Mid-Term #1 (125 points total)

Transcription

1 Ocean 520 Name: Chemical Oceanography 20 October 2009 Fall 2009 Points are in parentheses (show all your work) (use back if necessary) MidTerm #1 (125 points total) 1. Doney et al (2008) Ocean Acidification The surface ocean is becoming more acidic due to addition of anthropogenic CO 2. Preindustrial surface seawater ph was Currently the surface seawater ph is a) Calculate how much more [H + ] there is now relative to preindustrial times. Give the numerical concentration. (5) ph 8.21 : H + = = x 10 9 = 6.16 x 10 9 ph 8.10; H + = 7.94 x 10 9 The difference is ( ) x 10 9 = 1.78 x 10 9 ( mol H + / l) ( mol H + / l) = 1.78 x 10 9 mol H + / l = mol H + / l b) By what percentage has the [H + ] concentration changed since preindustrial times (relative to preindustrial times)? (5) [( mol H + / l) / ( mol H + / l)] * 100% = 1.78/6.16 = 28.8% 2. Ocean carbonate reactions Ocean acidification will make calcification by organisms more difficult. a) The concentration of bicarbonate ion (HCO 3 ) is important to know for solubility reactions. Calculate the concentration of HCO 3 in surface seawater if you know the DIC and ph. (10) Assume the apparent equilibrium constants for seawater of K 1 = and K 2 = Assume DIC = x 10 3 and ph = 8.0 DIC = [H 2 CO 3 ] + [HCO3 ] + [CO3 2 ] [H + ][HCO 3 ] = K 1 [H + ][CO 3 2 ] = K 2 [H 2 CO 3 ] [HCO 3 ] [H 2 CO 3 ] = [H + ][HCO 3 ] [CO 3 2 ] = K 2 *[HCO 3 ] K 1 [H + ] DIC = [H + ][ HCO 3 ] + [HCO 3 ] + K 2 *[ HCO 3 ] K 1 [H + ] x 10 3 = (10 8 )*[ HCO 3 ] + [HCO 3 ] + (10 9 )*[ HCO 3 ] x 10 3 = 0.01*[ HCO 3 ] + [HCO 3 ] + 0.1*[ HCO 3 ] = 1.11*[ HCO 3 ] [HCO 3 ] = (2.000 x 10 3 ) / 1.11 = 1.8 x 10 3 mol / l = x 10 3 = or use HCO 3 = C T α 1

2 b) We have learned the following equilibrium reactions and constants for seawater. 1. CO 2 (g) + H 2 O H 2 CO 3 K H = H 2 CO 3 H + + HCO 3 K 1 = HCO 3 H CO 3 K 2 = H 2 O OH + H + K w = CaCO 3 (aragonite) Ca CO 3 K sp = We can write the solubility of CaCO 3 in terms of HCO 3 and CO 2 (g) as follows: CaCO 3 + CO 2 (g) + H 2 O = Ca HCO 3 Derive the equilibrium constant (K ) for this reaction using the values for K H, K 1, K 2 and K s. (10) CO 2 (g) + H 2 O H 2 CO 3 K H = H 2 CO 3 H + + HCO 3 K 1 = H + + CO 2 3 HCO 3 1/K 2 = CaCO 3 (aragonite) Ca +2 + CO 2 3 K sp = CaCO 3(aragonite) + CO 2 (g) + H 2 O = Ca HCO 3 K = K H * K 1 * K sp *(1/K 2 ) K = ( )*( )*( )*( ) = = x 10 5 = 2.04 x 10 5 c) Calculate the equilibrium concentration of HCO 3 in equilibrium with aragonite in seawater assuming we know the concentration of Ca 2+ in seawater and P CO2 in the atmosphere. (10) Assume that the concentration of Ca 2+ = 10.0 x 10 3 M and P CO2 = 380 ppm CaCO 3(aragonite) + CO 2 (g) + H 2 O = Ca HCO 3 K = K = [Ca 2+ ][ HCO 3 ] 2 [HCO 3 ] 2 = K* P CO2 380ppm = 380 x 10 6 P CO2 [Ca 2+ ] = x 10 4 [HCO 3 ] 2 = ( M 2 atm 1 )*( 380 x 10 6 atm) = 7.76 x 10 7 M (10.0 x 10 3 M ) [HCO 3 ] = 8.80 x 10 4 M = x 10 4 = =

3 d) Is surface seawater supersaturated, at equilibrium or undersaturated with respect to aragonite. Explain using the Ω ratio (degree of saturation). (10) From our calculation in part a we know that the concentration of HCO 3 is 1.8 x 10 3 mol / l. In part c we calculated the concentration of HCO 3 necessary for equilibrium with aragonite in seawater which gave us an equilibrium concentration of HCO 3 equal to 8.80 x 10 4 M. HCO 3 in sw HCO 3 in equil (1.8 x 10 3 mol / l) > (8.80 x 10 4 mol / l) Therefore the surface water is super saturated with respect to aragonite. In other words our ion activity product is larger than our K value and Ω is greater than 1. IAP = [Ca 2+ ][ HCO 3 ] 2 Ω = IAP = = = 4.07 P CO2 K = 10 2 (1.8 x 10 3 ) 2 / Ω > 1 so seawater is supersaturated = 10 2 ( ) 2 / = / =

4 3. Gas Exchange Is beer carbonated? a) The ph of your favorite, frosty, carbonated beverage is 5.8 and the HCO 3 content is 2.0 x 10 3 mole liter 1. What is the partial pressure of CO 2 in the can before you open it? Let K 1 = and K H = (10) Write a reaction including both CO 2, HCO 3 and H +. CO 2 (g) + H 2 O H 2 CO 3 K H = H 2 CO 3 H + + HCO 3 K 1 = CO 2 (g) + H 2 O H + + HCO 3 K = K H * K 1 = K = [H + ][ HCO 3 ] rearrange to: P CO2 = [H + ][ HCO 3 ] P CO2 K P CO2 = ( M)*( 2.0 x 10 3 M) = 100 x 10 3 atm = atm M 2 atm 1 b) Calculate the flux of CO 2 (in mol s 1 m 2 ) for your favorite, frosty, carbonated beverage. Assume the surface of the beverage is in equilbrium with the P CO2 of the atmosphere (385 x 10 6 atm.). Let the molecular diffusion coefficient be D CO2 = 2 x 10 9 m 2 s 1 and let the stagnant boundary layer thickness be Z film = 5 x 10 5 m. Which way does the CO 2 flux go? (10) [H 2 CO 3 ] surface = K H P CO2 = M atm 1 * 385 x 10 6 atm = x 10 5 M [H 2 CO 3 ] beer = K H P CO2 = M atm 1 * 100 x 10 3 atm = 3.16 x 10 3 M d[h 2 CO 3 ] = [H 2 CO 3 ] surface [H 2 CO 3 ] beer = 3.15 x 10 3 M F = D(d[H 2 CO 3 ]) / dz F = (2 x 10 9 m 2 s 1 )*( 3.15 x 10 3 M) / (5 x 10 5 m) = 1.26 x 10 7 m mol s 1 l 1 F = (1.26 x 10 7 m mol s 1 l 1 )*(1000 l m 3 ) = 1.26 x 10 4 mol s 1 m 2 The flux is positive so it is out of the beer and into the atmosphere.

5 4. Calcite (CaCO 3 (s)) is the main carbonate mineral in modern marine sediments. Ancient marine rocks contain mostly dolomite (CaMg(CO 3 ) 2 (s)). The reaction of calcite to dolomite is written as: 2CaCO 3 (s) + Mg 2+ = CaMg(CO 3 ) 2 (s) + Ca 2+ a) Calculate the equilibrium constant for this reaction at 25 C given the following free energies of formation ( G f ) and using G r = 2.3 RT log K = log K at 25 C (10 pts) Species G f (kj mol 1 ) CaCO 3 (s) CaMg(CO 3 ) 2 (s) Mg Ca Gr = ( ) ((2 x ) ) = 281 kj mol 1 Use Gr = log K Log K = 284 / = b) Write the corresponding equilibrium constant expression ( 5 pts) K = (CaMg(CO 3 ) 2 (s) (Ca 2+ ) / CaCO 3 (s) 2 (Mg 2+ ) = c) Assuming that seawater contains total concentrations of [Ca] = 1.0 x 10 2 mol/l and [Mg] = 5.3 x 10 2 mol/l what is the stable mineral phase, calcite or dolomite. (10 pts) Q = [Ca 2+ ] %free γ Ca2+ / [Mg 2+ ] %free γ Mg2+ = (10 2 ) (0.25) / 5.3 x 10 2 ) (0.31) = = so Q < K The reaction will go to the right. Dolomite is favored as the stable mineral phase.

6 5. The Table below gives the world average composition of river water. a) Correct the concentrations of Na, Mg, SO4, K and Ca for seasalt aerosols using Cl as a tracer? What assumptions do you need to make? (5) To correct the concentrations for sea salt aerosols, we need to subtract the amount of each ion contributed by deposition of sea salt aerosols, since these do not derive directly from weathering of rocks on land. If we assume that: (1) all Cl in rivers comes from sea salt aerosols, and (2) that sea salt aerosols have a composition with proportions identical to seawater, then we can use Cl as a tracer of the ions contributed by deposition of sea salt aerosols by multiplying the river concentration of Cl by the [X]/[Cl] ratio in seawater, where [X] is the concentration of Na, Mg, SO4, K or Ca: [X] aerosol = ([X] seawater /[Cl ] seawater )*[Cl ] river To get the corrected concentration [X*], we just need to subtract the aerosol component we calculated above from the total river concentration of each ion: [X*] = [X] river [X] aerosol Corrected concentrations: Element [X] river (mm) [X] seawater (mm) [X]/[Cl] seawate [X] aerosol [X]* (mm) r (mm) Cl Na Mg SO K Ca

7 Assumptions: We have assumed that: (1) all Cl in rivers comes from sea salt aerosols (this is probably not the case, as some Cl is contributed by weathering of evaporite rocks; see McDuff and Morel, p in coursepack). (2) sea salt aerosols have a composition with proportions identical to seawater (probably a good assumption). Propose an approach to calculate how much CO 2 (atm) is consumed by weathering of b) carbonate rocks (CaCO 3 ) (5 points) c) silicate rocks (5 points) hint: Assume all Si comes from silicate rocks. Calculate the HCO 3 that weathering produces. Then calculate the HCO 3 that comes from weathering of carbonate rocks. As a check, if all the Ca comes from weathering CaCO 3 then Ca should equal 0.5 x HCO 3 from carbonate weathering. You will find that it doesn t. The reason is that some carbonate rocks are MgCO 3 or CaMg(CO 3 ) 2. Approach: Start with the two weathering reactions we are considering: Carbonate rock weathering: CaCO 3 (s) + CO 2 (g) + H 2 O = Ca HCO 3 Note: 2 moles of HCO 3 and 1 mole of Ca 2+ are produced by the reaction of 1 mole of atmospheric CO 2 and 1 mole of CaCO 3. Silicate mineral weathering (using Kfeldspar as an example): KAlSi 3 O 8 (s) + CO 2 (g) + 5/2 H 2 O = ½ Al 2 Si 2 O 5 (OH) 4 (s) + K + + HCO3 o + 2H 4 SiO 4 (Kfeldspar) (kaolinite) Note: 1 mole of HCO 3 and 2 moles of H 4 SiO 4 o is produced by the reaction of 1 mole of atmospheric CO 2 (g) and 1 mole of Kfeldspar. (H 4 SiO 4 o is equivalent to SiO 2 reported in the table) Step 1. Assume all silica comes from silicate rocks. Then we can use the given concentration of SiO 2 (0.218 mmol/l) to calculate the riverine flux of silica: River Flux SiO 2 = River flow rate*[sio 2 ] river = (3.7 x L y 1 )*(0.218 mmol/l) River Flux SiO 2 = mmol/y According to the equation above, 1 mole atmospheric CO 2 is consumed for every 2 moles of silica produced, so: (River SiO 2 flux)/2 = Atmospheric CO 2 consumed by silica weathering (0.807 x mmol/y)/2 = x mmol/y Step 2. Determine the bicarbonate flux due to silicate weathering. Note 1 mole equivalent of HCO 3 is produced by the silicate weathering reaction, so we know that the bicarbonate flux due to silicate weathering is also equal to x mmol/y. Step 3. Assume all river bicarbonate is due to silicate and carbonate weathering combined. If this is true, then the bicarbonate flux due to carbonate weathering is equal to the total river bicarbonate flux minus the bicarbonate flux due to silicate weathering.

8 Total river bicarbonate flux = River flow rate*[hco 3 ] river = = (3.7 x L y 1 )*(0.958 mmol/l) = x mmol/yr HCO 3 flux due to carbonate weathering = (Total river HCO 3 flux) (HCO 3 flux due to silicate weathering) = x mmol/yr x mmol/y = x mmol/y 4. From the bicarbonate flux due to carbonate weathering, calculate the atmospheric CO 2 flux due to carbonate weathering. From the equation above, we can see that for every 2 moles of bicarbonate produced by carbonate weathering, 1 mole of atmospheric CO 2 is consumed. So the flux of CO 2 due to bicarbonate weathering is half that of the HCO 3 flux: Atm. CO 2 consumed by carbonate weathering = (HCO 3 flux due to carbonate weathering)/2 = (3.141 x mmol/y)/2 = x mmol/y 5. Check: Does the corrected Ca 2+ flux equal 0.5 x HCO 3 from carbonate weathering? Ca flux = River flow rate*[ca 2+ ] river = 3.7 x L y 1 * mmol/l = x mmol/y The Ca 2+ flux, at x mmol/y, is smaller than 0.5 x HCO 3 flux due to CaCO 3 weathering, at x mmol/y. The 0.2 mmol/y difference is due to some carbonate rocks being MgCO 3 or CaMg(CO 3 ) 2. d) From your model, what is the total amount of CO 2 consumed by weathering (in PgC y 1 )? Use a total river flow of 3.7 x L y 1. (5 points) To find the total atmospheric CO 2 consumed by weathering, add the carbonate and silicate weathering fluxes together: Atm. CO 2 consumed by carbonate + silicate weathering = x mmol/y x mmol/y = x mmol/y = x mol/y Convert this value to Pg/y: 1 Pg = g x mol CO 2 /y * 1 mol C * g C = x g x 1 Pg = 0.24 Pg/y 1 mol CO2 1 mol C g e) How does your rate compare with that of Sarmiento and Gruber (2002) as shown in their Figure 1. (5 points) Sarmiento and Gruber give 0.2 Pg/y as their estimate of atmospheric CO2 consumed by weathering. The rate calculated by this model is larger by 20% at 0.24 Pg/y, but on the same order of magnitude. f) How does the weathering ratio of CaCO 3 /silicate rocks compare with the outcrop abundance of carbonate sedimentary rocks to silicate rocks. See attached table from Meybeck (1987, American Journal of Science, 287, ). If the weathering ratio differs from the outcrop ratio, suggest why. (5 points) The weathering ratio of CaCO 3 /silicate rocks is: 1.571/0.404 = ~4:1 In the attached table, CaCO 3 composes ~16% of outcrop abundance. Rocks with silicate minerals compose nearly all the rest of rocks (except for evaporites) with an abundance of ~83%. Therefore the outcrop ratio of CaCO 3 /silicate rocks = ~1:5, approximately the

9 inverse of the weathering ratio. Silicate rocks are much more abundant than carbonate rocks. This is because carbonate rocks weather more quickly than silicate rocks. From Meybeck (1987) American Journal of Science, 287,