1. DISSOLUTION

2. The simplest weathering reaction is dissolution of soluble salts, e.g. CaSO 4 (anhydrite)  Ca 2+ + SO 4 2- Solubility - The total amount of a substance that will dissolve in a solution at equilibrium.

3. SOLUBILITY PRODUCT The solubility of a mineral is governed by the solubility product, the equilibrium constant for a reaction such as: CaSO 4 (anhydrite)  Ca 2+ + SO 4 2- The solubility product is given by: If anhydrite is a pure solid, then a CaSO 4 = 1. and in dilute solutions: a Ca 2+  (Ca 2+ ) and a SO 4 2-  (SO 4 2- ).

4. So we may write: K SP = 10 -4.5 = [Ca 2+ ][SO 4 2- ] What is the solubility of anhydrite in pure water? If anhydrite dissolution is the only source of both Ca 2+ and SO 4 2-, then by stoichiometry: [Ca 2+ ] = [SO 4 2- ] = x x 2 = 10 -4.5 x = 10 -2.25 = 5.62x10 -3 mol L -1 MW anhydrite = 136.14 g mol -1 solubility = (5.62x10 -3 mol L -1 )(136.14 g mol -1 ) = 0.765 g L -1

5. A MORE COMPLICATED SALT Al 2 (SO 4 ) 3 (s)  2Al 3+ + 3SO 4 2- Assume a Al 2 (SO 3 ) 3 = 1, a Al 3+  [Al 3+ ] and a SO 4 2-  [SO 4 2- ]. Then: K SP = 69.19 = [Al 3+ ] 2 [SO 4 2- ] 3 Let x = the total number of moles of Al 2 (SO 4 ) 3 (s) dissolved. Then, by stoichiometry: [Al 3+ ] = 2xand[SO 4 2- ] = 3x

6. 69.19 = (2x) 2 (3x) 3 69.19 = 108x 5 x = 0.9147 mol L -1 [Al 3+ ] = 2x = 1.829 mol L -1 [SO 4 2- ] = 3x = 2.744 mol L -1 x = (0.9147 mol L -1 )(342 g mol -1 ) = 312.8 g L -1 Al 2 (SO 4 ) 3 (s)

7. SATURATION INDEX In a natural solution, it is not likely that [Ca 2+ ] = [SO 4 2- ], 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. K SP = 10 -4.5 = [Ca 2+ ] eq [SO 4 2- ] eq IAP = [Ca 2+ ] act [SO 4 2- ] act Saturation index

8. Suppose a groundwater is analyzed to contain 5x10 -2 mol L -1 Ca 2+ and 7x10 -3 mol L -1 SO 4 2-. Is this water saturated with respect to anhydrite? K SP = 10 -4.5 mol 2 L -2 IAP = (5x10 -2 )(7x10 -3 ) = 3.5x10 -4 = 10 -3.45 mol 2 L -2  = 10 -3.45 /10 -4.5 = 10 1.05 = 11.22  > 1, i.e., IAP > K SP, so the solution is supersaturated and anhydrite should precipitate. If  = 1, i.e., IAP = K SP, the solution would be saturated. If  < 1, i.e., IAP < K SP, the solution would be undersaturated; the mineral should dissolve.

9. ANOTHER EXAMPLE Suppose the drainage from a tailings pile contains 5x10 -3 mol L -1 SO 4 2-, 10 -4 mol L -1 Fe 3+, 10 -3 mol L -1 K +, and has pH = 3. Is this water saturated with respect to the mineral jarosite, KFe 3 (SO 4 ) 2 (OH) 6 ? K SP = 10 -96.5 mol 12 L -12 IAP = [K + ][Fe 3+ ] 3 [SO 4 2- ] 2 [OH - ] 6 [OH - ] = K w /[H + ] = 10 -14 /10 -3 = 10 -11 mol L -1 IAP = (10 -3 )(10 -4 ) 3 (5x10 -3 ) 2 (10 -11 ) 6 = 10 -85.60 mol 12 L -12  = 10 -85.60 /10 -96.5 = 10 10.898 = 7.91x10 10 So the solution would be highly supersaturated.

10. HOW MUCH SALT SHOULD PRECIPITATE? Returning to the previous example, i.e., the groundwater with 5x10 -2 mol L -1 Ca 2+ and 7x10 -3 mol L -1 SO 4 2-, how much anhydrite should precipitate at equilibrium? If x mol L -1 of anhydrite precipitate, then at equilibrium: [Ca 2+ ] = 5x10 -3 - x; [SO 4 2- ] = 7x10 -3 - x and [Ca 2+ ][SO 4 2- ] = 10 -4.5 (5x10 -3 - x)(7x10 -3 - x) = 10 -4.5 x 2 - (5.7x10 -2 )x + (3.249x10 -4 ) = 0 x 1 = 5.05x10 -2 mol L -1 ; x 2 = 6.45x10 -3 mol L -1

11. The first root is unreal because it would cause both final concentrations to be negative. So, 6.45x10 -3 mol L -1 of anhydrite precipitates, or (6.45x10 -3 mol L -1 )(136.1 g mol -1 ) = 0.878 g L -1 and [Ca 2+ ] = (5x10 -2 ) - (6.45x10 -3 ) = 4.36x10 -2 mol L -1 [SO 4 2- ] = (7x10 -3 ) - (6.45x10 -3 ) = 5.5x10 -4 mol L -1 Ca : SO 4 2- ratio increases with increasing precipitation of anhydrite Normally gypsum would precipitate instead of anhydrite!

12. Figure 10.1 from Faure (1991) Initial [Ca 2+ ] = 5x10 -2 and [SO 4 2- ] = 7x10 -3 M. Ca 2+ /SO 4 2- is initially > 1 and always increases upon evaporation and precipitation. Precipitation of gypsum - CaSO 4 2H 2 O

13. THE COMMON-ION EFFECT Natural waters are very complex and we may have saturation with respect to several phases simultaneously. Example: What are the concentrations of all species in a solution in equilibrium with both barite and gypsum? 1) Species: Ca 2+, Ba 2+, SO 4 2-, H +, OH - 2) Mass action expressions: CaSO 4 ·2H 2 O  Ca 2+ + SO 4 2- + 2H 2 O K SP = [Ca 2+ ][SO 4 2- ] = 10 -4.6

14. BaSO 4  Ba 2+ + SO 4 2- K SP = [Ba 2+ ][SO 4 2- ] = 10 -10.0 H 2 O  H + + OH - K w = [H + ][OH - ] = 10 -14 3) Mass-balance: [Ba 2+ ] + [Ca 2+ ] = [SO 4 2- ] 4) Charge-balance: 2[Ba 2+ ] + 2[Ca 2+ ] + [H + ] = 2[SO 4 2- ] + [OH - ]

15. 10 -4.6 + 10 -10.0 = [SO 4 2- ] 2 [SO 4 2- ] = (10 -4.6 ) 1/2 = 10 -2.3 mol L -1 [Ca 2+ ] = 10 -4.6 /10 -2.3 = 10 -2.3 mol L -1 [Ba 2+ ] = 10 -10.0 /10 -2.3 = 10 -7.7 mol L -1 The least soluble salt (barite), contributes a negligible amount of sulfate to the solution. The more soluble salt decreases the solubility of the less soluble salt (common-ion effect).

16. The solubility of gypsum is hardly affected by the presence of barite. Solubility of barite alone: [Ba 2+ ][SO 4 2- ] = 10 -10.0 [Ba 2+ ] 2 = 10 -10.0 [Ba 2+ ] = 10 -5.0 mol L -1 Solubility of gypsum alone: [Ca 2+ ][SO 4 2- ] = 10 -4.6 [Ca 2+ ] 2 = 10 -2.3 [Ca 2+ ] = 10 -2.3 mol L -1