Fall 2014 Due: 10:30 a.m., Tuesday, Sept 16

Transcription

1 GEOL330/634 Fall 2014 Due: 10:30 a.m., Tuesday, Sept 16 Problem Set #1 Key 1. What determines if an ion is a major ion in seawater? (5) Major ions are those that contribute to salinity. If salinity can be determined to parts per thousand (g/kg) or 1 mg/kg (ppm) a major element is any ion with a concentration greater than 1 ppm. 2. Why do chemical oceanographers prefer to use mole units rather than weight units (like grams)? (5) Because when thinking about the stoichiometry of reactions it is atoms or moles that react not weight or grams 3. Global Methane reservoirs and fluxes (from Bill Reeburgh; a) Is the concentration of atmospheric methane (CH4) at steady state? (5) Sources = 500 Tg CH4 y - 1 sinks = 460 Tg CH4 y - 1 So the sources and sinks are not in balance thus the atmospheric CH4 is not at steady state. b) How does the flux of CH4 (as C) from the ocean to the atmosphere compare with that of CO2 from Sarmiento and Gruber? (5) The flux from the ocean = 10 Tg CH4 y - 1. That is 12/16 x 10 = 7.5 TgC y - 1 or gc y - 1. The total pre- industrial flux of C in S&G is 70.6 Pg C y - 1 or 70.6 x gc y - 1 so the C from CH4 is 0.01% of the total PI C Flux.

2 4. Some elements in seawater are considered to be conservative. (15) a) Propose a definition that would apply to conservative elements in seawater. (5) Conservative elements in seawater are those whose distributions are controlled by physical processes only (like mixing), and are not affected by chemical and biological processes. Also acceptable: elements that follow the law of conservative ratios. b) Which of the major elements in seawater are considered conservative? (5) Na +, K +, Cl -, SO 2-, Br-, B(OH)3 (or B), and F - c) Vanadium (V) is a trace element that has been proposed to be conservative in seawater. How would you design a study to show that this is true? (5) Collect seawater samples from a variety of locations and depths across a known salinity gradient. Measure concentrations of V in each sample and plot versus salinity, potential temperature or some other conservative property. If the points fall on a straight line, V could be considered conservative. d) If Vanadium is conservative does it have a constant concentration throughout the oceans? Explain (5) No. Salinity varies throughout the ocean, and the concentrations of conservative elements vary with it, such that their ratio is always constant. 3. Why do chemical oceanographers prefer to use concentration units of moles /kg rather than moles/ltr? (5) Water is compressible causing the amount of material in a liter of seawater to be dependent on pressure. Likewise, temperature influences the volume of a given amount of seawater. So the mass in a liter varies with depth but a kg is always a kg!

3 5. Trace Metal profiles (15) Two part answers for each. Draw example profiles for each type of controlling process listed below. Put full water column profiles for the Atlantic and Pacific in the same graph. Label depth (0 to 5 km) and concentration scales. a. What would be the general distribution characteristics of a trace metal influenced by: b. How would the profiles differ from the Atlantic to the Pacific influenced by: i) nutrient like (5) Similar to that of Cadmium (below). Concentrations are low in the euphotic zone where primary production of organic material consumes most available nutrients. Concentrations increase at depth as nutrients are remineralized. Concentrations are higher in the Pacific than the Atlantic since Pacific deep water has been out of contact with the surface ocean for longer than that in the Atlantic, and has thus had more time to accumulate nutrients. ii) hydrothermal inputs (5) Any profile showing a deep peak around 2500 m or the depth of the nearby crest of mid-ocean ridge Differences in profiles wouldn t really result from basin to basin, but rather with proximity to vents.

4 iii) scavenging by particles settling through the water column (5) Profile should be similar to that of Aluminum (below). Profiles are relatively enriched in the surface water and decrease with depth as the elements are adsorbed to particles that sink through the water column. In this case, relatively old deep water in the Pacific results in lower concentrations than in the Atlantic.

5 6. A chemical oceanographer is sitting home on Saturday night and decides to entertain herself by making synthetic seawater with chemicals she has around the house (being a chemist, her available household chemicals are wide ranging). She wants to make a solution containing the six most abundant ions in close approximation to their natural seawater concentrations. Her plan is to weigh out the necessary chemicals into a laundry tub and then to add Philadelphia tap water (which is so dilute that it can be considered like distilled water) until the final weight of the solution is 1 kilogram. Her goal is to have the following ionic concentrations: Ion mmol kg -1 (seawater solution) Na Mg Ca K + 10 Cl SO 4 30 She has the following chemicals NaCl MgSO 4 CaCl 2 HCl KOH Mg(OH) 2 common spice (Morton Table Salt) fertilizer, foot soaks (Epsom salts) drying agent, preservative (Dri-Z-Air) liquid, stolen from lab lye laxative (Milk of Magnesia) a) How many grams of each chemical should she add to get the correct ionic concentrations (note: H + and OH - will end up balancing, making H + + OH - = H 2 O)? (15) First you need to find the molar mass of each of the salts. You can look this up for the compounds themselves, or add up atomic masses from a periodic table of the elements: eg for NaCl = g/mol Na g/mol Cl = g/mol NaCl NaCl(s) g/mol or mg/mmol MgSO4(s) CaCl2(s) KOH(s) Mg(OH)2 (s) HCl Now you want to find the mass of each of these compounds needed to make a solution with the following components: Ion Concentration (mmol/kg) Start with the ions that are only in one of the salts: Na, Ca, K, SO4 (470 mmol Na + )(58.44 mg NaCl/mmol) = mg NaCl = g NaCl

6 (10 mmol Ca 2+ )( mg CaCl2/mmol) = mg CaCl2 = 1.11 g CaCl2 (10 mmol K + )(56.10 mg KOH /mmol) = mg KOH = 0.56 g KOH (30 mmol SO4 2- )( mg MgSO4/mmol) = mg MgSO4= 3.61 g MgSO4 Next, you make up the difference in the remaining two ions by accounting for what has already been added with the other salts: Add: 50 mmol Mg mmol Mg 2+ from MgSO4 = 20 mmol Mg 2+ (20 mmol Mg 2+ )(58.33 mg Mg(OH)2/mmol) = mg = 1.17 g Mg(OH)2 Add: 540 mmol Cl mmol Cl - from NaCl 2(10 mmol Cl - from CaCl2) = 50 mmol Cl- (50mmol Cl- )(36.46 mg HCl /mmol) = mg = 1.82 g HCl b) Assuming complete dissolution, what will be the molarity (mol/ltr) of each ionic species (use a density of g cm - 3 for pure water at room temperature) (10) The most accurate way to calculate this: Total mass of solution = 1000 g Total mass of salts = 27.47gNaCl gCaCl gKOH gMgSO g Mg(OH) g HCl = g salt Mass of water = Mass of solution mass of salt = 1000 g solution g salt = g water Density of pure water = g cm - 3 = g/ml Density = mass/volume, so Volume water = Mass/density = g water / g/ml = ml water * 1 L/1000 ml = L water Molarity = mols of ion/l solution If you assumed that your solution was the same density of SW at surface and about 25 degrees C, full credit. If you assumed that your solution was the same density of the DI water you added, - 2 points