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Dissolved Inorganic Carbon (DIC)

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

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Dissolved Inorganic Carbon (DIC)

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  1. Dissolved Inorganic Carbon (DIC) • Initially, DIC in groundwater comes from CO2 • CO2(g) + H2O ↔ H2CO3° • PCO2: partial pressure (in atm) • PCO2 of soil gas can be 10-100 times the PCO2 of atmosphere • In groundwater, CO2 usually increases along a flow path due to biodegradation in a closed system • CH2O + O2 CO2 + H2O • CH2O = “generic” organic matter

  2. Dissolved Inorganic Carbon (DIC) • Equilibrium expression with a gas is known as Henry’s Law: • CO2 + H2O  H2CO3; KCO2 = 10-1.47 • H2CO3 HCO3- + H+; Ka1 = 10-6.35 • HCO3- CO32- + H+; Ka2 = 10-10.33

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

  4. Alkalinity • Alkalinity = acid neutralizing capability (ANC) of water • Total effect of all bases in solution • Typically assumed to be directly correlated to HCO3-concentration in groundwater • HCO3- = alkalinity 0.82

  5. Salts (Electrolytes)

  6. Salts • When you mix an acid + base, H+ and OH-form H2O • The remaining anionand cationcan form a salt • e.g., mix H2SO4 + CaOH, make CaSO4 • 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. Solubility of Salts • Remember: A saturated solution of a salt is in a state of equilibrium • Al2(SO4)3(s) 2Al3+ + 3SO42- • Can write our familiar equilibrium expression with an equilibrium constant • Ksp = ([Al3+]2)([SO42-]3) • Ksp= solubility product constant • Activities of solids = 1 by definition • Kspvalues can be calculated (or looked up) • Kspfor Al2(SO4)3(s) = 69.19 (at 25°C) • Very large Ksp which means the salt is very soluble

  8. Solubility • Al2(SO4)3(s) 2Al3+ + 3SO42- • What is the solubility of Al2(SO4)3? • What are the activities of Al3+ and SO42- in a saturated solution of Al2(SO4)3? • Thus in a saturated solution of Al2(SO4)3 • [Al3+] = 2x = 1.829 mol/L • [SO42-] = 3x = 2.744 mol/L • Solubility of Al2(SO4)3 = 0.915 x the molecular weight (342.148 g/mol) = 3.13 x 102 g/L

  9. 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 Ca2+ and 7 x 10-3 mol/L SO42- saturated with respect to gypsum (CaSO42H2O)?

  10. Gypsum Solubility • CaSO4 2H2O  Ca2+ + SO42- • Ksp = [Ca2+] [SO42-] = 10-4.6 • If the solution is saturated, the product of the activities would = 10-4.6 • (Note that [Ca2+] and [SO42-] don’t have to be equal) • (Ca2+)(SO42-) = (5x10-2)(7x10-3) = IAP = 10-3.45 • SI = -3.45 – (-4.6) = 1.15 • Because SI > 0, gypsum predicted to precipitate

  11. How much gypsum would precipitate to reach equilibrium (saturation)? • CaSO42H2O  Ca2+ + SO42- + 2H2O • Ksp = [Ca2+] [SO42-] = 10-4.6 • As gypsum precipitates (reverse reaction), the IAP will decrease because [Ca2+] and [SO42-] are being used up • Once the IAP = Ksp, the solution will be in equilibrium with respect to gypsum

  12. How much gypsum would precipitate to reach equilibrium (saturation)? • CaSO42H2O  Ca2+ + SO42- + 2H2O • Ksp = [Ca2+] [SO42-] = 10-4.6 • The solution initially has 5x10-2 mol/L Ca2+ and 7x10-3 mol/L SO42- • To reach equilibrium, x moles precipitate: • [Ca2+] = 5x10-2 - x; [SO42-] = 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 = Ksp and the solution is saturated with respect to gypsum, and no more will precipitate • Equilibrium has been reached

  13. Changing solution composition due to precipitation of gypsum • As gypsum precipitates, the [Ca2+] / [SO42-] 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, [SO42-] will be used up, and none will remain in solution

  14. (Equilibrium reached) Ca2+ SO42- = 79.2 Ca2+ SO42- = 7.1

  15. Geochemical Divide • The initial ratio of species can affect which minerals precipitate • GEOCHEMICAL DIVIDE • If [Ca2+] / [SO42-] had been < 1 instead of > 1, then [SO42-] would have become concentrated relative to [Ca2+] • 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. 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 • CaSO42H2O  Ca2+ + SO42- + 2H2O • Ksp (gypsum) = [Ca2+] [SO42-] = 10-4.6 • BaSO4 Ba2+ + SO42- • Ksp (barite) = [Ba2+] [SO42-] = 10-10.0 • Barite is much less soluble than gypsum

  17. Precipitation of Salts in Natural Waters • [SO42-] has the same value in both equilibria: • [Ba2+] 10-4.6 / [Ca2+] = 10-10.0 • [Ba2+] / [Ca2+] = 10-5.4 • [Ca2+] is 250,000 x [Ba2+] when the solution is saturated with respect to both minerals

  18. Gypsum and Barite equilibrium • [SO42-] has the same value in both equilibria: • [Ba2+] 10-4.6 / [Ca2+] = 10-10.0 • [Ba2+] / [Ca2+] = 10-5.4 • [Ca2+] is 250,000 x [Ba2+] when the solution is saturated with respect to both minerals • Solve for [SO42-] using simultaneous equations • [SO42-]2 = 10-4.6 + 10-10.0 • [SO42-] = 10-2.3 mol/L • Note that barite only contributes a negligible amount of [SO42-]

  19. 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, [SO42-] will increase • CaSO42H2O  Ca2+ + SO42- + 2H2O • The increase in [SO42-] can cause the solution to become supersaturated with respect to barite, which is less soluble than gypsum

  20. Gypsum and Barite equilibrium • CaSO42H2O  Ca2+ + SO42- + 2H2O • Ba2+ + SO42-  BaSO4 • Barite precipitates as gypsum dissolves until [Ca2+] / [Ba2+] 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. 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 lesssoluble • Makes sense: less soluble happier as solid, more soluble happier dissolved (relatively)

  22. Supersaturation • Solutions in nature become supersaturatedwith 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., CaCO3)

  23. Calcite Solubility • CaCO3 Ca2+ + CO32-; Ksp = 10-8.35 (1) • HCO3- CO32- + H+; Ka2 = 10-10.33(2) • H2CO3 HCO3- + H+; KH2 = 10-6.35(3) • CO2(g) + H2O  H2CO3; K = 10-1.47(4) • If open to atmosphere • H2O  H+ + OH-(5) • 7 ions/molecules, need 2 more equations or to fix something (make constant)

  24. Calcite Solubility • Fix PCO2 at 10-3.5atm • (4)[H2CO3] = 10-1.47 x 10-3.5 = 10-4.97 • [H2CO3] = 1.07 x 10-5mol/L • (6) charge balance: • 2(Ca2+) + (H+) = 2(CO32-) + (HCO3-) + (OH-) • Now have 6 equations and 6 unknowns

  25. Calcite Solubility • After some algebra: • (Ca2+) = 5.01 x 10-4mol/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 CO2, at 25°C • pH = 8.30

  26. H2CO3

  27. Calcite Solubility • The reaction we just used for calcite dissolution generally doesn’t occur in nature • CaCO3 Ca2+ + CO32- • Dissolution of calcite done primarily by acid • In natural systems, primary acid is • CaCO3+ H2CO3 Ca2+ + 2HCO3- CO2

  28. Calcite Solubility • Let us consider CaCO3 solubility as affected by variations in PCO2, pH, and T • CO2(g) CO2(aq) • CO2(aq) + H2O  H2CO3 • CaCO3+ H2CO3 Ca2+ + 2HCO3- • Predict changes in solubility from these reactions

  29. Calcite Solubility • CO2(g) CO2(aq) • CO2(aq) + H2O  H2CO3 • CaCO3 + H2CO3 Ca2+ + 2HCO3- • Increase PCO2? • Increases (H2CO3), which increases amount of CaCO3 dissolved (at constant T) • Decreasing PCO2? • Decreases (H2CO3), causes saturated solution to become supersaturated and precipitate CaCO3 until equilibrium restored

  30. CaCO3 + H2CO3 Ca2+ + 2HCO3- What if we increase PCO2 from atmospheric by 10x? At Saturation ● ~25 mg/L calcite could be precipitated ●

  31. Calcite Solubility • Why does the pH decrease as PCO2 increases? • CO2(g) CO2(aq) • CO2(aq) + H2O  H2CO3 • CaCO3 + H2CO3 Ca2+ + 2HCO3- • Increasing PCO2 increases H2CO3, which dissociates: • H2CO3 HCO3- + H+ • Increasing H+ in solution decreases pH • And what happens to calcite as we decrease pH?

  32. H2CO3

  33. H2CO3

  34. PCO2 • What can affect PCO2? • May decrease due to photosynthesis of aquatic plants; may allow algae to precipitate CaCO3 • Degradation of organic matter in soil zones can increase PCO2 • CH2O + O2 → CO2 + H2O • Caves, PCO2 exolves in caves forming speleothems

  35. Falling Springs St. Clair County

  36. Stalactite Stalagmite

  37. Calcite Solubility and pH • Solubility increases very significantly with increasing acidity of solution (lower pH) • [Ca2+] = 1013.30 [H+]2 • log [Ca2+] = 13.30 – 2 pH • Solubility changes 100x with 1 pH unit change • Calcite cannot persist in even mildly acidic waters

  38. No calcite at pH < ~5.5 H2CO3

  39. 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 CaCO3 dissolves; none reaches the deep sea bottom • Calcite compensation depth (CCD) • 4.2 – 5.0 km deep

  40. 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. Incongruent Dissolution • KAlSi3O8 + 9H2O + 2H+ Al2Si2O5(OH)4 + 2K+ + 4H4SiO4 • Let’s predict how reaction responds to changes in environmental parameters • What if K+and/or H4SiO4 removed by flowing groundwater? • What if there’s an abundance of H2O? • If these particular conditions persist, achieving equilibrium (saturation) may not be possible

  42. 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. Geochemical Cycles and Kinetics (reaction rates)

  44. 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. Hydrologic (Water) Cycle

  46. Carbon Cycle

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

  48. Transfers between reservoirs d4,1n dt Reservoir 1 d1,2n dt Reservoir 4 Reservoir 2 d3,4n dt d2,3n dt Reservoir 3 n = component concentration t = time dn = rate of transfer of a component dt from one reservoir to another = Flux

  49. 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

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