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Mineral Solubility Dissolution Reactions Activity-Ratio Diagrams

Mineral Solubility Dissolution Reactions Activity-Ratio Diagrams Phosphorus Fertilizer Reactions in Calcareous Soil Gypsum in Acid Soil Clay Mineral Weathering. Dissolution Reactions Solubility depends on relative strength of bonds in mineral compared with bonds in solvation complex

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Mineral Solubility Dissolution Reactions Activity-Ratio Diagrams

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  1. Mineral Solubility Dissolution Reactions Activity-Ratio Diagrams Phosphorus Fertilizer Reactions in Calcareous Soil Gypsum in Acid Soil Clay Mineral Weathering

  2. Dissolution Reactions Solubility depends on relative strength of bonds in mineral compared with bonds in solvation complex CaSO4 • 2H2O, solvation complexes energetically favorable Little covalent character to bonds Al(OH)3, solvation complexes not energetically favorable Much higher extent of covalent character

  3. To solubilize latter, must destabilize bonds in lattice as with attack by H+ or ligand exchange

  4. Kinetics of dissolution of easily dissolved solids controlled by film diffusion Kinetics for clay minerals and hydrous oxides controlled by surface reaction Zeroth order d[M] / dt = k k depending on surface area, temperature, pressure, and concentrations of H+ and strongly complexing ligands  d[M] / dt = k* [H+]N

  5. However, equilibrium is achieved and solution concentrations of constituent ions are set by thermodynamic equilibrium constant Al(OH)3 (s) = Al3+(aq) + (OH-)3(aq) (Al3+)(OH-)3 / (Al(OH)3) = Kdis If minerals are pure and crystalline, activities of solid minerals = 1 Otherwise (solid solution or poorly crystalline), activities different from 1

  6. Ksp = (Al3+)(OH-)3 = Kdis (Al(OH)3) Where reaction involves OH-, commonly this species is formally replaced by H+ Al(OH)3(s) + 3H+(aq) = Al3+(aq) + 3H2O For which *Ksp = *Kdis (Al(OH)3) / (H2O)3 = (Al3+) / (H+)3

  7. The products (Al3+)(OH-)3 or (Al3+) / (H+)3 called ion activity products Generally, MaLb(s) = aMm+(aq) + bLl-(aq) IAP = (Mm+)a (Ll-)b

  8. If one compares measured to equilibrium IAPs, can say whether equilibrium exists Relative saturation • = IAP / Ksp • < 1, undersaturated • = 1, equilibrium  > 1, supersaturated

  9. d[Al(OH)3] / dt = k(Ω – 1) Do problem 4.

  10. Near equilibrium, the rate of precipitation of calcite is proportional to Ω - 1, where Ω = IAP / Kso. How many times larger is the rate of precipitation of calcite when the IAP = 10-7 than when it is 10-8? d[Ca(CO)3] / dt = k(Ω – 1) (Ω – 1) = 10-7/10-8.48 -1 = 101.48 -1 = 29.20 (Ω – 1) = 10-8/10-8.48 -1 = 100.48 -1 = 2.02 Therefore ~ 15x

  11. Activity-Ratio Diagrams Does a particular solid phase control solution concentrations of certain ions and, if so, which solid phase? • Guess set of solids and write appropriate dissolution reactions • Express Kdis equation in log form and rearrange to give form log [ (solid phase) / (ion of interest) ] = -log Kdis +  log [ (solution activities)] 3. Plot log [ (solid phase) / (ion of interest) ] versus a log [(solution activity)] Commony, -log(H+) = pH To construct linear plots, must arbitrarily set other log [(solution activity)]

  12. Example calculation Consider what Ca-mineral may be controlling (Ca2+) in a arid region soil Compare anhydrite (CaSO4), gypsum (CaSO4 • 2H2O) and calcite (CaCO3) CaSO4(s) = Ca2+(aq) + SO42-(aq) log Kdis = -4.38 (25 C and 1 atm) CaSO4 • 2H2O(s) = Ca2+(aq) + SO42-(aq) + 2H2O log Kdis = -4.62 and

  13. CaCO3(s) = Ca2+(aq) + CO32-(aq) log Kdis = 1.93 One could also write calcite dissolution as an acidic hydrolysis reaction CaCO3(s) + 2H+(aq) = Ca2+(aq) + CO2(g) + H2O log Kdis = 9.75

  14. For anhydrite, log Kdis = log (Ca2+) + log (SO42-) – log (CaSO4) From which by rearrangement log [ (anhydrite) / (Ca2+) ] = - log Kdis + log (SO42-) Similarly, log [ (gypsum) / (Ca2+) ] = -log Kdis + log (SO42-) + 2 log (H2O) log [ (calcite) / (Ca2+) ] = -log Kdis + 2pH + log (CO2) + log (H2O)

  15. Various choices for independent variable pH, (SO42-), (CO2) = partial pressure in atm, or (H2O) = relative humidity For H2O, activity typically = 1 but may be less under arid conditions Let’s use pH and set (SO42-) = 0.003, PCO2 = 0.0003 and (H2O) = 1 Substituting, log [ (anhydrite) / (Ca2+) ] = 1.86 log [ (gypsum) / (Ca2+) ] = 2.10 log [ (calcite) / (Ca2+) ] = -13.27 + 2pH

  16. Interpret this figure

  17. According to this approach The solid phase that controls solubility is the one that produces the largest activity ratio for the free ionic species in solution Largest log [ (solid) / (Ca2+) ] at certain pH pH < 7.8, gypsum controls but pH > 7.8, calcite controls Anhydrite doesn’t come into play under these conditions

  18. Note that if PCO2 > 0.00032 atm (say, 0.003) log [ (calcite) / (Ca2+) ] = -11.27 + 2pH

  19. Note if (H2O) < 1 (say, 0.60) log [ (anydrite) / (Ca2+) ] = 2.38 log [ (gypsum) / (Ca2+) ] = 2.18 Clearly, predictions depend on accurate Kdis values and accuracy of assumed conditions (e.g., PCO2)

  20. Other Approaches

  21. Phosphate Fertilizer Reactions in Calcareous Soil CaHPO4 • 2H2O = Ca2+ + HPO42- + 2H2O logKdis = -6.67 DCPDH CaHPO4 = Ca2+ + HPO42- logKdis = -6.90 DCP 1/6 Ca8H2(PO4)6 • 5H2O + 2/3 H+ = 4/3Ca2+ + HPO42- + 5/6H2O logKdis = -3.28 OCP 1/6 Ca10(OH)2(PO4)6 + 4/3H+ = 5/3 Ca2+ + HPO42- + 1/3H2O logKdis = -2.28 HA CaCO3 + 2H+ = Ca2+ + CO2 + H2O logKdis = 9.75 calcite

  22. Problem Assume dissolution of calcite controls calcium concentration Develop activity / ratio diagrams for DCPDH DCP OCP HA -6.57 = log(Ca2+) + log(HPO42-) – log(DCPDH) = 9.75 - 2pH – log(CO2) – log[(DCPDH) / (HPO42-)] log[(DCPDH) / (HPO42-)] = 16.32 – log PCO2 – 2pH

  23. log[(DCP) / (HPO42-)] = 16.65 – log PCO2 – 2pH log[(OCP) / (HPO42-)] = 17.59 – log PCO2 – 2pH log[(HA) / (HPO42-)] = 21.24 – log PCO2 – 2pH Some questions before proceeding. How does one arrive at 1/6 Ca8H2(PO4)6 • 5H2O + 2/3 H+ = 4/3Ca2+ + HPO42- + 5/6H2O from Ca8H2(PO4)6 • 5H2O = 8Ca2+ + 2HPO42- + 4PO43- + 5H2O?

  24. Ca8H2(PO4)6 • 5H2O = 8Ca2+ + 2HPO42- + 4PO43- + 5H2O 4PO43- + 4H+ = 4HPO42- 1/6 Ca8H2(PO4)6 • 5H2O + 2/3 H+ = 4/3Ca2+ + HPO42- + 5/6H2O for which log Kdis = -3.28 From this, how does one arrive at log[(OCP) / (HPO42-)] = 17.59 – log PCO2 – 2pH using (Ca2+) = Kdis(CaCO3)(H+)2/PCO2(H2O) or log(Ca2+) = 9.75 – log PCO2 – 2pH

  25. 1/6 Ca8H2(PO4)6 • 5H2O + 2/3 H+ = 4/3Ca2+ + HPO42- + 5/6H2O log[1/6(OCP)/(HPO42-)] - 2/3pH = 3.28 + 4/3(Ca2+) Substituting log(Ca2+) = 9.75 – log PCO2 – 2pH log[1/6(OCP)/(HPO42-)] - 2/3pH = 3.28 + 13.00 – 4/3log PCO2 – 8/3pH log[1/6(OCP)/(HPO42-)] = 16.28 - log PCO2 – 2pH – 1/3 log PCO2 What about hydroxyapatite? 1/6 Ca10(OH)2(PO4)6 + 4/3H+ = 5/3 Ca2+ + HPO42- + 1/3H2O log[(HA) / (HPO42-)] = 18.63 – log PCO2 – 2pH – 2/3log PCO2

  26. Since HA gives the smallest (HPO42-), it should control phosphate solubility. However, all solids can coexist and there (experimentally) is a stepwise transformation from DCPDH to HA.

  27. If the initial state of a soil is such that several solid phases can form potentially with a given ion, the solid phase that forms first will be the one for which the activity ratio is nearest above the initial value in the soil. Thereafter, the remaining accessible solid phases will form in order of increasing activity ratio, with the rate of formation of a solid phase in sequence decreasing as its activity ratio increases. In an open system, any one of the solid phases may be maintained indefinitely. Gay-Lussac-Ostwald (GLO) Step Rule If d[solid] / dt = k(Ω – 1) near equilibrium, apparently d[solid] / dt << k(Ω – 1) if Ω >> 1

  28. Gypsum in Acid Soils AlOHSO4$ 5H2O jurbanite logKdis = -3.8 Al4(OH)10SO4$ 5H2O basaluminite logKdis = 5.63 KAl3(OH)6(SO4)2 alunite logKdis = 0.2 For acidic dissolution, AlOHSO4$ 5H2O + H+ = Al3+ + SO42- + 6H2O log[(AlOHSO4$ 5H2O) / (Al3+)] = 3.8 – pH + log(SO42-) + 6log(H2O) Set pH = 4.5, (H2O) = 1 and (K+) = 0.0001 log[(jurbanite) / (Al3+) = 8.30 + log(SO42-)

  29. log[(gibbsite) / (Al3+)] = -8.11 + 3pH + 3log(H2O) = 5.39 = -8.77 + 3pH + 3log(H2O) = 4.73

  30. Clay Mineral Weathering Inferences on stability in different weathering environments Gibbsite Kaolinite Smectite

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