Lecture 9: Ocean Carbonate Chemistry: Carbonate Reactions Reactions and equilibrium constants (K) Solutions – numerical graphical What can you measure? Theme 1 (continuation) – Interior Ocean Carbon Cycle Theme 2 – Ocean Acidification (man’s alteration of the ocean)

Sarmiento and Gruber (2002) Sinks for Anthropogenic Carbon Physics Today August 2002 30-36 1Pg = 1015g

Oh Oh Chemistry!

Weathering and River Flux Atmospheric CO2 is converted to HCO3- in rivers and transported to the ocean Examples: Weathering of CaCO3 CaCO3(s) + CO2(g) + H2O = Ca2+ + 2 HCO3- 1 + 1 2 Weathering of alumino-silicate minerals to clay minerals. silicate minerals + CO2(g) + H2O == clay minerals + HCO3- + 2 H4SiO4 + cation 1 1 2 A specific example of orthoclase to kaolinite KAlSi3O8(s) + CO2(g) + 11/2H2O = 1/2 Al2Si2O5(OH)4(s) + K+ + HCO3- + 2H4SiO4 Calculate Global Flux:: Global River Flux = River Flow x global average HCO3 concentration Global 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-1 S&G (2002) give 0.8 Pg y-1 with 0.4 Pg y-1 from weathering

River Flux Gas Exchange BorgC BCaCO3 Biological Pump CO2 + rocks = HCO3- + clays CO2 River Flux Gas Exchange Atm Ocn CO2 → H2CO3 → HCO3- → CO32- Upwelling/ Mixing + H2O = CH2O + O2 + Ca2+ = CaCO3 CO2 BorgC BCaCO3 Biological Pump Controls: pH of ocean Sediment diagenesis

CO2 reacts with H2O to make H2CO3 CO2 (g) + H2O = H2CO3 K’H H2CO3 is a weak acid H2CO3 = H+ + HCO3- K’1 HCO3- = H+ + CO32- K’2 H2O is also a weak acid H2O = H+ + OH- K’W Species n = 6 CO2(g) H2CO3 = carbonic acid HCO3- = bicarbonate CO32- = carbonate H+ = proton or hydrogen ion OH- = hydroxyl

Equilibrium 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-] [ ] Concentration

Values of K’ The values here are for S = 35, T = 25C and P = 1 atm. Constant Apparent Seawater Constant (K') Freshwater Constants (K) K’H 10-1.53 10-1.5 K’1 10-6.00 10-6.3 K’2 10-9.10 10-10.3 K’w 10-13.9 10-14.0 K’ vary with T, S and P See Appendix 4.2 in Emerson and Hedges

Total CO2 (SCO2 or CT ) – Dissolved Inorganic carbon (DIC) pH H+ from pH = -log H+ at pH = 6; [H+] = 10-6 OH- from OH- = KW / H+ at pH = 6; [OH-] = 10-8 Total 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 = 10-6.0 plus HCO3- = 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 = 10-6.0 H+ + CO32- = HCO3- K = 10+9.1 -------------------------------------------------------------------- H2CO3 + CO32- = 2HCO3- K = 103.1 

Calculations: Graphical Approach Algebraic Approach

Construct a Distribution Diagram for H2CO3 – Closed System a. 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 CT e. Sketch the lines for the species (not open to the atmosphere)

Table of acids in seawater pK = -logK Element Reaction mol kg-1 -logC pK' H2O H2O = H+ + OH- 13.9 C H2CO3 = HCO3- + H+ 2.4 x 10-3 2.6 6.0 HCO3- = CO32- + H+ 9.1 B B(OH)3 + H2O = B(OH)4- + H+ 4.25 x 10-4 3.37 8.7 Mg Mg2+ + H2O = MgOH+ + H+ 5.32 x 10-2 1.27 12.5 Si H4SiO4 = SiO(OH)3- + H+ 1.5 x 10-4 3.82 9.4 P H3PO4 = H2PO4- + H+ 3.0 x 10-6 5.52 1.6 H2PO4- = HPO 2- + H+ 6.0 HPO42- = PO43- + H+ 8.6 S(VI) HSO4- = SO42- + H+ 2.82 x 10-2 1.55 1.5 F HF = F- + H+ 5.2 x 10-5 4.28 2.5 Ca Ca2+ + H2O = CaOH+ + H+ 1.03 x 10-2 1.99 13.0 And in anoxic systems N NH4+ = NH3 + H+ 10 x 10-6 5.0 9.5 S(-II) H2S = HS- + H+ 10 x 10-6 5.0 7.0 HS- = S2- + H+ 13.4 (e.g. K’ = 10-13.9) Q. Which is larger? pK = 6.0 or 9.1 Q. If K is larger, what does that mean?

Carbonic Acid – 6 unknowns Carbonic 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 Gas H2CO3* Carbonic Acid (H2CO3* = CO2 (aq) + H2CO3) HCO3- Bicarbonate CO32- Carbonate H+ Proton OH- Hydroxide To 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: 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 DIC DIC = 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-] + other bases present (e.g. DOC) Use Carbonate Alkalinity for calculations (AC). 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 at about pH = 4.3. Filter samples. Cells contribute alkalinity. d) PCO2 The PCO2 in a sample is the PCO2 that a water would have if it were in equilibrium with a gas phase.

Carbonate System Calculations pH and CT Alkalinity and PCO2 A 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 CT HCO3- = a1 CT CO32- = a2 CT The 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 kg-1 [CO32-] = 10-4 mol kg-1 Alk = HCO3 + 2 CO3 + OH - H For this problem neglect H and OH (a good assumption ), then: Alk = CT a1 + 2 CT a2 Alk = CT (a1 + 2a2) We can use this equation if we have a closed system and we know 2 of the 3 variables (Alk, CT or pH). For an open system we can express CT in terms of PCO2 as follows: We know that H2CO3* = CT ao ( you can also use this equation if you know pH and PCO2) But H2CO3 can be expressed in terms of the Henry's Law: KH PCO2 = CT ao So CT = KH PCO2 / ao Now: Alk = (KH PCO2 / ao ) ( a1 + 2a2) Alk = KH PCO2 ( (a1 + 2 a2 ) / ao )) Alk = KH PCO2 ( (HK1 + 2 K1K2 )/ H2 ) Assume: Alk = 10-3 PCO2 = 10-3.5 pK1' = 6.0 pK2' = 9.0 Then: pH = 8.3

CaCO3 solubility calculations CaCO3 = Ca2+ + CO32- K’s0 (calcite) = 4.26 x 10-7 = 10-6.37 K’s0 (aragonite) = 6.46 x 10-7 = 10-6.19 Ion Concentration Product = ICP = [Ca2+][CO32-] or CaCO3 + CO2(g) + H2O = Ca2+ + 2 HCO3- Ion Concentration Product = ICP = [Ca2+][HCO3-]2 / CO2(g) Degree of Saturation (Ω): Omega = Ω = ICP / K’s0 If water at equilibrium (saturation) Ω = 1 If water oversaturated Ω > 1 CaCO3 precipitates There are more products than there should be If water undersaturated Ω <1 CaCO3 dissolves There are not enough products than there should be

What controls the pH of seawater? pH and PCO2 are not conservative. DIC and Alk are capacity factors. pH in seawater is controlled by alkalinity and DIC and can be calculated from these two parameters as shown below. Alk  HCO3- + 2 CO32- Alk  CT a1 + 2 CT a2 = CT (a1 + 2a2 ) Alk = CT (HK1' + 2 K1' K2' ) / (H2 + H K1' + K1'K2') Rearranging, we can calculate pH from Alk and CT (use the quadratic equation) (H+) = -K1' (Alk-CT) + [(K1')2 (Alk-CT)2 - 4 Alk K1' K2' (Alk - CT)]  / 2 Alk So the question boils down to what controls alkalinity and total CO2. Internal variations of pH in the ocean and controlled by internal variations in DIC and alkalinity which are controlled by photosynthesis, respiration and CaCO3 dissolution and precipitation. The long term controls on alkalinity and DIC are the balance between the sources and sinks and these are the weathering (sources) and burial (sinks) of silicate and carbonate rocks and organic matter.

For laptops For iPhone http://cdiac.ornl.gov/oceans/co2rprt.html Program Developed for CO2 System Calculations CO2calc: A User-Friendly Seawater Carbon Calculator for Windows, Mac OS X, and iOS (iPhone) By L.L. Robbins, M.E. Hansen, J.A. Kleypas, and S.C. Meylan http://cdiac.ornl.gov/oceans/co2rprt.html

Guam Shinjo et al Marine Geology

Shinjo et al (2013) Marine Geology

The Proton Balance The balance of species that have excess protons to species deficient in protons relative to a stated reference species. Reference species Proton Balance For H2CO3: H+ = HCO3- + 2 CO32- + OH- For HCO3-: H+ + H2CO3 = CO32- + OH- For CO32-: H+ + 2 H2CO3 + HCO3- = OH- The proton conditions define three equivalence points on the graph and these are used to define 6 capacity factors for the solution. You can approach each equivalence point from either the acid or base direction. If you add strong acid (e,g, HCl ) it is represented as CA Strong base (e.g. NaOH) is represented as CB. For Example: For a pure solution of H2CO3: CB + H+ = HCO3- + 2 CO32- + OH- + CA Then: CB – CA = HCO3- + 2CO32- + OH- - H+ = Alkalinity CA – CB = H+ - HCO3- + 2CO32- + OH- = H+-Acidity

Open System - Gas Solubility – Henry’s Law The exchange or chemical equilibrium of a gas between gaseous and liquid phases can be written as: A (g) ===== A (aq) At equilibrium we can define K = [A(aq)] / [A(g)] Henry's Law: We can express the gas concentration in terms of partial pressure using the ideal gas law: PV = nRT so that [A(g)] is equal to the number of moles n divided by the volume n/V = [A(g)] = PA / RT where PA is the partial pressure of A Then K = [A(aq)] / PA / RT or [A(aq)] = (K/RT) PA [A(aq)] = KH PA units for K are mol kg-1 atm-1; for PA are atm in mol kg-1 Henry's Law states that the solubility of a gas is proportional its overlying partial pressure.

Example: Gas concentrations in equilibrium with the atmosphere Atmosphere Composition Gas Mole Fraction in Dry Air (fG) (where fG = moles gas i/total moles) N2 0.78080 O2 0.20952 Ar 9.34 x 10-3 CO2 3.3 x 10-4 Gas Pi KH (0C , S = 35) Ci (0C, S = 35; P = 1 Atm N2 0.7808 0.80 x 10-3 62.4 x 10-3 mol kg-1 O2 0.2095 1.69 x 10-3 35.4 x 10-3 Ar 0.0093 1.83 x 10-3 0.017 x 10-3 CO2 0.00033 63 x 10-3 0.021 x 10-3

Open System Distributions Assume equilibrium with a constant composition gas phase with PCO2 = 10-3.5