OCN400 Problem Set #1 Winter 2015 Due: 9:30 a.m., Monday, Jan 12

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1 OCN400 Problem Set #1 Winter 2015 Due: 9:30 a.m., Monday, Jan 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. Why do chemical oceanographers prefer to use concentration units of moles /kg rather than moles/ltr? (5) Water is compressible (the amount of material in a liter of seawater is dependent on pressure and temperature). The mass of water in a liter varies with depth but a kg is always a kg. 4. Some elements in seawater are considered to be conservative. 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 -, SO4 2-, Br-, B(OH)3 (or B), and F - Approx. 1 point off for a missing or incorrect species. c) Molybdenum (Mo) 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 Mo in each sample and plot versus salinity, potential temperature or some other conservative property. If the points fall on a straight line, Mo could be considered conservative.

2 5. Methane is an important greenhouse gas. Global methane reservoirs and fluxes are shown below (from Bill Reeburgh; a) Is the concentration of atmospheric methane (CH 4 ) at steady state? (5) No. The sources and sinks of methane to the atmosphere are not in balance. Sources = 500 Tg CH 4 y -1 Sinks = 460 Tg CH 4 y -1 (There is a net increase of 40 Tg methane per year in the atmosphere.) b) How does the flux of CH 4 (as Pg C y -1 ) from the ocean to the atmosphere compare with that of CO 2 from Sarmiento and Gruber? (5) Total Methane flux from the ocean = 10 Tg CH4 y-1 (10 Tg CH4 y-1)(12 mass units C/16 mass units CH4) = 7.5 Tg C y-1 (0.001 Pg/1 Tg) = PgC per year Total pre-industrial flux of C in S&G is 70.6 Pg C y-1 The C from CH4 is 0.01% (100*( PgC y-1/70.6 PgC y-1)) of the total preindustrial C flux.

3 6. Trace Metal profiles Two part answers for each. 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) a. Concentrations are depleted in the surface and enriched in the deep ocean. (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.) b. Concentrations are greater in the Pacific (deep water has been out of contact with the surface ocean for a longer amount of time, allowing more time for nutrients to accumulate). ii) hydrothermal inputs (5) a. There would be a mid-depth concentration maximum (around approximately 2500m, or the depth of a nearby ocean ridge crest) b. Differences would not arise strictly due to basin differences or water age, but rather with proximity to hydrothermal vents. iii) scavenging by particles settling through the ocean water column (5) a. Concentrations are high in the surface (have a near surface maximum) and are depleted in the deep ocean. (Elements are adsorbed onto particles that sink through the water column.) b. Concentrations are greater in the Atlantic (deep water has had more time to be scavenged, and therefore concentrations are depleted as water ages).

4 7. 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 Seattle tap water (which is so dilute that it can be considered like distilled water) until the final weight of the solution is 1 kilogram. She had recently taken OCN400 thus her goal is to have the following ionic concentrations based on Slide 4 of the Lecture 2 pwerpoint: Ion mmol kg -1 (seawater solution) Na Mg Ca K + 10 Cl SO 4 28 She has the following chemicals at home. NaCl common spice (Morton Table Salt) MgSO 4 fertilizer, foot soaks (Epsom salts) CaCl 2 drying agent, preservative (Dri-Z-Air) HCl liquid, stolen from lab KOH lye Mg(OH) 2 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 (e.g. for NaCl = (22.99 g/mol Na) + (35.45 g/mol Cl) = g/mol NaCl): NaCl(s) g/mol (or mg/mmol) MgSO4(s) g/mol CaCl2(s) g/mol KOH(s) g/mol Mg(OH)2 (s) g/mol HCl g/mol Now you want to find the mass of each of these compounds needed to make a solution with the given ion concentrations (mmol/kg). Start with the ions that are only in one of the salts: Na, Ca, K, SO4 (469 mmol Na+)(1 mol Na+/1000 mmol)(1mol NaCl/1 mol Na+)(58.44 mg NaCl/mmol) = mg NaCl = g NaCl

5 As a shortcut, we can use equivalent units of mg/mmol for each molecular weight, but the main unit conversion concepts remain: (10 mmol Ca 2+)(110.98mg CaCl2/mmol) = mg CaCl2 = 1.11 g CaCl2 (10 mmol K+)(56.10 mg KOH /mmol) = mg KOH = 0.56 g KOH (28 mmol SO42-)( mg MgSO4/mmol) = mg MgSO4= 3.37 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: 53 mmol Mg2+ 28 mmol Mg2+ from MgSO4 = 25 mmol Mg2+ (25 mmol Mg2+)(58.32 mg Mg(OH)2/mmol) = 1458 mg = 1.46 g Mg(OH)2 Add: 546 mmol Cl- 469 mmol Cl- from NaCl 2(10 mmol Cl- from CaCl2) = 57mmol Cl- (57mmol Cl-)(36.46 mg HCl /mmol) = mg = 2.08 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) Total mass of solution = 1000 g Total mass of salts = 27.41gNaCl 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 g water(1 ml /0.993 g) (1 L/1000 ml) = L Molarity = mols of solute/l solution (469 mmol Na+)(1 mol Na+/1000 mmol)/ L = M (53 mmol Mg2+)(1 mol Mg2+/1000 mmol) )/ L = M (10 mmol Ca2+)(1 mol Ca2+/1000 mmol) )/ L = M (10 mmol K+)(1 mol K+/1000 mmol) )/ L = M (546 mmol Cl-)(1 mol Cl-/1000 mmol) )/ L = M (28 mmol SO42-)(1 mol SO42-/1000 mmol) )/ L = M Including water formed from H and OH is okay. You do end up with excess OH (slightly basic, like seawater) If you assumed that your solution was the same density of the DI water you added, -2 points.