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Mineral Stability

Mineral Stability. saturation. equilibrium. Na +. disequilibrium. 0. Time. Law of Mass Action. ( ) = conc. at equilibrium. Activity is the effective concentration a = g c. = activity coefficient c = concentration. Rules when dealing with equilibrium constants.

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Mineral Stability

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  1. Mineral Stability saturation equilibrium Na+ disequilibrium 0 Time

  2. Law of Mass Action ( ) = conc. at equilibrium Activity is the effective concentration a = gc • = activity coefficient • c = concentration

  3. Rules when dealing with equilibrium constants • when adding equations, their equilibrium constants are multiplied • when subtracting equations, their equilibrium constants are divided • when reversing an equation, its equilibrium constant is inverted

  4. Mineral Dissociation Cd(OH)2(s) is amphoteric: Combine (add) equations, multiply K: KT = K1 x K2

  5. Mineral Dissociation Reverse the reaction, invert K: 1/Kaq Subtracting equations, divide K: K1/Kaq

  6. Brucite dissociation — Solubility of a sparingly soluble base Mg(OH)2(s) Mg(OH)+(aq) + OH-K1 Mg(OH)+(aq) Mg2+(aq) + OH-K2 pK1 = 8.6, pK2 = 2.6 1 — Weak acids and bases do not control the pH of natural environments but rather respond to it 2 — The solubility of sparingly soluble bases is pH dependent Log of Mg ions Dominantly Mg+ when pH <11.4 (most geologic environments) Faure p. 123, Fig. 9.2

  7. Weak acids and bases — pH control of dissociation H2S(aq) HS- + S2- pH < 7.0 pH = 7.0-12.9 pH > 12.9 These relationships occur in all weak acids and bases because their dissociation into ions is controlled by the pH of the environment Faure p. 125, Fig. 9.3

  8. Silicic acid dissociation — solubility of amorphous silica SiO2(amorphous) + H2O H4SiO4 Solubility of amorphous silica rises rapidly at pH ≥8 because it is based on the sum of the concentrations of the Si-bearing ions in solution Agate Opal A decrease of 0.1 pH from 8.5 to 8.4 of a saturated solution can deposit 1.37 mg of amorphous SiO2 per liter of solution SiO2•nH2O Faure p. 127, Fig. 9.4

  9. Dissolution and the solubility of salts The simplest weathering reaction is dissolution of soluble salts, for example: CaSO4(anhydrite)  Ca2+ + SO42- Solubility - The total amount of a substance that will dissolve in a solution at equilibrium.

  10. Solubility Product The solubility of a mineral is governed by the solubility product, the equilibrium constant for a reaction such as: CaSO4(anhydrite)  Ca2+ + SO42- The solubility product is given by: If anhydrite is a pure solid, then aCaSO4 = 1. and in dilute solutions: aCa2+  (Ca2+) and aSO42-  (SO42-).

  11. KSP [Ca2+][SO42-] = 10-4.5 What is the solubility of anhydrite in pure water? If anhydrite dissolution is the only source of both Ca2+ and SO42-, then: [Ca2+] = [SO42-] = x x2 = 10-4.5 x = 10-2.25 = 5.62x10-3 mol/L MWanhydrite = 136.14 g/mol Solubility = (5.62x10-3 mol/L)(136.14 g/mol) = 0.765 g/L

  12. A more complicated salt… Al2(SO4)3(s)  2Al3+ + 3SO42- Assume aAl2(SO3)3 = 1, aAl3+  [Al3+] and aSO42-  [SO42-] KSP  [Al3+]2[SO42-]3 = 69.19 Let x = the total number of moles of Al2(SO4)3(s) dissolved. [Al3+]= 2x and [SO42-] = 3x

  13. Al2(SO4)3(s)  2Al3+ + 3SO42- = 69.19 69.19 = (2x)2(3x)3 69.19 = 108x5 x = 0.9147 mol/L [Al3+]= 2x = 1.829 mol/L [SO42-] = 3x = 2.744 mol/L x = (0.9147 mol/L)(342 g/mol) x = 312.8 g/LAl2(SO4)3(s)

  14. Saturation index In a natural solution, it is not likely that [Ca2+] = [SO42-], for example, because there will be more than one source of each of these ions. In this case we use saturation indices to determine if the water is saturated with respect to anhydrite. KSP = 10-4.5 [Ca2+]eq[SO42-]eq IAP = [Ca2+]act[SO42-]act Saturation index

  15. CaSO4 Ca2+ + SO42- Suppose a groundwater is analyzed to contain 5x10-2 mol/L Ca2+ and 7x10-3 mol/L SO42-. Is this water saturated with respect to anhydrite (CaSO4(s))? KSP = 10-4.5 IAP = (5x10-2)(7x10-3) = 3.5x10-4 = 10-3.45 • = 10-3.45/10-4.5 = 101.05 = 11.22 • > 1, i.e., IAP > KSP, so the solution is supersaturated and anhydrite should precipitate If  = 1, i.e., IAP = KSP, the solution would be saturated (equilibrium conditions) If  < 1, i.e., IAP < KSP, the solution would be undersaturated; the mineral should dissolve

  16. Another example… Suppose the drainage from a tailings pile contains 5x10-3 mol/L SO42-, 10-4 mol/L Fe3+, 10-3 mol/L K+, and has pH = 3. Is this water saturated with respect to the mineral jarosite, KFe3(SO4)2(OH)6? KSP = 10-96.5 IAP = [K+][Fe3+]3[SO42-]2[OH-]6 [OH-] = Kw/[H+] = 10-14/10-3 = 10-11 mol/L IAP = (10-3)(10-4)3(5x10-3)2(10-11)6 = 10-85.60 • = 10-85.60/10-96.5 = 1010.898 = 7.91x1010 So the solution would be highly supersaturated and jarosite should precipitate

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