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Phase Change Reactions Precipitation-Dissolution of Inorganic Species

Phase Change Reactions Precipitation-Dissolution of Inorganic Species

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Phase Change Reactions Precipitation-Dissolution of Inorganic Species

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  1. Phase Change ReactionsPrecipitation-Dissolution of Inorganic Species Bruce Herbert Geology & Geophysics

  2. Precipitation-Dissolution and Metal-Ligand Properties • Generally, species exhibit similar precipitation-dissolution reactions as complexation reactions in that • The more stable solid phases of a hard metal will be precipitates with a hard base (all other factors being equal). • The more stable solid phases of a soft metal will be precipitates with a soft base (all other factors being equal).

  3. Thermodynamics of Precipitation-Dissolution • General formula for two component dissolution (6.1) • where M is a metal, L a ligand, a and b are stoichiometric coefficients, m and n are the charges of the ions, and Kdis is the equilibrium dissolution constant.

  4. Thermodynamics of Precipitation-Dissolution • The solubility product constant, Kso is defined as (6.2) • If the solid is in its Standard State, as is commonly assumed, then Kdis=Kso. If the solid is not in its Standard State, the IAP will be a function of all of the thermodynamic variables that affect the activity of the solid. • Precipitation-dissolution reactions often occur over much longer time scales than complexation reactions in solution. Species in the solution phase will come to equilibration among themselves before they reach equilibration with the solid phase.

  5. Thermodynamics of Precipitation-Dissolution • We can use this fact to define two useful criteria for precipitation-dissolution reactions: • The ion activity product, IAP is defined as (6.3) • The relative saturation, , is defined as (6.4)

  6. Thermodynamics of Precipitation-Dissolution • The relative saturation can be monitored over time to assess the degree of equilibration in a system. • If < 1, then the system is undersaturated with respect to the solid phase as defined by the reaction in 6.1. • If > 1, then the system is oversaturated with respect to the solid phase as defined by the reaction in 6.1. • If = 1, then the system is in equilibration with respect to the solid phase as defined by the reaction in 6.1.

  7. Thermodynamics of Precipitation-Dissolution • MINTEQA2 and PHREEQ calculates a saturation index, SI SI = log [IAP/Kso] (6.5) • If the SI < 0, then the system is undersaturated • If the SI > 0, then the system is oversaturated • IF the SI ≈ 0, then the system is at equilibrium

  8. Thermodynamics of Precipitation-Dissolution • If SI ≠ 0, then we can make one of three conclusions concerning the system of interest: • The reaction is not in equilibrium • No solid phase corresponding to the reaction as written exists in the system • The reaction is at (possibly metastable) equilibrium, but the solid phase is not in the Standard State assumed in computing Kso

  9. Example: The Dissolution of Gibbsite • The value of Kdis can be calculated with Standard-State chemical potentials. • Assuming that gibbsite is in its Standard State, then Kdis= Ksoand Al(OH)3(s) = Al3+(aq) + 3OH-(aq) (6.6) Check calcs

  10. Thermodynamics of Precipitation-Dissolution • The central problem with precipitation-dissolution reactions, as environmental geologist, is to predict which solid phase controls aqueous activities of a metal or ligand. • The controlling phase at equilibrium will be the one which results in the smallest value of the aqueous activity of the ion. • The corollary is also true: the chemical potential, m(aq), is smallest whenever the aqueous activity is at a minimum. At that time the chemical potential of a species in the solid and aqueous phases will be equal.

  11. Example: Does Cd(OH)2 or CdCO3 control (Cd2+) in solution? • Background Data: pH = 7.6; (HCO3-) = 10-3 • For the Hydroxide Phase: • *Kso is the dissolution equilibrium constant for cadmium hydroxide by adding the ionization of water. • Assume both Cd(OH)2(s) and H2O(l) are in their Standard States. Then: • log (Cd2+) = log *Kso + 2 log (H+) = log *Kso -2 pH • since *Kso = (Cd2+)/(H+)2 • Then log (Cd2+) = -1.59 if Cd(OH)2(s) is the controlling phase

  12. Example: Does Cd(OH)2 or CdCO3 control (Cd2+) in solution? • Background Data: pH = 7.6; (HCO3-) = 10-3 • For the Carbonate Phase: • Assume CdCO3(s) is in its Standard State. Then: • log (Cd2+(aq)) = log Kso - pH - log (HCO3-) • then log (Cd2+) = -5.47 • therefore CdCO3(s) is the controlling phase

  13. Reverse experimental procedure • Determine (Cd2+) in solution using a Cd-sensitive electrode • Determine (CO32-) and calculate IAP • Compare IAP to published values of Kso. • If the two are not equal then either: • Equilibrium with the solid phase does not exist • Solid phase controlling ion activity is not the one suspected • Solid is not in the Standard State.

  14. Solubility of Oxides and Hydroxides • Oxides and hydroxides are often the most common precipitates of trace metals. Their precipitation-dissolution is strongly affected by pH. • Solubility of oxides and hydroxides can be expressed as: • M(OH)2(s) = M2+(aq) + 2OH-(aq) Kso = (M2+)(OH-)2 • MO(s) + H20(l) = M2+(aq) + 2OH-(aq) Kso = (M2+)(OH-)2

  15. Solubility of Oxides and Hydroxides • We can rewrite the reaction in order to include protons M(OH)2(s) + 2H+(aq) = M2+(aq) + 2H2O(l) *Kso = (M2+)/(H+)2 = Kso/Kw2 (6.15) MO(s) + 2H+(aq) = M2+(aq) + H2O(l) *Kso = (M2+)/(H+)2 = Kso/Kw2 (6.16) • where Kw is the hydrolysis constant for water H2O(l) = H+(aq) + OH-(aq) Kso = (H+)(OH-)/(H2O) (6.17) • This gives log[Mz+] = log Kso + z pKw - z pH (6.18)

  16. Graphical representation of ZnO (s) • The dissolution of zinc oxide as a function of pH is governed by the following reactions • Reaction, Logarithmic Form, log *Kso ZnO(s) + 2H+(aq) = Zn2+(aq) + 2H2O(l) log (Zn2+) = log *Kso - 2pH, log *Kso = 11.2 ZnO(s) + H+(aq) = ZnOH+(aq) log (ZnOH+) = log *Kso - pH, log *Kso = 2.2 ZnO(s) + 2H2O(l) = Zn(OH)3-(aq)+ H+(aq) log (Zn(OH)3-) = log *Kso + pH, log *Kso = -16.9 ZnO(s) + 3H2O(l) = Zn(OH)42-(aq)+ 2H+(aq) log (Zn(OH)42-) = log *Kso + 2pH, log *Kso = -29.7

  17. Graphical representation of ZnO (s) • The logarithmic equations are equations of straight lines and can be plotted (using Excel) where pH forms the independent variable:

  18. Graphical representation of ZnO (s)

  19. Log Activity Zn Species Concentrations of dissolved Zn species in equilibrium with ZnO as a function of pH.

  20. Log Activity Al Species Concentrations of dissolved Al species in equilibrium with gibbsite as a function of pH.

  21. Log Activity Fe Species Concentrations of dissolved Fe species in equilibrium with Fe(OH)3 as a function of pH.

  22. Log activity dissolved Si Species Activities of dissolved silica species in equilibrium with quartz and amorphous silica at 25°C. Note that silica solubility is pH-independent at pH < 9, but increases dramatically with increasing pH at pH >9.

  23. Case Study: Cotter U Mill Site The Cotter/Lincoln Park site consists of a uranium processing mill located adjacent to the unincorporated community of Lincoln Park. The mill operated continuously from 1958 until 1979, and intermittently since that time. Mill operations released radioactive materials and metals into the environment. These releases contaminated soil and groundwater around the mill and the Lincoln Park area. For more info: http://www.antenna.nl/wise/uranium/umopcc.html Davis, A., and D.D. Runnells. 1987. Geochemical Interactions between acidic tailings fluid and bedrock: Use of the computer model MINTEQ. Applied Geochemistry 2: 231-241.

  24. Case Study: Cotter U Mill Site http://www.epa.gov/region08/superfund/co/lincolnpark/

  25. Case Study: Cotter Uranium Mill Site The contaminants of most concern at the site are molybdenum and uranium. The primary exposure pathways would be drinking contaminated water and inhaling contaminated dust. Radon, a decay product in the uranium chain, is also of potential concern. • Major cleanup activities performed since 1988 include: • Connecting Lincoln Park residents to city water; Constructing a ground-water barrier at the Soil Conservation Service (SCS) dam to minimize migration of contaminated ground water into Lincoln Park; Moving tailings and contaminated soils into a lined impoundment to eliminate them as a source of contamination; and Excavating contaminated stream sediments

  26. Case Study: Cotter U Mill Site

  27. Case Study: Cotter U Mill Site

  28. Case Study: Cotter U Mill Site

  29. Case Study: Cotter U Mill Site

  30. Case Study: Cotter U Mill Site

  31. Case Study: Cotter U Mill Site

  32. Appendix • Understand the principles governing the solubility of quartz. • Understand the principles governing the solubility of Al- and Fe-oxyhydroxides.

  33. SILICA SOLUBILITY - I • In the absence of organic ligands or fluoride, quartz solubility is relatively low in natural waters. • Below pH 9, the dissolution reaction is: SiO2(quartz) + 2H2O(l)  H4SiO40 for which the equilibrium constant at 25°C is: • At pH < 9, quartz solubility is independent of pH. • Quartz is frequently supersaturated in natural waters because quartz precipitation kinetics are slow.

  34. SILICA SOLUBILITY - II • Thus, quartz saturation does not usually control the concentration of silica in low-temperature natural waters. Amorphous silica can control dissolved Si: SiO2(am) + 2H2O(l)  H4SiO40 for which the equilibrium constant at 25°C is: • Quartz is formed diagenetically through the following sequence of reactions: opal-A (siliceous biogenic ooze)  opal-A’ (nonbiogenic amorphous silica)  opal-CT  chalcedony  microcrystalline quartz

  35. SILICA SOLUBILITY - III At pH > 9, H4SiO40 dissociates according to: H4SiO40  H3SiO4- + H+ H3SiO4-  H2SiO42- + H+ The total solubility of quartz (or amorphous silica) is:

  36. SILICA SOLUBILITY - IV The equations for the dissociation constants of silicic acid can be rearranged (assuming a = M ) to get: We can now write:

  37. Log activity dissolved Si Species Activities of dissolved silica species in equilibrium with quartz and amorphous silica at 25°C. Note that silica solubility is pH-independent at pH < 9, but increases dramatically with increasing pH at pH >9.

  38. SILICA SOLUBILITY - V An alternate way to understand quartz solubility is to start with: SiO2(quartz) + 2H2O(l)  H4SiO40 Now adding the two reactions: SiO2(quartz) + 2H2O(l)  H4SiO40Kqtz H4SiO40  H3SiO4- + H+ K1 SiO2(quartz) + 2H2O(l)  H3SiO4- + H+K

  39. SILICA SOLUBILITY - VI Taking the log of both sides and rearranging we get: Finally adding the three reactions: SiO2(quartz) + 2H2O(l)  H4SiO40Kqtz H4SiO40  H3SiO4- + H+ K1 H3SiO4-  H2SiO42- + H+ K2 SiO2(quartz) + 2H2O(l)  H2SiO42- + 2H+K

  40. SILICA SOLUBILITY - VII SUMMARY • Silica solubility is relatively low and independent of pH at pH < 9 where H4SiO40 is the dominant species. • Silica solubility increases with increasing pH above 9, where H3SiO4- and H2SiO42- are dominant. • Fluoride, and possibly organic compounds, may increase the solubility of silica. • Saturation with quartz does not control silica concentrations in low-temperature natural waters; saturation with amorphous silica may.

  41. Appendix • Understand the principles governing the solubility of quartz. • Understand the principles governing the solubility of Al- and Fe-oxyhydroxides.

  42. SOLUBILITY OF GIBBSITE - I • We will use gibbsite to illustrate principles of the solubility of Al-bearing minerals; the solubility of such minerals is highly pH-dependent. • The solubility product for gibbsite is given by: Al(OH)3(gibbsite) Al3+ + 3OH- • We can also write this in the alternate form: Al(OH)3(gibbsite) + 3H+ Al3+ + 3H2O(l)

  43. SOLUBILITY OF GIBBSITE - II • Use of the latter equation shows that the concentration of Al3+ will be very low in the pH range of most natural waters. • For example, at pH = 7, we calculate the concentration of Al3+ to be 2.2910-12 mol L-1! • However, Al3+ forms a series of hydroxide complexes which increase its solubility somewhat: Al3+ + H2O(l)  Al(OH)2+ + H+Kh,1 Al3+ + 2H2O(l)  Al(OH)2+ + 2H+Kh,2 Al3+ + 4H2O(l)  Al(OH)4- + 4H+Kh,4

  44. SOLUBILITY OF GIBBSITE - III • The mass action expressions for these reactions may be written: The total dissolved aluminum concentration is given by:

  45. SOLUBILITY OF GIBBSITE - IV • We now assume that activity coefficients are unity, so that activity equals concentration. • Next, we rewrite the solubility product of gibbsite to obtain: We see that the logarithm of the concentration of Al3+ in equilibrium with gibbsite is a straight line function of pH, with a slope of -3. In other words, the concentration of Al3+ decreases 3 log units for every unit increase in pH.

  46. SOLUBILITY OF GIBBSITE - V The concentration of Al(OH)2+ can be obtained from: but the concentration of Al3+ has already been calculated so: We see that the logarithm of the concentration of Al(OH)2+ in equilibrium with gibbsite is also a straight line function of pH, but with a slope of -2.

  47. SOLUBILITY OF GIBBSITE - VI Similarly for the other two species:

  48. SOLUBILITY OF GIBBSITE - VII Now, substituting into the mass-balance expression: we get and taking the logarithm of both sides and substituting the K values at 25°C:

  49. Log Activity Al Species Concentrations of dissolved Al species in equilibrium with gibbsite as a function of pH.

  50. SOLUBILITY OF ZINCITE (ZnO) - I The thermodynamic data for solubility problems can be presented in another way. At 25°C and 1 bar: ZnO(s) + 2H+  Zn2+ + H2O(l) log Ks0 = 11.2 ZnO(s) + H+  ZnOH+ log Ks1 = 2.2 ZnO(s) + 2H2O(l)  Zn(OH)3- + H+ log Ks3 = -16.9 ZnO(s) + 3H2O(l)  Zn(OH)42- + 2H+ log Ks4 = -29.7 The solubility of zincite is given by: