1 / 84

Unit 09a : Advanced Hydrogeology

Unit 09a : Advanced Hydrogeology. Chemical Reactions. Chemical Reactions. A wide variety of chemical reactions can take place between gases, solutes and solids in groundwater systems: Acid-base Solution-precipitation Complexation. Redox Hydrolysis Isotopic processes.

raine
Télécharger la présentation

Unit 09a : Advanced Hydrogeology

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Unit 09a : Advanced Hydrogeology Chemical Reactions

  2. Chemical Reactions • A wide variety of chemical reactions can take place between gases, solutes and solids in groundwater systems: • Acid-base • Solution-precipitation • Complexation. • Redox • Hydrolysis • Isotopic processes

  3. Hydrogen Ion Activity • [H+] represents the activity of hydrogen ions in solution: pH = - log[H+] = -log[H3O+] • since hydrogen ions exist in solution in the hydrated form as H3O+ • this allows us to distinguish between hydrogen ions and protons • pH is a “master variable” controlling chemical systems. • pH is controlled by acid-base reactions.

  4. Acids and Bases • An acid is a substance with a tendency to lose protons • A base is a substance with a tendency to gain protons. • Acids react with bases to transfer protons • In acid-base reactions, because no free protons are produced, there must be two acid-base systems involved: Acid1 + Base1 = Acid2 + Base2

  5. Bicarbonate Reaction HCO3- + H2O = H3O+ + CO32- • A proton is transferred (donated) from the bicarbonate ion to the water molecule to create an hydrogen ion. K = [H3O+ ] [CO32-] = [H+ ] [CO32-] [HCO3- ] [H2O] [HCO3- ] • assuming the activity of water is unity and abandoning our “hydrogen ion” distinction. • Remember the reaction is nevertheless an acid-base reaction with water as a base: HCO3- = H+ + CO32-

  6. Ammonia Reaction H2O + NH3 = NH4+ + OH- • Ammonia is a base that ionizes in water by accepting a proton. • In this case, water is the proton donor and water acts as an acid. • The concept of acid or base is simple: the proton donor is the acid the proton acceptor is a base. • In the first reaction water accepted a proton from the bicarbonate ion. In this reaction water donates a proton.

  7. Strong and Weak Acids HA + H2O = H3O+ + A- • The strength of an acid depends on its ability to drive the ionization reaction from left to right. • Strong acids donate protons freely in spite of the fact that water is a weak base (proton acceptor).

  8. Weak Acid-Base Reactions • There are a few weak acid reactions that are important in groundwater systems: • dissociation of water • dissociation of carbonic acid (dissolution of gaseous CO2) • dissociation of silicic acid (dissolution of silicate minerals)

  9. Dissociation of Water H20 + H2O = H3O+ + OH- • The equilibrium constant for this acid-base reaction, where water is both acid and base, is 10-14 Kw = [H+] [OH-] • again we assume [H2O] is unity and don’t bother to include it explicitly in the equations • Just as pH represents –log[H+], it is convenient to use pK to represent –log[K] • For the dissociation of water pKw = 14 • because [H+] = [OH-], the pH of pure water is 7.0

  10. Carbonic Acid CO2(g) + H20 = H2CO3* CO2(aq) • The equilibrium constant pKCO2 for the solution of carbon dioxide in water to produce carbonic acid is 1.46. H2CO3* = H+ + HCO3- • The first dissociation constant pK1 = 6.35 HCO3- = H+ + CO32- • The second dissociation constant pK2 = 10.33

  11. Carbonate Speciation Example p.1 • Calculate the distribution of mass between carbonate species at pH 7 given [CO2]T = 10-3 M. • Step 1: Identify the species. [CO2]T = [H2CO3*] + [HCO3-] + [CO32-] • Step 2: Calculate [H+] and [OH-] pH = 7 therefore [H+] = 10-7 [OH-] = Kw / [H+] = 10-14 / 10-7 = 10-7 • Step 3: Write [HCO3-] using K1 [HCO3-] = [H2CO3*] (K1 / [H+]) = [H2CO3*] (10-6.35 / 10-7) = [H2CO3*] (100.65) • Step 4: Write [CO32-] using K2 [CO32-] = [HCO3-] (K2 / [H+]) =[H2CO3*] (K1 / [H+]) (K2 / [H+]) = [H2CO3*] (100.65)(10-10.33 / 10-7) = [H2CO3*] (10-2.68)

  12. Carbonate Speciation Example p.2 • Step 5: Write [CO2]T in terms of [H2CO3*] to find [H2CO3*] [CO2]T = [H2CO3*] + [HCO3-] + [CO32-] [CO2]T = [H2CO3*] + [H2CO3*] (100.65) + [H2CO3*] (10-2.68) [CO2]T = [H2CO3*] ( 1 + 100.65 + 10-2.68 ) [CO2]T = [H2CO3*] (5.47) [H2CO3*] = [CO2]T / (5.47) [H2CO3*] = 10-3 / 5.47 = 10-3.74 • Step 6: Calculate [HCO3-] [HCO3-] = [H2CO3*] K1 / [H+] = 10-3.74(10-6.35 / 10-7) = 10-3.09 • Step 7: Calculate [CO32-] [CO32-] = [HCO3-] K2 / [H+] = 10-3.09(10-10.33 / 10-7) = 10-6.42 • The carbonate species [H2CO3*], [HCO3-] and [CO32-] have concentrations of 10-3.74, 10-3.09 and 10-6.42 M respectively at pH 7. Bicarbonate is the dominant ion.

  13. OH- H+ CO32- H2CO3* HCO3- HCO3- H+ OH- H2CO3* CO32- (CO2)T = 10-3 M Carbonate in Solution

  14. H+ OH- CO32- H2CO3* HCO3- HCO3- H+ OH- H2CO3* CO32- Carbonate in Solution (CO2)T = 10-4 M

  15. Carbonate System • Plotted for (CO2)T = 10-3 M or about 100 mg/L • Crossover points • pK1 = 6.35 where [HCO3-] = [H2CO3*] • pK2 = 10.33 where [HCO3-] = [CO32-] • Carbonate species H2CO3* is dominant for pH < 6.35 HCO3- is dominant for 6.35 > pH < 10.33 CO32- is dominant for pH > 10.33

  16. Silicic Acid • Carbon and silicon both form strong covalent bonds with oxygen. • Silicic acid is also a weak acid like carbonic acid. H2SiO3 = H+ + HSiO3- • The first dissociation constant pK1 = 9.86 HSiO3- = H+ + SiO32- • The second dissociation constant pK2 = 13.1

  17. OH- H+ SiO32- H2SiO3 HSiO3- HSiO3- H2SiO3 OH- H+ SiO32- (SiO2)T = 10-4 M Silica in Solution

  18. Silica System • Plotted for (SiO2)T = 10-4 M or about 10 mg/L • Crossover points • pK1 = 9.86 where [HSiO3-] = [H2SiO3] • pK2 = 13.1 where [HSiO3-] = [SiO32-] • Silicate species H2SiO3 is dominant for pH < 9.86 HSiO3- is dominant for 9.86 > pH < 13.1 SiO32- is dominant for pH > 13.1

  19. Alkalinity • The Bjerrum plots (log[C] against pH) show the effect of pH alone on speciation but in “real world” systems there are additional factors to consider. • The equilibria are influenced by strong bases added through the dissolution of carbonates and silicates. • Alkalinity is the net concentration of strong bases in excess of strong acids with a pure water – CO2 system as a reference point (zero alkalinity).

  20. Charge Balance • When CO2 is dissolved in water (at a fixed PCO2 ) the charge balance is: [H+] = [OH-] + [HCO3-] + 2[CO32-] • the cations must balance the anions • Adding a strong base (NaOH) and a strong acid (HCl) to the system will add Na+ and H+ cations and OH- and Cl- anions so the charge balance becomes: [Na+] + [H+] = [OH-] + [HCO3-] + 2[CO32-] + [Cl-]

  21. Alkalinity Defined • The net excess contribution of ions from the strong base is the alkalinity (for Na+ > Cl-)given by: [Na+] - [Cl-] = [OH-] + [HCO3-] + 2[CO32-] - [H+] • If we generalize to any base and any acid (rather than NaOH and HCl) we can write: Alkalinity = S [i+]sb – S [i-]sa = [OH-] + [HCO3-] + 2[CO32-] - [H+] • For the pure water-CO2 reference system, the alkalinity is zero.

  22. Net Alkalinity and Net Acidity • In most natural systems, the generation of net positive charges from dissolution of carbonates and silicates usually exceeds the contribution of negative charges from the ionization of strong acids. Most natural groundwaters are alkaline. • When strong acids are present (for example from pyrite oxidation) groundwaters may display net acidity. Such cases are often associated with sulphide mineralization or contamination by acid rock drainage (ARD).

  23. Mineral Dissolution • Increasing alkalinity results from an increase in positive ions on the LHS of the alkalinity equation from mineral dissolution. • An equal number of negative ions are added to the RHS to maintain neutrality and some of these ions come from ionization of H2CO3* to HCO3- and H2SiO3 to HSiO3-. • The equilibria for ionization of the weak acids removes hydrogen ions as HCO3- and HSiO3- are produced since the dissociation constant is invariant. • In most natural groundwater systems, pH increases along the flow path as minerals are dissolved.

  24. Solute Mass Loadings • Water is an excellent solvent. • Mineral dissolution is primarily responsible for the mass loadings in groundwater. • Other processes contribute to the reduction of solute mass loadings. These include: • gas exsolution • volatilization • precipitation

  25. Henry’s Law • Henry’s law does not strictly apply to gases (like CO2 and NH3) that react in solution. • Very little [CO2]aq reacts and H2CO3 concentrations are very low such that [CO2]aq [H2CO3*] • Henry’s law thus adequately approximates the distribution of CO2 between the aqueous and gaseous phases: KH = PCO2 [CO2]aq • KH is the Henry’s law constant with units of kPa.L.mol-1 or atm.L.mol-1.

  26. Carbon Dioxide Solution • Changes in PCO2 directly effect [CO2]aq and hence [H2CO3*] • Changes in [H2CO3*] through the dissociation constants pK1 and pK2 influence [HCO3-], [CO32-], [OH-] and pH. • Addition of CO2 to groundwater through the unsaturated zone increases [HCO3-] and pH and enhances the ability of the solution to dissolve silicate minerals.

  27. Volatilization • Volatilization is the process of liquid or solid phase evaporation at a liquid-gas or solid-gas interface. • The process of volatilization of solutes is controlled by Henry’s law. • NOTE: we are not discussing any non-aqueous phase liquids. • Volatilization can create problems in sampling. When samples have access to the atmosphere, loss of volatiles to the vapour phase can be significant.

  28. Dissolution and Precipitation • Dissolution and precipitation of solids are two of the most import processes controlling groundwater chemistry. • Groundwater systems evolve towards the equilibrium state either from undersaturation (most natural systems) or oversaturation (some contaminated systems). • In natural systems, dissolution proceeds and pH rises as waters evolve along the flow path. • In contaminated systems, precipitation can remove metals as pH rises along the flow path.

  29. Mineral Solubility • Solubility reflects the extent to which the reactant (mineral) and products (ions and/or secondary minerals) are favoured in a dissolution-precipitation reaction. • Because the activity of the reacting solid is taken to be unity, the magnitude equilibrium constant pK provides a relative measure of mineral solubility (in pure water). • When other ions are present, absolute and relative solubilities can change.

  30. Mineral IAP pK Halite [Na+][Cl-] -1.54 Sylvite [K+][Cl-] -0.98 Quartz [H2SiO3] 4.00 Gypsum [Ca2+][SO42-] 4.62 Magnesite [Mg2+][CO32-] 7.62 Aragonite [Ca2+][CO32-] 8.22 Calcite [Ca2+][CO32-] 8.35 Na-Montmorillonite 9.10 [Na+] [H4SiO4]4 [H4SiO4]2 Kaolinite 9.40 Siderite [Fe2+][CO32-] 10.70 Brucite [Mg2+][OH-]2 11.10 Ferrous Hydroxide [Fe2+][OH-]2 15.10 Dolomite [Ca2+][Mg2+][CO32-]2 16.70 Pyrrhotite [Fe2+][S2-] 18.10 Spalerite [Zn2+][S2-] 23.90 Galena [Pb2+][S2-] 27.50 Gibbsite [Al3+][OH-]3 33.50 Common Mineral Solubilities

  31. Ionic Strength Effect • Generally solubility increases with increasing ionic strength. • The presence of other ions reduces the activity of ions involved in the reaction. • This increases the number of ions needed in solution to achieve the equilibrium IAP.

  32. Common Ion Effect • When a solution contains the ion that is released when a solid dissolves, the presence of that ion means that less dissolution is required to reach the equilibrium IAP. • This phenomenon decreases the solubility and is called the common ion effect.

  33. Complexation • A complex is an ion that forms by combining simpler anions, cations and molecules. • The cation (or central atom) is typically a metal. • The anion (or ligand) is almost any simple anion (halide, sulphate, carbonate, phosphate, etc) • A simple complexation reaction involves a metal and a ligand: Zn2+ + Cl- = ZnCl-

  34. Complex Complexes • Sometimes complexes combine with ligands and metals are distributed among a large number of cation complexes. • For example, the hydrolysis of the trivalent chromium ion: Cr3+ + OH- = Cr(OH)2+ Cr(OH)2+ + OH- = Cr(OH)2+ Cr(OH)2+ + OH- = Cr(OH)30

  35. General Complexes • Most reactions involving complexes are “fast” in a kinetic sense. • A general complexation reaction involves a metal cation (M), b ligands (L) and c hydrogen ions (H): aM + bL + cH = MaLbHc • The stability constant for the complex KMLH is given by the association reaction: KMLH = [MaLbHc] [M]a[L]b[H]c or pKMLH = a.log[M] + b.log[L] + c.log[H] – log[MaLbHc] • The larger the value of KMLH, the more stable the complex

  36. More about Complexes • Most complexes involve a single cation: ZnCl+, Cr(OH)2+ • Polynuclear complexes are relatively unusual: Cr3(OH)45+, Cu2(OH)22+ • Complexation facilitates the transport of potentially toxic metals such as Cd, Cu, Cr, Mo, Pb, and U. For example, U forms complexes with many ligands including F, CO3, SO4 and PO4.

  37. Complex Speciation Example p.1 • A solution contains a trace of chromium (10-5 M) at a pH of 5. Determine the speciation among the Cr hydroxyl complexes given stability constants (where pK = -log K) pK2= -10.0, pK1 = -18.3 and pK0 = -24.0 • Step 1: Identify the species [Cr]T = [Cr3+] + [Cr(OH)2+] + [Cr(OH)2+] + [Cr(OH)30] • Step 2: Calculate [H+] and [OH-] pH = 5 therefore [H+] = 10-5 [OH-] = Kw / [H+] = 10-14 / 10-5 = 10-9 • Step 3: Use the association reactions to find [C] for complexes [Cr]T = [Cr3+] + K2[Cr3+][OH-] + K1[Cr3+][OH-]2 + K0[Cr3+][OH-]3

  38. Complex Speciation Example p.2 • Step 4: Solve for [Cr3+] [Cr]T = [Cr3+] + K2[Cr3+][OH-] + Ki[Cr3+][OH-]2 + K0[Cr3+][OH-]3 [Cr]T = [Cr3+] (1 + K2[OH-] + Ki[OH-]2 + K0[OH-]3) [Cr3+] = [Cr]T / (1 + K2[OH-] + Ki[OH-]2 + K0[OH-]3) [Cr3+] = 10-5 / (1 + 1010.10-9 + 1018.3.10-18 + 1024.10-27) [Cr3+] = 10-5 / (1 +101 + 100.3 + 10-3) [Cr3+] = 10-5 / (1 + 10 + 1.995 + 0.001) = 10-5 / (12.996) = 10-6.12 • Step 5: Solve for [Cr(OH)2+] [Cr(OH)2+] = K2[Cr3+][OH-] = 1010.10-6.12.10-9 = 10-5.12 • Step 6: Solve for [Cr(OH)2+] [Cr(OH)2+] = Ki[Cr3+][OH-]2 = 1018.3.10-6.12.10-18 = 10-5.82 • Step 7: Solve for [Cr(OH)2+] [Cr(OH)30] = K0[Cr3+][OH-]3 = 1024.10-6.12.10-27 = 10-9.12 • The chromium species [Cr3+], [Cr(OH)2+], [Cr(OH)2+], and [Cr(OH)30] have molar concentrations of 10-6.12, 10-5.12, 10-5.82 and 10-9.12 M respectively at pH 5. Cr(OH)2+ is the dominant ion.

  39. Major Ion Complexation • When we calculated IAP/K ratios and saturation indices earlier, we used total concentrations. • The true concentrations are significantly reduced by complexation. • Even at relatively low ionic strength >0.02 M, the error in determining mineral saturations can be substantial if complexation is neglected. • In seawater, for example, only 40% of the total SO4exists as SO42-, 37% exists as NaSO4+ and 19% as MgSO40

  40. Metal Mobility • Generally, in groundwater, metals are most mobile at low pH. • Ignoring surface reactions, metal concentrations begin to decline when pH increases to the point where equilibrium is reached with a solid phase. • The solid phases are usually metal-hydroxides, metal-sulphides or metal-carbonates.

  41. Uranium Complexes • Complexes can enhance the transport of metals at low concentrations. • Uranium is a good example, forming many uranyl (UO22+) complexes with ligands including F-, CO32-, SO42- and PO43-. • At pH 7, (UO2)(HPO4)22- is the dominant uranyl species and increases the solubility of some uranium minerals by several orders of magnitude (Langmuir, 1978).

  42. Surface Reactions • When water containing a trace constituent is mixed with a disseminated solid and allowed to equilibrate, mass partitions between the solution and the solid surface. S = (Co – C).V / Ms where S is the mass sorbed on the surface (M/M); Co is the initial concentration in solution (ML-3) and C is the equilibrium concentration in solution (ML-3), V is the solution volume (L3) and Ms is the mass of solid (M).

  43. 1.0 n = 0.1 0.9 n = 0.2 0.8 n = 0.5 0.7 0.6 0.5 Mass Sorption 0.4 n = 1 0.3 n = 2 0.2 n = 5 0.1 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Concentration Freundlich Isotherm • The function S(C) describing the surface sorption for various equilibrium concentrations is called a sorption isotherm. • Isotherms functions have no theoretical form and have be derived empirically. S = K.Cn is a form suggest by Freundlich where K and n are empirical constants.

  44. 1.0 K = 20 0.9 0.8 0.7 0.6 K = 0.5 0.5 Mass Sorption 0.4 0.3 0.2 0.1 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Concentration Langmuir Isotherm • The Langmuir isotherm has a more complex form: S = Q.K.C / (1 + K.C) where Q is the maximum sorptive capacity of the surface and K is a partition coefficient. • The value of K controls the extent of sorption. • The Langmuir isotherm limits the maximum mass sorption through the parameter Q.

  45. Linear Isotherm • The Freundlich isotherm for n =1 is called a linear isotherm S = Kd.C where Kd is the distribution coefficient. • This special case has been widely used to represent sorption of metals. • Finding a single value Kd to characterize the sorption process has proved difficult and more complex models are demanded by experience. 1.0 0.9 0.8 0.7 0.6 0.5 Mass Sorption 0.4 0.3 0.2 0.1 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Concentration

  46. Surface Reactions • For surface reactions, it is possible to account for the properties of the solution and the solid surfaces. • Cation exchange is the best-known surface reaction. • The process is driven by electrostatic attraction between charged cations and the surface charge on clay mineral and oxide/hydroxide surfaces. • Clay mineral surfaces have significant negative fixed charges due lattice substitutions and broken bonds at the edges of the minerals. • Cations bind to the surfaces to balance the charge.

  47. Cation Exchange Capacity • Cation exchange capacity (CEC) describes the quantity of exchangeable cations sorbed onto a surface. • CEC has units of meq per 100 g of sample. • CEC varies from one mineral to another but is strongly related to surface area.

  48. Clay Mineral CEC

  49. Cation Affinity • Clay minerals exhibit a preference for specific ions occupying exchange sites: Li+ < Na+ < H+ < K+ < NH4+ < Mg2+ < Ca2+ < Al3+ • In general, cation affinity for exchange sites increases with ionic charge. • At high concentrations, ion hydration and complexation can influence cation affinities. • In general, monovalent ions have hydration energies of around 100 kcal/mol compared with 400-500 kcal/mol for divalent ions and >1000 kcal/mol for Al3+ and Fe3+

  50. Exchange Reactions • The general form of a cation exchange reaction is: nMX + mNn+ = nMm+ + mNX where M and N are metal cations with charges m+ and n+ respectively and MX and NX are the corresponding metals sorbed on the solid phase. • For example: Na-clay + K+ = Na+ + K-clay n1Ca2+ + n2Mg2+ + n3Fe2+ + 2(n1+n2+n3)Na-clay = 2(n1+n2+n3)Na+ + Ca-Mg-Fe-clay Ca2+ + 2Mg2+ + Fe2+ + 8Na-clay = 8Na+ + Ca-Mg-Fe-clay

More Related