1. Lecture 8: Ocean Carbonate Chemistry: Carbonate Reactions Reactions Solutions – numerical graphical K versus K’ What can you measure? Theme 1 (continuation) – Interior Ocean Carbon Cycle Theme 2 – Ocean Acidification (man’s alteration of the ocean)

2. Sarmiento and Gruber (2002) Sinks for Anthropogenic Carbon Physics Today August 2002 30-36 1Pg = 10 15 g

3. CO 2 CO 2 → H 2 CO 3 → HCO 3 - → CO 3 2- + H 2 O = CH 2 O + O 2 B orgC + Ca 2+ = CaCO 3 B CaCO3 Atm Ocn Biological Pump Application: Paleooceanography Controls: pH of ocean Sediment diagenesis CO 2 Gas Exchange Upwelling/ Mixing River Flux CO 2 + rocks = HCO 3 - + clays

4. Examples: Weathering of CaCO 3 CaCO 3 (s) + CO 2 (g) + H 2 O = Ca 2+ + 2 HCO 3 - 1 2 Weathering of alumino-silicate minerals to clay minerals. silicate minerals + CO 2 (g) + H 2 O == clay minerals + HCO 3 - + 2 H 4 SiO 4  + cation 1 1 2 A specific example of orthoclase to kaolinite KAlSi 3 O 8 (s) + CO 2 (g) + 11/2H 2 O = 1/2 Al 2 Si 2 O 5 (OH) 4 (s) + K + + HCO 3 - + 2H 4 SiO 4  Weathering and River Flux Atmospheric CO 2 is converted to HCO 3 - in rivers and transported to the ocean Example: Global River Flux = River Flow x global average HCO 3 concentration Global River Flux = 3.7 x 10 16 l y -1 x 0.9 mM = 33.3 x 10 12 mol C y -1 x 12 g mol -1 = 396 x 10 12 g C y -1 = 0.4 Pg C y -1

5. What happens to the CO 2 that dissolves in water? CO 2 is taken up by ocean biology to produce a flux of organic mater to the deep sea (B orgC ) CO 2 + H 2 O = CH 2 O + O 2 Some carbon is taken up to make a particulate flux of CaCO 3 (B CaCO3 ) Ca 2+ + 2HCO 3 - = CaCO 3 (s) + CO 2 + H 2 O The biologically driven flux is called the “Biological Pump”. The sediment record of B orgC and B CaCO3 are used to unravel paleoproductivity. The flux of B orgC to sediments drives an extensive set of oxidation-reduction reactions that are part of sediment diagenesis. Carbonate chemistry controls the pH of seawater which is a master Variable for many geochemical processes.

6. Bronsted Concept of Acids Acids are compounds that can donate a proton to another substance which is a proton acceptor. Acid1 = Base1 + proton Proton + Base2 = Acid 2 --------------------------------------- Acid1 + Base2 = Acid2 + Base1 Examples: HCl + H 2 O = H 3 O + + Cl - H 2 O + H 2 O = H 3 O + + OH - H 2 CO 3 + H 2 O = H 3 O + + HCO 3 - To simplify we will write acids as Arrhenius Acids where acids react to produce excess H + in solution: HCl = H + + Cl - Free H + don’t exist in solution

7. Table of acids in seawater ElementReactionmol kg -1 -logCpK' H 2 OH 2 O = H + + OH - 13.9 CH 2 CO 3 = HCO 3 - + H + 2.4 x 10 -3 2.6 6.0 HCO 3 - = CO 3 2- + H + 9.1 BB(OH) 3 + H 2 O = B(OH) 4 - + H + 4.25 x 10 -4 3.37 8.7 MgMg 2+ + H 2 O = MgOH + + H + 5.32 x 10 -2 1.27 12.5 SiH 4 SiO 4 = SiO(OH) 3 - + H + 1.5 x 10 -4 3.82 9.4 PH 3 PO 4 = H 2 PO 4 - + H + 3.0 x 10 -6 5.52 1.6 H 2 PO 4 - = HPO 2- + H + 6.0 HPO 4 2- = PO 4 3- + H + 8.6 S(VI)HSO 4 - = SO 4 2- + H + 2.82 x 10 -2 1.55 1.5 FHF = F - + H + 5.2 x 10 -5 4.28 2.5 CaCa 2+ + H 2 O = CaOH + + H + 1.03 x 10 -2 1.99 13.0 And in anoxic systems NNH 4 + = NH 3 + H + 10 x 10 -6 5.0 9.5 S(-II)H 2 S = HS - + H + 10 x 10 -6 5.0 7.0 HS - = S 2- + H + 13.4 pK = -logK (e.g. K’ = 10 -13.9 ) Q. Which is larger? pK = 6.0 or 9.1

8. CO 2 reacts with H 2 O to make H 2 CO 3 CO 2 (g) + H 2 O = H 2 CO 3 K H H 2 CO 3 is a weak acid H 2 CO 3 = H + + HCO 3 - K 1 HCO 3 - = H + + CO 3 2- K 2 H 2 O is also a weak acid H 2 O = H + + OH - K W H 2 CO 3 = carbonic acid HCO 3 - = bicarbonate CO 3 2- = carbonate H + = proton or hydrogen ion OH - = hydroxyl

9. Equilibrium Constants on Different Scales Equilibrium constants can be defined on the infinite dilution activity scale and the ionic medium scale Equilibrium constants can be defined differently on these two scales. There are both advantages and disadvantages of the infinite dilution and ionic medium approaches.

10. Consider the following reaction: HA = H + + A - We define the equilibrium constants as follows. A.On the infinite dilution scale – K is expressed in terms of activities. Calculated from  G r  K = (H + )(A - ) / (HA) B. On the ionic medium scale the equilibrium constant (K') is defined in terms of concentrations in the ionic medium of interest. Measured. K' = [H + ] [A - ] / [HA] Activities = () Concentrations = [ ]

11. infinite dilution scale (K) pros K can be calculated from  G r  One K for all solutions cons Need to express concentration as activity using free ion activity coefficients (  i ) and %free (f i ) ionic medium scale (K’) pros Use concentrations. Don’t need to assume activity coefficients. When K’s have been determined, they are usually very well known. cons K’s need to have been determined experimentally as function of T, P and S.

12. When pH is measured as the activity of (H + ), as it is commonly done, the mixed constant is defined in terms of (H + ). K' = (H + ) [A - ] / [HA]

13. The difference between K and K' is the ratio of the total activity coefficients. K' = (H + ) [A - ] / [HA] K = (H + )(A - ) / (HA) K = (H + )[A - ]  i f i / [HA]  i f i K = (H + )[A - ]/[HA] x  i f i /  i f i K = K'  A /  HA

14. Values of K and K’ Representative values for these constants are given below. Equations are given in Millero (1995) with which you can calculate all K's for any salinity and T, P conditions. The values here are for S = 35, T = 25  C and P = 1 atm. Constant Thermodynamic Constant (K)Apparent Seawater Constant (K') K H 10 -1.47 10 -1.53 K 1 10 -6.35 10 -6.00 K 2 10 -10.33 10 -9.10 K w 10 -14.0 10 -13.9 S = 0 S = 35

15. Example of calculating  T from K and K’ The difference between K and K' can be illustrated by this simple example. K = (H + )(A - ) / (HA) = (H + )[A - ]  T,A- / [HA]  T,HA Rearrange to get: K = (H + ) [A - ]  T,A- = K'  T,A- [HA]  T,HA  T,HA The difference in magnitude of K and K' is the ratio of the total activity coefficients of the base to the acid. If you know both K and K', you can learn something about the activity corrections. For:H 2 CO 3 = HCO 3 - + H + K 1 = (HCO 3 - )(H + ) / (H 2 CO 3 ) or K 1 = [HCO 3 - ]  T,HCO3 (H + ) [H 2 CO 3 ]  T,H2CO3 or K 1 = [HCO 3 ](H + )  T,HCO3 = 10 -6.3 from tables of  G r  at 25  C and 1 atm [H 2 CO 3 ]  T,H2CO3 The value of K' has been determined for the same reaction. At S = 35, 25  C and 1 atm K1' = [HCO 3 -] (H + ) = 10 -6.0 [H 2 CO 3 ] If we set K 1 = K 1 '  T,HCO3 /  T, H2CO3 We can solve for  T,HCO3 /  T, H2CO3 = K 1 / K 1 ' = 10 -6.3 / 10 -6.0 = 10 -0.3 = 5.0 x 10 -1

16. We can compare this ratio with that obtained from the Garrels and Thompson (1962) speciation model of surface seawater where:  T,HCO3 /  T, H2CO3 = f HCO3  HCO3 = (0.69)(0.68) = 4.1 x 10 -1 f H2CO3  H2CO3 (1.00)(1.13) Not too bad. Not too good. These two estimates differ by about 20%. Is that good enough? The comparison is not so good for K 2 and K 2 ’ K vs K’ Example, continued

17. Apparent Equilibrium Constants: 4 equilibrium constants in seawater = K’ = f (S,T,P) These are expressed as K'. 1. CO 2 (g) + H 2 O = H 2 CO 3 * (Henry's Law) K’ H = [H 2 CO 3 * ] / P CO2 (note that gas concentrations are given as partial pressure; e.g. atmospheric P CO2 = 10 -3.5 ) 2. H 2 CO 3 * = H + + HCO 3 - K’ 1 = [HCO 3 - ](H + ) / [H 2 CO 3 * ] 3. HCO 3 - = H + + CO 3 2- K’ 2 = (H + )[CO 3 2- ] / [HCO 3 - ] 4. H 2 O = H + + OH - K’ w = (H + )(OH - ) ( ) Activity = effective concentration [ ] Concentration

18. Values of K’ The values here are for S = 35, T = 25  C and P = 1 atm. Constant Apparent Seawater Constant (K') K’ H 10 -1.53 K’ 1 10 -6.00 K’ 2 10 -9.10 K’ w 10 -13.9

19. pH H + from pH = -log H+ at pH = 6; H + = 10 -6 OH - from OH - = K W / H + at pH = 6; OH - = 10 -8 Total CO 2 (  CO 2 or C T ) – Dissolved Inorganic carbon (DIC) DIC = [H 2 CO 3 ] + [HCO 3 - ] + [CO 3 2- ] defined as concentrations!

20. Example: If you add reactions what is the K for the new reaction? H 2 CO 3 = H + + HCO 3 - K 1 = 10 -6.0 plus HCO 3 - = H + + CO 3 2- K 2 = 10 -9.1 ------------------------------------------------ H 2 CO 3 = 2H + + CO 3 2- K 12 = 10 -15.1 Example: Say we want the K for the reaction CO 3 2- + H 2 CO 3 = 2 HCO 3 - Then we have to reverse one of the reactions. Its K will change sign as well!! So: H 2 CO 3 = H + + HCO 3 - K = 10 -6.0 H + + CO 3 2- = HCO 3 - K = 10 +9.1 -------------------------------------------------------------------- H 2 CO 3 + CO 3 2- = 2HCO 3 - K = 10 3.1 

21. Calculations: Graphical Approach Algebraic Approach

22. Construct a Distribution Diagram for H 2 CO 3 – Closed System a. First specify the total CO 2 (e.g. C T = 2.0 x 10 -3 = 10 -2.7 M) b. Locate C T on the graph and draw a horizontal line for that value. c. Locate the two system points on that line where pH = pK 1 and pH = pK 2. d. Make the crossover point, which is 0.3 log units less than C T e. Sketch the lines for the species

23. The Proton Balance The balance of species that have excess protons to species deficient in protons relative to a stated reference level. Reference Levels Proton Balance For H 2 CO 3 /H 2 O H + = HCO 3 - + 2 CO 3 2- + OH - For HCO 3 - /H 2 O H + + H 2 CO 3 = CO 3 2- + OH - For CO 3 2- /H 2 O H + + 2 H 2 CO 3 + HCO 3 - = OH - The proton conditions define three equivalance points on the graph (see circles) and these are used to define 6 capacity factors for the solution. You can titrate to each equivalence point from either the acid or base direction. If you add strong acid (e,g, HCl ) it is represented as C A Strong base (e.g. NaOH) is represented as C B. For Example: For a pure solution of H 2 CO 3 : C B + H + = HCO 3 - + 2 CO 3 2- + OH - + C A Then: C B – C A = HCO 3 - + 2CO 3 2- + OH - - H + = Alkalinity C A – C B = H + - HCO 3 - + 2CO 3 2- + OH - = H + -Acidity

24. 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)] = P A / RT where P A is the partial pressure of A Then K = [A(aq)] / P A / RT or [A(aq)] = (K/RT) P A [A(aq)] = K H P A units for K are mol kg -1 atm -1 ; for P A are atm in mol kg -1 Henry's Law states that the solubility of a gas is proportional its overlying partial pressure.

25. Open System Distributions Assume equilibrium with a constant composition gas phase with P CO2 = 10 -3.5 H 2 CO 3 = K H x P CO2 = 10 -1.5 x 10 -3.5 = 10 -5.0

26. Carbonic Acid – 6 unknowns Carbonic acid is the classic example of a diprotic acid 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: CO 2 (g)Carbon Dioxide Gas H 2 CO 3 * Carbonic Acid (H 2 CO 3 * = CO 2 (aq) + H 2 CO 3 ) HCO 3 - Bicarbonate CO 3 2- Carbonate H + Proton OH - Hydroxide To solve for six unknowns we need six equations

27. 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 of H + or as pH = -log (H + ) b) Total CO 2 C T = [H 2 CO 3 ] + [HCO 3 - ] + [CO 3 2- ] c) Alkalinity Alkalinity = [HCO 3 - ] + 2[CO 3 2- ] + [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. HCO 3 -, CO 3 2-, B(OH) 4 - ) to the alkalinity endpoint which occurs where (H + ) = (HCO 3 - ) (see graph) d) P CO2 The P CO2 in a sample is the P CO2 that a water would have if it were in equilibrium with a gas phase

28. Carbonate System Calculations A useful shorthand is the alpha notation, where the alpha (  ) express the fraction each carbonate species is of the total DIC. These  values are a function of pH only for a given set of acidity constants. Thus: H 2 CO 3 =  o C T HCO 3 - =  1 C T CO 3 2- =  2 C T The derivations of the equations are as follows:  o = H 2 CO 3 / C T = H 2 CO 3 / (H 2 CO 3 + HCO 3 + CO 3 ) = 1 / ( 1+ HCO 3 / H 2 CO 3 + CO 3 /H 2 CO 3 ) = 1 / ( 1 + K 1 /H + K 1 K 2 /H 2 ) = H 2 / ( H 2 + HK 1 + K 1 K 2 ) The values for  1 and  2 can be derived in a similar manner.  1 = HK 1 / (H 2 + H K 1 + K 1 K 2 )  2 = K 1 K 2 / ( H 2 + H K 1 + K 1 K 2 ) For example: Assume pH = 8, C T = 10 -3, pK 1 ' = 6.0 and pK 2 ' = 9.0 [H 2 CO 3 *] = 10 -5 mol kg -1 (note the answer is in concentration because we used K') [HCO 3 - ] = 10 -3 mol kg -1 [CO 3 2- ] = 10 -4 mol kg -1 pH and C T Alkalinity and P CO2 Alk = HCO 3 + 2 CO 3 + OH - H For this problem neglect H and OH (a good assumption ), then: = C T  1 + 2 C T  2 = C T (  1 + 2  2 ) We can use this equation if we have a closed system and we know 2 of the 3 variables (Alk, C T or pH). For an open system we can express C T in terms of P CO2 as follows: We know that H 2 CO 3 * = C T  o ( you can also use this equation if you know pH and P CO2 ) But H 2 CO 3 can be expressed in terms of the Henry's Law: K H P CO2 = C T  o So C T = K H P CO2 /  o Now: Alk = (K H P CO2 /  o ) (  1 + 2  2 ) Alk = K H P CO2 ( (  1 + 2  2 ) /  o ) Alk = K H P CO2 ( HK 1 + 2 K 1 K 2 / H 2 ) Assume: Alk = 10 -3 P CO2 = 10 -3.5 pK 1 ' = 6.0 pK 2 ' = 9.0 Then: pH = 8.3

29. CaCO 3 solubility calculations CaCO 3 = Ca 2+ + CO 3 2- K s0 (calcite) = 4.26 x 10 -7 = 10 -6.37 K s0 (aragonite) = 6.46 x 10 -7 = 10 -6.19 or CaCO 3 + CO 2 + H 2 O = Ca 2+ + 2 HCO 3 - Ion Concentration Product = ICP = [Ca 2+ ][CO 3 2- ] Omega = Ω = ICP / K’ s0 If water at equilibrium (saturation) Ω = 1 If water oversaturated Ω > 1 CaCO 3 precipitates If water undersaturated Ω <1 CaCO 3 dissolves

30. What controls the pH of seawater? pH in seawater is controlled by alkalinity and DIC and can be calculated from these two parameters as shown below. Alk  HCO 3 - + 2 CO 3 2- Alk  C T  1 + 2 C T  2 Alk = C T (HK 1 ' + 2 K 1 ' K 2 ' ) / (H 2 + H K 1 ' + K 1 'K 2 ') Rearranging, we can calculate pH from Alk and C T. (H+) =  -K 1 ' (Alk-C T ) + [(K 1 ') 2 (Alk-C T ) 2 - 4 Alk K 1 ' K 2 ' (Alk - 2C T )]  1/2 / 2 Alk So the question boils down to what controls alkalinity and total CO 2. Internal variations of pH in the ocean and controlled by internal variations in DIC and alkalinity which are controlled by photosynthesis, respiration and CaCO 3 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.

33. GasP i K H (0  C, S = 35)C i (0  C, S = 35; P = 1 Atm N 2 0.78080.80 x 10 -3 62.4 x 10 -3 mol kg-1 O 2 0.20951.69 x 10 -3 35.4 x 10 -3 Ar0.00931.83 x 10 -3 0.017 x 10 -3 CO 2 0.00033 63 x 10 -3 0.021 x 10 -3 GasMole Fraction in Dry Air (f G ) (where f G = moles gas i/total moles) N 2 0.78080 O 2 0.20952 Ar9.34 x 10 -3 CO23.3 x 10 -4 Example: Gas concentrations in equilibrium with the atmosphere Atmosphere Composition

34. Algebraic Approach – Monoprotic Acid Let acetic acid (CH 3 COOH) = HA. The base form (CH 3 COO - ) = A - We need to determine the concentrations of 4 species = HA, A -, H + and OH -. The 4 key equations are: 1. The reaction HA = H + + A - Acid Hydrogen Anion (or base) The Equilibrium Constant = K = (H + )(A - ) / (HA) 2. The reaction H 2 O = H + + OH - K w = (H + )(OH - ) 3. Mass balance on A C A = [HA] + [A - ] 4. Charge Balance[H + ] = [A - ] + [OH - ] By combining equations 2 and 3 given above we can write algebraic expressions to solve for the main species of acetic acid (HA) and acetate (A - ). [HA] = C A H + / K + H + [A - ] = C A K / K + H + The master variable is pH – How do these species vary with pH?

35. Graphical Approach (log – log diagram) There are three regions for these graphs ; pH = pK ( system point ) ; pH >> pK; pH << pK

36. K versus K’ - pros and cons Oceanographers frequently use an equilibrium constant defined in terms of concentrations. These are called apparent or operation equilibrium constants. We use the symbol K' to distinguish them from K. Formally they are equilibrium constants determined on the seawater activity scale. Apparent equilibrium constants (K') are written in the same form as K except that all species are written as concentrations. The exception is H+ which is usually written as the activity (H+). There are pros and cons for both the K and K' approaches. K the pro is that we can calculate the K from  G r and one value of K can be used for all problems in all solutions (one K fits all). The cons are that to use K we need to obtain values for the free ion activity coefficients (  i ) and the %free (f i ) for each solution. For K‘, the con is that there needs to have been experimental determination of this constant for enough values of S, T and P that equations can be derived to calculate K' for the S, T and P of interest. The pro is that when this has been done the values of K' are usually more precise than the corresponding value of K. The other pro is that we do not need values of  i and f i when we use K'.