] carbonate = µmol L - 1 % carbonate = µmol L - 1 /957.05µmol L - 1 (100) = 88.61% **More weathering due to silicate in the global average river The percent of bicarbonate from weathering silicate rocks is significantly higher in the global average river. The silicate concentration in the Fraser River is very low as compared to that of the average river, whereas the carbonate concentrations from the Fraser and the global average are more comparable. 2 points for correctly stated comparison w/out percentages Lose a point for not subtracting off aerosol concentrations from total calcium/silicate when determining percent weathering. 9