1. 1

2. Dissolved Inorganic Carbon (DIC) Initially, DIC in groundwater comes from CO 2 – CO 2(g) + H 2 O ↔ H 2 CO 3 ° – P CO2 : partial pressure (in atm) – P CO2 of soil gas can be 10-100 times the P CO2 of atmosphere In groundwater, CO 2 usually increases along a flow path due to biodegradation in a closed system – CH 2 O + O 2  CO 2 + H 2 O CH 2 O = “generic” organic matter 2

3. Dissolved Inorganic Carbon (DIC) Equilibrium expression with a gas is known as Henry’s Law: CO 2 + H 2 O  H 2 CO 3 ; K CO2 = 10 -1.47 H 2 CO 3  HCO 3 - + H + ; K a1 = 10 -6.35 HCO 3 -  CO 3 2- + H + ; K a2 = 10 -10.33 3

4. 4 Total DIC = 10 -1 M pH = 6.35 pH = 10.33 Common pH range in natural waters

5. Alkalinity Alkalinity = acid neutralizing capability (ANC) of water – Total effect of all bases in solution – Typically assumed to be directly correlated to HCO 3 - concentration in groundwater – HCO 3 - = alkalinity 0.82 5

6. Salts (Electrolytes) 6

7. Salts When you mix an acid + base, H + and OH - form H 2 O The remaining anion and cation can form a salt – e.g., mix H 2 SO 4 + CaOH, make CaSO 4 – mix HCl + NaOH, make NaCl Salts are named after the acid they come from – e.g., chlorides, carbonates, sulfates, etc. All minerals are salts except oxides, hydroxides, and native elements 7

8. Solubility of Salts Remember: A saturated solution of a salt is in a state of equilibrium Al 2 (SO 4 ) 3(s)  2Al 3+ + 3SO 4 2- – Can write our familiar equilibrium expression with an equilibrium constant – K sp = ([Al 3+ ] 2 )([SO 4 2- ] 3 ) K sp = solubility product constant Activities of solids = 1 by definition K sp values can be calculated (or looked up) – K sp for Al 2 (SO 4 ) 3(s) = 69.19 (at 25°C) Very large K sp which means the salt is very soluble 8

9. Solubility Al 2 (SO 4 ) 3(s)  2Al 3+ + 3SO 4 2- – What is the solubility of Al 2 (SO 4 ) 3 ? What are the activities of Al 3+ and SO 4 2- in a saturated solution of Al 2 (SO 4 ) 3 ? – – Thus in a saturated solution of Al 2 (SO 4 ) 3 [Al 3+ ] = 2x = 1.829 mol/L [SO 4 2- ] = 3x = 2.744 mol/L – Solubility of Al 2 (SO 4 ) 3 = 0.915 x the molecular weight (342.148 g/mol) = 3.13 x 10 2 g/L 9

10. Solubility We often want to know whether a solution is saturated with respect to a mineral e.g., Is a solution with 5 x 10 -2 mol/L Ca 2+ and 7 x 10 -3 mol/L SO 4 2- saturated with respect to gypsum (CaSO 4  2H 2 O)? 10

11. Gypsum Solubility CaSO 4  2H 2 O  Ca 2+ + SO 4 2- – K sp = [Ca 2+ ] [SO 4 2- ] = 10 -4.6 – If the solution is saturated, the product of the activities would = 10 -4.6 – (Note that [Ca 2+ ] and [SO 4 2- ] don’t have to be equal) – (Ca 2+ )(SO 4 2- ) = (5x10 -2 )(7x10 -3 ) = IAP = 10 -3.45 – SI = -3.45 – (-4.6) = 1.15 – Because SI > 0, gypsum predicted to precipitate 11

12. How much gypsum would precipitate to reach equilibrium (saturation)? CaSO 4  2H 2 O  Ca 2+ + SO 4 2- + 2H 2 O – K sp = [Ca 2+ ] [SO 4 2- ] = 10 -4.6 – As gypsum precipitates (reverse reaction), the IAP will decrease because [Ca 2+ ] and [SO 4 2- ] are being used up – Once the IAP = K sp, the solution will be in equilibrium with respect to gypsum 12

13. How much gypsum would precipitate to reach equilibrium (saturation)? CaSO 4  2H 2 O  Ca 2+ + SO 4 2- + 2H 2 O – K sp = [Ca 2+ ] [SO 4 2- ] = 10 -4.6 – The solution initially has 5x10 -2 mol/L Ca 2+ and 7x10 -3 mol/L SO 4 2- – To reach equilibrium, x moles precipitate: [Ca 2+ ] = 5x10 -2 - x; [SO 4 2- ] = 7x10 -3 - x; Substitute into eq. above: [5x10 -2 - x] [7x10 -3 - x] = 10 -4.6 Eventually get x = 6.45 x 10 -3 – Amount of gypsum that will precipitate in this solution is 6.45 x 10 -3 x 172.17 (mc. wt.) = 1.11 g/L At this point, IAP = K sp and the solution is saturated with respect to gypsum, and no more will precipitate Equilibrium has been reached 13

14. Changing solution composition due to precipitation of gypsum As gypsum precipitates, the [Ca 2+ ] / [SO 4 2- ] ratio increases from 7.1 to 79.2 – The precipitation of a salt reduces the concentrations of ions and changes the chemical composition of remaining solution – In our example, if precipitation continues, [SO 4 2- ] will be used up, and none will remain in solution 14

15. Ca 2+ SO 4 2- Ca 2+ SO 4 2- = 7.1 = 79.2 (Equilibrium reached) 15

16. Geochemical Divide The initial ratio of species can affect which minerals precipitate GEOCHEMICAL DIVIDE – If [Ca 2+ ] / [SO 4 2- ] had been 1, then [SO 4 2- ] would have become concentrated relative to [Ca 2+ ] – End up with a different final solution – May lead to precipitation of different minerals – This is important during the evolution of brines by evaporative concentration 16

17. Precipitation of Salts in Natural Waters Natural waters are complex, may have more than 1 salt precipitating Let’s consider 2 sulfate minerals, gypsum and barite – CaSO 4  2H 2 O  Ca 2+ + SO 4 2- + 2H 2 O K sp (gypsum) = [Ca 2+ ] [SO 4 2- ] = 10 -4.6 – BaSO 4  Ba 2+ + SO 4 2- K sp (barite) = [Ba 2+ ] [SO 4 2- ] = 10 -10.0 – Barite is much less soluble than gypsum 17

18. Precipitation of Salts in Natural Waters [SO 4 2- ] has the same value in both equilibria: – [Ba 2+ ] 10 -4.6 / [Ca 2+ ] = 10 -10.0 – [Ba 2+ ] / [Ca 2+ ] = 10 -5.4 – [Ca 2+ ] is 250,000 x [Ba 2+ ] when the solution is saturated with respect to both minerals 18

19. Gypsum and Barite equilibrium [SO 4 2- ] has the same value in both equilibria: – [Ba 2+ ] 10 -4.6 / [Ca 2+ ] = 10 -10.0 – [Ba 2+ ] / [Ca 2+ ] = 10 -5.4 – [Ca 2+ ] is 250,000 x [Ba 2+ ] when the solution is saturated with respect to both minerals Solve for [SO 4 2- ] using simultaneous equations – [SO 4 2- ] 2 = 10 -4.6 + 10 -10.0 – [SO 4 2- ] = 10 -2.3 mol/L – Note that barite only contributes a negligible amount of [SO 4 2- ] 19

20. Gypsum and Barite equilibrium Suppose a saturated solution of barite comes into contact with gypsum – It is likely that the solution is undersaturated with respect to gypsum, which is much more soluble than barite – If gypsum dissolves, [SO 4 2- ] will increase CaSO 4  2H 2 O  Ca 2+ + SO 4 2- + 2H 2 O – The increase in [SO 4 2- ] can cause the solution to become supersaturated with respect to barite, which is less soluble than gypsum 20

21. Gypsum and Barite equilibrium CaSO 4  2H 2 O  Ca 2+ + SO 4 2- + 2H 2 O Ba 2+ + SO 4 2-  BaSO 4 – Barite precipitates as gypsum dissolves until [Ca 2+ ] / [Ba 2+ ] approaches 250,000 Then replacement of gypsum by barite stops because solution is saturated with respect to both minerals – This is called the common ion effect 21

22. Precipitation of Salts in Natural Waters Replacement of 1 mineral by another is common in geology – Introduction of a common ion causes solution to become supersaturated with respect to the less soluble compound – Thus the more soluble compound is always replaced by less soluble – Makes sense: less soluble happier as solid, more soluble happier dissolved (relatively) 22

23. Supersaturation Solutions in nature become supersaturated with respect to a mineral by: – Introduction of a common ion – Change in pH – Evaporative concentration – Temperature variations In general solubilities increase with increasing T, but not always (e.g., CaCO 3 ) 23

24. Calcite Solubility CaCO 3  Ca 2+ + CO 3 2- ; K sp = 10 -8.35 (1) HCO 3 -  CO 3 2- + H + ; K a2 = 10 -10.33 (2) H 2 CO 3  HCO 3 - + H + ; K H2 = 10 -6.35 (3) CO 2(g) + H 2 O  H 2 CO 3 ; K = 10 -1.47 (4) – If open to atmosphere H 2 O  H + + OH - (5) 7 ions/molecules, need 2 more equations or to fix something (make constant) 24

25. Calcite Solubility Fix P CO2 at 10 -3.5 atm (4) [H 2 CO 3 ] = 10 -1.47 x 10 -3.5 = 10 -4.97 – [H 2 CO 3 ] = 1.07 x 10 -5 mol/L (6) charge balance: – 2(Ca 2+ ) + (H + ) = 2(CO 3 2- ) + (HCO 3 - ) + (OH - ) Now have 6 equations and 6 unknowns 25

26. Calcite Solubility After some algebra: (Ca 2+ ) = 5.01 x 10 -4 mol/L (20.1 mg/L) Solubility (S) of calcite = 5.01 x 10 -4 x 100.0787 (MW) = 5.01 x 10 -2 g/L – For calcite in water in equilibrium with CO 2, at 25°C – pH = 8.30 26

27. 27 H 2 CO 3

28. Calcite Solubility The reaction we just used for calcite dissolution generally doesn’t occur in nature – CaCO 3  Ca 2+ + CO 3 2- Dissolution of calcite done primarily by acid – In natural systems, primary acid is – CaCO 3 + H 2 CO 3  Ca 2+ + 2HCO 3 - 28 CO 2

29. Calcite Solubility Let us consider CaCO 3 solubility as affected by variations in P CO2, pH, and T – CO 2(g)  CO 2(aq) – CO 2(aq) + H 2 O  H 2 CO 3 – CaCO 3 + H 2 CO 3  Ca 2+ + 2HCO 3 - Predict changes in solubility from these reactions 29

30. Calcite Solubility CO 2(g)  CO 2(aq) CO 2(aq) + H 2 O  H 2 CO 3 CaCO 3 + H 2 CO 3  Ca 2+ + 2HCO 3 - Increase P CO2 ? – Increases (H 2 CO 3 ), which increases amount of CaCO 3 dissolved (at constant T) Decreasing P CO2 ? – Decreases (H 2 CO 3 ), causes saturated solution to become supersaturated and precipitate CaCO 3 until equilibrium restored 30

31. CaCO 3 + H 2 CO 3  Ca 2+ + 2HCO 3 - 31 At Saturation ~25 mg/L calcite could be precipitated ● ● What if we increase P CO2 from atmospheric by 10x?

32. Calcite Solubility Why does the pH decrease as P CO2 increases? – CO 2(g)  CO 2(aq) – CO 2(aq) + H 2 O  H 2 CO 3 – CaCO 3 + H 2 CO 3  Ca 2+ + 2HCO 3 - Increasing P CO2 increases H 2 CO 3, which dissociates: – H 2 CO 3  HCO 3 - + H + – Increasing H + in solution decreases pH – And what happens to calcite as we decrease pH? 32

33. 33 H 2 CO 3

34. 34 H 2 CO 3

35. P CO2 What can affect P CO2 ? – May decrease due to photosynthesis of aquatic plants; may allow algae to precipitate CaCO 3 – Degradation of organic matter in soil zones can increase P CO2 CH 2 O + O 2 → CO 2 + H 2 O – Caves, P CO2 exolves in caves forming speleothems 35

36. 36 Falling Springs St. Clair County

37. 37 Stalactite Stalagmite

38. Calcite Solubility and pH Solubility increases very significantly with increasing acidity of solution (lower pH) – [Ca 2+ ] = 10 13.30 [H + ] 2 – log [Ca 2+ ] = 13.30 – 2 pH Solubility changes 100x with 1 pH unit change – Calcite cannot persist in even mildly acidic waters 38

39. 39 No calcite at pH < ~5.5 H 2 CO 3

40. Calcite Solubility and T Solubility also affected by T, because equilibrium constants change – Solubility of calcite decreases with increasing temperature – As particles sink in the oceans, the water gets colder, and CaCO 3 dissolves; none reaches the deep sea bottom Calcite compensation depth (CCD) 4.2 – 5.0 km deep 40

41. Chemical Weathering Calcite dissolution is a form of chemical weathering Congruent dissolution: no new solid phases formed Incongruent dissolution: new solid formed – Al silicates usually dissolve incongruently Products of chemical weathering – New minerals (clays, oxides, …) – Ions/molecules dissolved; help determine water quality – Unreactive mineral grains (e.g., quartz, garnet, muscovite) are major source of “sediment” 41

42. Incongruent Dissolution KAlSi 3 O 8 + 9H 2 O + 2H +  Al 2 Si 2 O 5 (OH) 4 + 2K + + 4H 4 SiO 4 Let’s predict how reaction responds to changes in environmental parameters – What if K + and/or H 4 SiO 4 removed by flowing groundwater? – What if there’s an abundance of H 2 O? – If these particular conditions persist, achieving equilibrium (saturation) may not be possible 42

43. Reaction Equilibrium Can a chemical reaction achieve equilibrium in nature? Water/rock ratio is a key variable – The higher the water/rock ratio, the more likely the reaction goes to completion, not equilibrium Products removed – If the ratio is small, the reactions can control the environment and equilibrium is possible 43

44. Geochemical Cycles and Kinetics (reaction rates) 44

45. Geochemical Cycles Material is being cycled continuously in the Earth’s surface system We can think of the Earth’s surface as consisting of several reservoirs connected by “pipes” through which matter moves – Crust, hydrosphere, atmosphere All chemical elements are cycled 45

46. Hydrologic (Water) Cycle 46

47. Carbon Cycle 47

48. Rock Cycle Sediment Metamorphic Rocks Igneous Rocks Sedimentary Rocks Magma formation Intrusion Crystallization Weathering Transport Deposition Burial Diagenesis Lithification Deformation Recrystallization Segregation Weathering, etc. Subduction Weathering, etc. Subduction 48

49. Transfers between reservoirs Reservoir 2 Reservoir 1 Reservoir 3 d 4,1 n dt d 1,2 n dt d 3,4 n dt d 2,3 n dt Reservoir 4 dn = rate of transfer of a component dt from one reservoir to another = Flux n = component concentration t = time 49

50. Steady State At Earth’s creation, there was a finite amount of each element Very little input of material since then (meteorites, extraterrestrial dust) Since these cycles have been going on for a very long time, we assume they are essentially at steady state 50

51. Steady State Steady state means that the composition of reservoirs in a cycle does not change over time – No accumulation or loss of the material of interest – Input + any production in the reservoir = outflow + any consumption in the reservoir – The mass balance = 0; no creation or loss of material In the Earth surface environment, especially the oceans and atmosphere, cycling of materials has been occurring at near steady-state 51

52. Steady State Reservoir 2 Reservoir 1 Reservoir 3 d 1,2 n dt d 2,3 n dt d 1,2 n dt d 2,3 n dt = d 2 n dt = 0 52

53. Residence Time Residence time is the average time a molecule spends in a reservoir between the time it arrives and the time it leaves Determined by dividing the amount in the reservoir by the flux in (or out) 2 n d 1,2 n dt T = 53

54. Residence time (example) Suppose we have a 50 L tank of water, with a flux in = flux out = 5 L/min; what is the residence time? 50 L 5 L/min 54

55. Residence time (example) Because the water flux in = flux out, it’s at steady state, so dn H2O /dt = 0 T = 50L = 10 minutes 5L/min This is the time required to add or subtract 50 L of water 50 L 5 L/min 55

56. Residence time (example) However, since there is mixing in the reservoir, not every molecule of H 2 O is replaced every 10 minutes Some molecules will remain in longer, some exit more quickly 5 L/min 56

57. Cycles and reaction rates As materials cycle through the Earth, they are moved and transformed at various rates (the “pipes” connecting reservoirs) Transport processes and chemical reactions take time Kinetics is the study of reaction rates 57