Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Lecture 9: Ocean Carbonate Chemistry:Carbonate ReactionsReactions and equilibrium constants (K)Solutions – numerical graphicalWhat can you measure?Theme 1 (continuation) – Interior Ocean Carbon CycleTheme 2 – Ocean Acidification (man’s alteration of the ocean)Sarmiento and Gruber (2002) Sinks for Anthropogenic CarbonPhysics Today August 2002 30-361Pg = 1015gExamples:Weathering of CaCO3CaCO3(s) + CO2(g) + H2O = Ca2+ + 2 HCO3- 1 + 1 2Weathering of alumino-silicate minerals to clay minerals.silicate minerals + CO2(g) + H2O == clay minerals + HCO3- + 2 H4SiO4 + cation 1 1 2A specific example of orthoclase to kaoliniteKAlSi3O8(s) + CO2(g) + 11/2H2O = 1/2 Al2Si2O5(OH)4(s) + K+ + HCO3- + 2H4SiO4Weathering and River FluxAtmospheric CO2 is converted to HCO3- in rivers and transported to the oceanExample:Global River Flux = River Flow x global average HCO3 concentrationGlobal River Flux = 3.7 x 1016 l y-1 x 0.9 mM = 33.3 x 1012 mol y-1 x 12 g/mol = 0.4 x 1015 g y-1 = 0.4 Pg y-1S&G (2002) give 0.8 Pg y-1 with 0.4 Pg y-1 from weatheringCO2CO2 → H2CO3 → HCO3- → CO32-+ H2O = CH2O + O2BorgC+ Ca2+ = CaCO3BCaCO3AtmOcnBiological PumpControls:pH of oceanSediment diagenesisCO2Gas ExchangeUpwelling/MixingRiver FluxCO2 + rocks = HCO3- + claysOh OhChemistry!CO2 reacts with H2O to make H2CO3CO2 (g) + H2O = H2CO3K’H H2CO3 is a weak acidH2CO3 = H+ + HCO3-K’1 HCO3- = H+ + CO32-K’2 H2O is also a weak acidH2O = H+ + OH-K’W Species n = 6 CO2(g)H2CO3 = carbonic acidHCO3- = bicarbonateCO32- = carbonateH+ = proton or hydrogen ionOH- = hydroxylEquilibrium Constants:4 equilibrium constants in seawater = K’ = f (S,T,P)These are expressed as K'.1. CO2(g) + H2O = H2CO3* (Henry's Law)K’H = [H2CO3*] / PCO2(note that gas concentrations are given as partial pressure; e.g. atmospheric PCO2 = 10-3.5)2. H2CO3* = H+ + HCO3-K’1 = [HCO3-][H+] / [H2CO3*]3. HCO3- = H+ + CO32-K’2 = [H+][CO32-] / [HCO3-]4. H2O = H+ + OH-K’w = [H+][OH-][ ] ConcentrationValues of K’The values here are for S = 35, T = 25C and P = 1 atm.Constant Apparent Seawater Constant (K')K’H10-1.53K’110-6.00K’210-9.10K’w10-13.9K’ vary with T, S and PpHH+ from pH = -log H+ at pH = 6; [H+] = 10-6 OH- from OH- = KW / H+ at pH = 6; [OH-] = 10-8Total CO2 (SCO2 or CT ) – Dissolved Inorganic carbon (DIC)DIC = [H2CO3] + [HCO3-] + [CO32-]Example: If you add reactions what is the K for the new reaction?H2CO3 = H+ + HCO3- K1 = 10-6.0plusHCO3- = H+ + CO32- K2 = 10-9.1------------------------------------------------H2CO3 = 2H+ + CO32- K12 = 10-15.1 Q. At what pH does H2CO3 = CO32-?Example: Say we want the K for the reaction CO32- + H2CO3 = 2 HCO3-Then we have to reverse one of the reactions. Its K will change sign as well!!So:H2CO3 = H+ + HCO3-K = 10-6.0H+ + CO32- = HCO3-K = 10+9.1--------------------------------------------------------------------H2CO3 + CO32- = 2HCO3- K = 103.1  2322 3 3( )( )( )HCOH CO CO--Calculations:Graphical ApproachAlgebraic ApproachConstruct a Distribution Diagram for H2CO3 – Closed Systema. First specify the total CO2 (e.g. CT = 2.0 x 10-3 = 10-2.7 M)b. Locate CT on the graph and draw a horizontal line for that value.c. Locate the two system points on that line where pH = pK1 and pH = pK2.d. Make the crossover point, which is 0.3 log units less than CTe. Sketch the lines for the species (not open to the atmosphere)Table of acids in seawaterElement Reaction mol kg-1-logC pK'H2O H2O = H+ + OH-13.9C H2CO3 = HCO3- + H+2.4 x 10-32.6 6.0HCO3- = CO32- + H+ 9.1B B(OH)3 + H2O = B(OH)4- + H+4.25 x 10-43.37 8.7Mg Mg2+ + H2O = MgOH+ + H+5.32 x 10-21.27 12.5Si H4SiO4 = SiO(OH)3- + H+1.5 x 10-43.82 9.4P H3PO4 = H2PO4- + H+3.0 x 10-65.52 1.6H2PO4- = HPO 2- + H+ 6.0HPO42- = PO43- + H+ 8.6S(VI) HSO4- = SO42- + H+2.82 x 10-21.55 1.5F HF = F- + H+ 5.2 x 10-54.28 2.5Ca Ca2+ + H2O = CaOH+ + H+1.03 x 10-21.99 13.0And in anoxic systemsN NH4+ = NH3 + H+10 x 10-65.0 9.5S(-II)H2S = HS- + H+10 x 10-65.0 7.0HS- = S2- + H+ 13.4pK = -logK(e.g. K’ = 10-13.9)Q. Which is larger? pK = 6.0 or 9.1Q. If K is larger, what does that mean?Carbonic Acid – 6 unknownsCarbonic acid is the classic example of a diprotic acid (it has two H+)and it can have a gaseous form. It also can be expressed as open or closed to the atmosphere (or a gas phase)There are 6 species we need to solve for:CO2(g) Carbon Dioxide GasH2CO3* Carbonic Acid (H2CO3* = CO2 (aq) + H2CO3)HCO3-BicarbonateCO32-CarbonateH+ProtonOH-HydroxideTo solve for six unknowns we need six equations Four of the equations are equilibrium constants!What can you measure?We can not measure these species directly. What we can measure are:a) pH pH is defined in terms of the activity or concentration of H+. Depends on calibration.Written as pH = -log (H+) b) Total CO2 or DICDIC = CT = [H2CO3] + [HCO3-] + [CO32-] c) Alkalinity (defined by the proton balance for a pure solution of CO2)Alkalinity = [HCO3-] + 2[CO32-] + [OH-] - [H+] + [B(OH)4-] + any other bases present The alkalinity is defined as the amount of acid necessary to titrate all the weak bases in seawater (e.g. HCO3-, CO32-, B(OH)4-, NH3) to the alkalinity endpoint which occurs where (H+) = (HCO3-) (see graph). This is about pH = 4.3.d) PCO2The PCO2 in a sample is the PCO2 that a water would have if it were in equilibrium with a gas phase.Carbonate System CalculationsA useful shorthand is the alpha notation, where the alpha (a) express the fraction each carbonate species is of the total DIC. These a values are a function of pH only for a given set of acidity constants. Thus:H2CO3 = ao CTHCO3- = a1 CTCO32- = a2 CTThe derivations of the equations are as follows:ao = H2CO3 / CT = H2CO3 / (H2CO3 + HCO3 + CO3) = 1 / ( 1+ HCO3 / H2CO3 + CO3/H2CO3) = 1 / ( 1 + K1/H + K1K2/H2) = H2 / ( H2 + HK1 + K1K2)The values for a1 and a2 can be derived in a similar manner.a1 = HK1 / (H2 + H K1 + K1K2)a2 = K1K2 / ( H2 + H K1 + K1K2)For example:Assume pH = 8, CT = 10-3, pK1' = 6.0 and pK2' = 9.0[H2CO3*] = 10-5 mol kg-1(note the answer is in concentration because we used K')[HCO3-] = 10-3 mol