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PRECIPITATION AND DISSOLUTION

2. EXTENT OF DISSOLUTION. We can assess the extent of dissolution or precipitation using the equilibrium constant, e.g.:SiO2(s) 2H2O(l) ? H4SiO40Assuming SiO2(s) is a pure solid and solution is dilute so that water is nearly pure:Ks0 = (H4SiO40) = ?H4SiO40[H4SiO40]for neutral species, ? ? 1, s

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PRECIPITATION AND DISSOLUTION

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    1. 1 PRECIPITATION AND DISSOLUTION

    2. 2 EXTENT OF DISSOLUTION We can assess the extent of dissolution or precipitation using the equilibrium constant, e.g.: SiO2(s) + 2H2O(l) ? H4SiO40 Assuming SiO2(s) is a pure solid and solution is dilute so that water is nearly pure: Ks0 = (H4SiO40) = ?H4SiO40[H4SiO40] for neutral species, ? ? 1, so Ks0 = [H4SiO40] If a solution has an actual [H4SiO40] > Ks0, it is supersaturated and SiO2(s) should precipitate if kinetics allow; if [H4SiO40] < Ks0, the solution is undersaturated.

    3. 3 SOLUBILITY PRODUCTS Note that the concentration of H4SiO40 does not depend on the total amount of SiO2(s), as long at least some solid is present upon attainment of saturation. A generalized salt dissolves according to: AmBn(s) ? mA+n + nB-m and has a solubility product given by: Ks0 = [A+n]m[B-]n again, assuming the solid is pure.

    4. 4 OTHER REACTIONS ALSO AFFECT SOLUBILITY The solubility of FeS(s) depends not only on: FeS(s) ? Fe2+ + S2- but also on: Fe2+ + H2O ? FeOH+ + H+ S2- + H2O ? HS- + OH- HS- + H2O ? H2S0 + OH- Fe2+ + HS- ? FeHS+ FeS(s) + S2- ? FeS22-, etc. Solubility ? ?Fe = [Fe2+] + [FeOH+] + [FeHS+] + [FeS22-]

    5. 5 SOLUBILITY COMPLICATED BY METASTABILITY A metastable compound is one that may be at equilibrium, but is not the most stable in the system. An active, metastable compound may form first from a highly supersaturated solution. The active form may convert to the stable form, or age only very slowly. Metastable or active forms have higher solubilities than stable forms. Many experimental studies deal with active form, but in nature the stable form may be more important!

    6. 6

    7. 7 ION ACTIVITY PRODUCT AND SATURATION INDEX CaCO3(s) ? Ca2+ + CO32-

    8. 8 SOLUBILITIES OF SIMPLE SALTS Consider the mineral anhydrite: CaSO4(s) ? Ca2+ + SO42- What is the solubility of anhydrite in pure water?

    9. 9 Ks0 = [Ca2+][SO42-] = x2 = 10-4.5 x = 10-2.25 x = 5.62x10-3 mol L-1 5.62 x 10-3 mol L-1 136.14 g/mole = 0.766 g L-1 Now consider the salt Al2(SO4)3(s): Al2(SO4)3(s) ? 2Al3+ + 3SO42- Let x = mol L-1 of Al2(SO4)3(s) dissolved, then on complete, congruent dissolution we have: [Al3+] = 2x; [SO42-] = 3x Ks0 = [Al3+]2[SO42-]3 = (2x)2(3x)3 = (4x2)(27x3) = 108x5 = 69.19

    10. 10 x = 0.9148 [Al3+] = 2(0.9148) = 1.830 mol L-1 [SO42-] = 3(0.9148) = 2.744 mol L-1 Solubility of Al2(SO4)3(s) = 0.9148 mol L-1 x 342.1478 g mol-1 = 313 g L-1 Now consider wulfenite (PbMoO4). A solution contains 2x10-8 mol L-1 Pb2+ and 3x10-7 mol L-1 MoO42-. Should wulfenite precipitate from this solution, and if so, how much wulfenite will form? IAP = [Pb2+][MoO42-] = (2x10-8)(3x10-7) = 6x10-15 M2 Ks0 = 10-16.0 so ? = 6x10-15/10-16.0 = 60

    11. 11 Thus, this solution is supersaturated with respect to wulfenite by a factor of 60, and it should precipitate. How much? Let y = mol L-1 of wulfenite that precipitates, then [Pb2+]eq = 2x10-8 - y; [MoO42-]eq = 3x10-7 - y Ks0 = [Pb2+]eq[MoO42-]eq = (2x10-8 - y)(3x10-7 - y) = 6x10-15 - 3.2x10-7y + y2 = 10-16 5.90x10-15 - 3.2x10-7y + y2 = 0

    12. 12 y = 3.004x10-7 or 1.964x10-8 mol L-1 But the first root is impossible because it would make [Pb2+]eq and [MoO42-]eq less than zero, so the true root is the second one. Thus, 1.964x10-8 mol L-1 x 303.264 g mol-1 = 5.96 x 10-6 g L-1 = 5.96 ppb wulfenite would precipitate! Check of calculations: Ks0 = [Pb2+]eq[MoO42-]eq = (2x10-8 - y)(3x10-7 - y) Ks0 = (2x10-8 - 1.964x10-8)(3x10-7 - 1.964x10-8) = (3.60x10-10)(2.8036x10-7) = 1.0093x10-16 = 10-15.996 ? 10-16.0 so the calculations check.

    13. 13 GEOCHEMICAL DIVIDES In the last problem the ratio [MoO42-]/[Pb2+] changed from 3x10-7/2x10-8 = 15 to 2.804x10-7/3.60x10-10 = 779. If wulfenite continued to precipitate from this solution, as a result of evaporation for example, the ratio [MoO42-]/[Pb2+] would tend towards infinity! This is an example of a geochemical divide. This solution is enriched in molybdate and depleted in lead because the initial ratio [MoO42-]/[Pb2+] > 1. Thus, precipitation can only cause this ratio to increase. It can never decrease, so the ratio [MoO42-]/[Pb2+] = 1 is a divide we can never cross! If we started with [MoO42-]/[Pb2+] < 1, precipitation would never result in a ratio > 1!

    14. 14

    15. 15 COMMON-ION EFFECT Natural solutions of even simple salts are more complex than the examples we have looked at so far. Consider simultaneous saturation of a solution with both wulfenite (PbMoO4) with pKs0 = 16.0 and powellite (CaMoO4) with pKs0 = 7.94. PbMoO4 ? Pb2+ + MoO42- CaMoO4 ? Ca2+ + MoO42- Both reactions contribute MoO42- ions to the solution, but the final activity (concentration) of MoO42- must be the same for both equilibria. The solubility product expressions become:

    16. 16 [Pb2+][MoO42-] = 10-16 [Ca2+][MoO42-] = 10-7.94 and eliminating [MoO42-] we obtain: [Pb2+]/[Ca2+] = 10-16/10-7.94 = 10-8.06 so the concentration of [Pb2+] will be ~8.7x10-9 times less than the concentration of [Ca2+]. To determine actual concentrations, we start with the expression: [Pb2+] + [Ca2+] = [MoO42-] which is valid if wulfenite and powellite are the only sources of Pb2+, Ca2+ and MoO42-. Now, [Pb2+] = 10-16/[MoO42-] [Ca2+] = 10-7.94/[MoO42-]

    17. 17 But the first term is negligible compared to the second term, so we get [MoO42-]2 = 10-7.94 [MoO42-] = 10-3.97 = 1.072x10-4 M [Pb2+] = 10-16/10-3.97 = 9.33x10-13 M [Ca2+] = 10-7.94/10-3.97 = 10-3.97 = 1.072x10-4 M so, [Ca2+] is much larger than [Pb2+]. Also note that, in the absence of powellite, [Pb2+] = [MoO42-] = y, so y2 = Ks0 = 10-16.0 y = 10-8.0

    18. 18 So [Pb2+] = 10-8.0 M, or about 4 orders of magnitude more than in the presence of powellite. However, in the absence of wulfenite, [Ca2+] = [MoO42-] = y so y2 = 10-7.94 y = 10-3.97 So [Ca2+] = 10-3.97 M, or the same as in the presence of wulfenite! The common-ion effect occurs when the presence of a more soluble salt depresses the solubility of a less soluble salt containing a common ion (can be a cation or anion). This effect can be used in environmental remediation.

    19. 19 REPLACEMENT We have established that a solution in equilibrium with powellite and wulfenite at 25C will have [Pb2+]/[Ca2+] = 10-8.06 If a solution for which this ratio is greater than the equilibrium value (e.g., 10-4) encounters a powellite- bearing rock, powellite will dissolve, and wulfenite will precipitate, leading to the replacement of powellite by wulfenite. This is how some ore deposits form. This phenomenon can also be used in remediation.

    20. 20 SOLUBILITY OF OXIDES AND HYDROXIDES Governing reactions for divalent metals are: Me(OH)2(s) ? Me2+ + 2OH- MeO(s) + H2O(l) ? Me2+ + 2OH- cKs0 = [Me2+][OH-]2 Sometimes it is more appropriate to write: Me(OH)2(s) + 2H+ ? Me2+ + 2H2O(l) MeO(s) + 2H+ ? Me2+ + H2O(l)

    21. 21 TRIVALENT METALS For a trivalent metal oxide, e.g., goethite FeOOH(s) + 3H+ ? Fe3+ + 2H2O(l)

    22. 22

    23. 23 NEED TO INCLUDE HYDROXIDE COMPLEXES Need also to consider the formation hydroxide complexes, i.e., hydrolysis. For example: Zn2+ + H2O(l) ? ZnOH+ + H+ Al(OH)2+ + H2O(l) ? Al(OH)2+ + H+ In general, the total solubility of a metal oxide or hydroxide in the absence of complexing ligands is:

    24. 24 GRAPHICAL REPRESENTATION OF ZnO SOLUBILITY At 25C 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 ?ZnT = [Zn2+] + [ZnOH+] + [Zn(OH)3-] + [Zn(OH)42-] ?ZnT = *Ks0[H+]2 + *Ks1[H+] + *Ks3[H+]-1 + *Ks4[H+]-2

    25. 25 Another reaction is also possible: ZnO(s) + H2O(l) ? Zn(OH)20 but *Ks2 is poorly known because solubility is a minimum here. The above reactions demonstrate that ZnO(s) is an amphoteric substance. To calculate solubility diagram: [Zn2+] = *Ks0[H+]2 log [Zn2+] = log *Ks0 - 2pH [ZnOH+] = *Ks1[H+] log [ZnOH+] = log *Ks1 - pH

    26. 26 [Zn(OH)20] = *Ks2 log [Zn(OH)20] = log *Ks2 [Zn(OH)3-] = *Ks3[H+]-1 log [Zn(OH)3-] = log *Ks3 + pH [Zn(OH)42-] = *Ks4[H+]-2 log [Zn(OH)42-] = log *Ks4 + 2pH A U-shaped curve results with solubilities high at low and high pH, and lower in the middle. This is typical of all amphoteric oxides and hydroxides.

    27. 27

    28. 28 CHECK OF CALCULATIONS A check of the above calculations can be made by calculating the pH of the crossover points. Zn2+ + H2O(l) ? ZnO(s) + 2H+ log *Ks0 = -11.2 ZnO(s) + H+ ? ZnOH+ log *Ks1 = 2.2 Zn2+ + H2O(l) ? ZnOH+ + H+ log Kh1 = -9.0

    29. 29 ZnOH+ ? ZnO(s) + H+ log *Ks1 = -2.2 ZnO(s) + 2H2O(l) ? Zn(OH)3- + H+ log *Ks3 = -16.9 ZnOH+ + 2H2O(l) ? Zn(OH)3- + 2H+ log K = -19.1

    30. 30 Zn(OH)3- + H+ ? ZnO(s) + 2H2O(l) log *Ks3 = 16.9 ZnO(s) + 3H2O(l) ? Zn(OH)42- + 2H+ log *Ks4 = -29.7 Zn(OH)3- + H2O(l) ? Zn(OH)42- + H+ log K = -12.8

    31. 31

    32. 32

    33. 33 SOLUBILITY OF CARBONATES Closed System a) CT is constant. What is the maximum dissolved Ca2+ concentration as a function of CT and pH? CaCO3(s) ? Ca2+ + CO32- Ks0 = [Ca2+][CO32-] [Ca2+] = Ks0/[CO32-] = Ks0/(CT?2) ?2 is known as a function of pH, so this equation yields the desired result. Where pH > pK2 for H2CO3*, ?2 ? 1.0 and CT ? [CO32-]. Solubility is controlled by the above equation and is pH-independent.

    34. 34 Where pK1 < pH < pK2, ?1 ? 1.0 and CT ? [HCO3-] CaCO3(s) + H+ ? Ca2+ + HCO3-

    35. 35

    36. 36 DISSOLUTION OF CALCITE IN PURE WATER Species: Ca2+, H2CO3*, HCO3-, CO32-, H+, OH- Mass action expressions: pK1, pK2, pKs0, pKw Mass balance (by stoichiometry): [Ca2+] = CT = [H2CO3*] + [HCO3-] + [CO32-] Electroneutrality: 2[Ca2+] + [H+] = [HCO3-] + 2[CO32-] + [OH-] From the solubility product we have: [Ca2+] = Ks0/[CO32-] [CO32-] = CT?2; [HCO3-] = CT?1; [H2CO3*] = CT?0

    37. 37 [Ca2+] = Ks0/(?2CT) but [Ca2+] = CT so [Ca2+] = Ks0/(?2[Ca2+]) [Ca2+]2 = Ks0/?2 [Ca2+] = CT = (Ks0/?2)1/2 [CO32-] = ?2(Ks0/?2)1/2; [HCO3-] = ?1(Ks0/?2)1/2; [H2CO3*] = ?0(Ks0/?2)1/2 now substitute into the charge balance 2(Ks0/?2)1/2 + [H+] = ?1(Ks0/?2)1/2 + 2?2(Ks0/?2)1/2 + Kw/[H+] (Ks0/?2)1/2(2 - ?1- 2?2) + [H+] - Kw/[H+] = 0 solving by trial and error we get pH = 9.9; log [Ca2+] = -3.9; log [HCO3-] = -4.05; log [CO32-] = -4.40; log [Alk] = -3.62; log [OH-] = -4.1

    38. 38 GRAPHICAL SOLUTION TO SOLUBILITY OF CALCITE IN PURE WATER Start with charge balance: 2[Ca2+] + [H+] = [HCO3-] + 2[CO32-] + [OH-] Because calcite is a base, [OH-] >> [H+] and [H2CO3*] is negligible so CT = [Ca2+] ? [HCO3-] + [CO32-] and 2[Ca2+] = [Ca2+] + [CO32-] + [OH-] [Ca2+] = [CO32-] + [OH-] For calcite, [CO32-] is almost negligible, but for other more insoluble carbonates, it is negligible, so [Ca2+] ? [OH-]

    39. 39

    40. 40 DISSOLUTION OF CALCITE IN A SOLUTION OF STRONG ACID OR BASE All equations are same as in previous example, but must modify the charge balance to: CA - CB = (Ks0/?2)1/2(2 - ?1- 2?2) + [H+] - Kw/[H+] solve the above for pH by trial and error and then: [Ca2+] = CT = (Ks0/?2)1/2 [CO32-] = ?2(Ks0/?2)1/2 [HCO3-] = ?1(Ks0/?2)1/2 [H2CO3*] = ?0(Ks0/?2)1/2

    41. 41 OPEN SYSTEM: CaCO3(s)-CO2-H2O Because the system is open to the atmosphere, [Ca2+] ? CT, but [H2CO3*] = pCO2KH. We also have the charge-balance expression: 2[Ca2+] + [H+] = CT(?1 + 2?2) + [OH-] and [Ca2+] = Ks0/[CO32-] and CT = pCO2KH/?0 and [CO32-] = CT ?2

    42. 42 By trial and error we get pH = 8.4 and log [Ca2+] = -3.3 log [HCO3-] = -3.0 log [CO32-] = -5.0 log [Alk] = -3.0 log [OH-] = -5.6

    43. 43 A SIMPLIFICATION AND A GRAPHICAL SOLUTION Starting with the charge-balance expression: 2[Ca2+] + [H+] = [HCO3-] + 2[CO32-] + [OH-] The terms [H+], [OH-] and [CO32-] are all negligible, leaving: 2[Ca2+] ? [HCO3-] So the solution can be found by substituting to get:

    44. 44

    45. 45 OPEN SYSTEM: CaCO3(s)-CO2-H2O WITH STRONG ACID OR BASE Again we modify the charge-balance: CB + 2[Ca2+] + [H+] = CT(?1 + 2?2) + [OH-] + CA and we obtain by substitution:

    46. 46 CALCITE SOLUBILITY IN SEAWATER Is seawater saturated with respect to calcite? In seawater at 25?C, the solubility of calcite is given by: cKs0 = [CaT][CO32-T] = 5.94 x 10-7 where [CaT] = total concentration of soluble Ca(II) = [Ca2+] + [Ca-complexes] [CO32-T] = total concentration of soluble carbonate = [CO32-] + [CO32- - complexes] also pcKH = 1.53; pcK1 = 6.00; pcK2 = 9.11

    47. 47 Again starting with the charge-balance: 2[Ca2+] + [H+] = CT(?1 + 2?2) + [OH-] and making the substitutions:

    48. 48 SO, AFTER ALL THAT, IS SEAWATER SATURATED WITH RESPECT TO CALCITE? The actual concentrations of CaT and CO3,T can be calculated from pH = 8.2 and [Alk] = 2.4 x 10-3 eq/L for seawater. [CO32-T] = CT?2 = ([Alk]?2)/(?1 + 2?2) = 3.87x10-4 M [CaT] = 1.06x10-2 M [CaT]act[CO32-T]act = (1.06x10-2)(3.87x10-4) = 4.1x10-6 M2 ? = IAP/cKs0 = 4.1x10-6 M2/ 5.94 x 10-7 M2 = 6.9 Thus, surface seawater is supersaturated with respect to calcite by a factor of ~7 and should precipitate.

    49. 49 EFFECT OF PRESSURE AND TEMPERATURE ON THE SOLUBILITY OF CALCITE IN SEAWATER How do pH, [CO32-T], [Ca2+T] and ? vary when seawater is cooled to 5?C and subjected to PT = 1000 atm? Assumptions: 1) CaCO3(s) does not precipitate or dissolve 2) Water initially calcite supersaturated 3) Borate does not affect calculations [Ca2+T], CT and [Alk] are conservative properties and remain independent of pressure and temperature.

    50. 50 At any P and T [Alk] ? CT(?1 + 2?2) ?1 and ?2 can be calculated with cK1 and cK2 valid for seawater at the appropriate pressure and temperature. The pressure correction is:

    51. 51 With the value of pH obtained we can now calculate [CO32-T] = CT?2 ?? = IAP/cpKs0 and we find that, [Ca2+T], [CO32-T] and, therefore, IAP, do not change much with P and T. However, pH and cpKs0 change considerably, so ? changes. Calcite solubility increases significantly with decreasing T and increasing P. Calcite displays retrograde solubility, i.e., solubility decreases with increasing temperature.

    52. 52

    53. 53 REACTION PATHS FOR CALCITE DISSOLUTION IN GROUNDWATER How does groundwater composition change as rainwater reacts with calcite? Two endmember cases: 1) Water in contact with a large reservoir at constant pCO2 2) Water becomes isolated We need to calculate the reaction progress, i.e., the compositional changes as a function of CaCO3 dissolution. At T = 10C and I = 4x10-3 M we have: pcKH = 1.27; pcK1 = 6.43; pcK2 = 10.38; pcKs0 = 7.95.

    54. 54 1) Reservoir with constant pCO2 Same equations as used previously [HCO3-] = (?1/?0)KHpCO2 which provides a linear relationship between log [HCO3-] and pH. 2) Closed system When the water becomes separated, dissolved CO2 is consumed according to: H2CO3* + CaCO3(s) ? Ca2+ + 2HCO3- when the system is closed, acidity does not change with the extent of CaCO3 dissolution.

    55. 55 [Acy] = 2[H2CO3*] + [HCO3-] + [H+] - [OH-] [Acy] ? CT(2?0 + ?1) = constant For the initial conditions we can calculate [Acy] using: CT = KHpCO2/?0 Knowing [Acy], we can calculate CT and [HCO3-] = CT?1 at selected pH values. This gives the curve in the following figure.

    56. 56

    57. 57 MIXING OF GROUNDWATERS Suppose we have two groundwaters, both exposed to and in equilibrium with pCO2 = 10-2 atm. Groundwater I equilibrates with a formation containing siderite (FeCO3) and no calcite. Groundwater II equilibrates with a formation containing only calcite and no siderite. Groundwaters I and II are then mixed in equal proportions. 1) Compute [Me2+], [HCO3-] and [H+] for each groundwater. 2) Compute the composition of the mixture. 3) Is the mixture stable with respect to precipitation of FeCO3?

    58. 58 1) Compute [Me2+], [HCO3-] and [H+] for each groundwater. Derive Kps0 for the reaction: MeCO3(s) + CO2(g) + H2O(l) ? Me2+ + 2HCO3- We need to add the following reactions: MeCO3(s) ? Me2+ + CO32- Ks0 H2CO3* ? H+ + HCO3- K1 CO2(g) + H2O(l) ? H2CO3* KH CO32- + H+ ? HCO3- 1/K2 MeCO3(s) + CO2(g) + H2O(l) ? Me2+ + 2HCO3- Kps0 = Ks0K1KH/K2

    59. 59 Electroneutrality is approximated by: 2[Me2+] ? [HCO3-]

    60. 60 [H+] = (10-6.3)(10-1.5)(10-2)/10-3.15 = 10-6.65 M pH = 6.65 Groundwater II (calcite): log Kps0 = -8.42 - 6.3 - 1.5 - (-10.3) = -5.92 [Ca2+] = 10-2.84 M; [HCO3-] = 10-2.54 M; pH = 7.26

    61. 61 2) Compute the composition of the mixture. [Ca2+], [Fe2+] and [HCO3-] of the mixture can be obtained according to the relations: [Ca2+]mix = ([Ca2+]I + [Ca2+]II)/2 = (10-2.84 + 0)/2 = 10-2.54 M [Fe2+]mix = ([Fe2+]I + [Fe2+]II)/2 = (0 + 10-3.45)/2 = 10-3.75 M [HCO3-]mix = ([HCO3-]I + [HCO3-]II)/2 = (10-2.54 + 10-3.15)/2 = 10-2.75 M

    62. 62 SUMMARY OF RESULTS

    63. 63 3) Is the mixture stable with respect to precipitation of FeCO3? FeCO3(s) ? Fe2+ + CO32- log Ks0 = -10.24 H+ + CO32- ? HCO3- -log K2 = 10.3 FeCO3(s) + H+ ? Fe2+ + HCO3- log Ks1 = 0.06 IAP = [Fe2+][HCO3-]/[H+] = (10-3.75)(10-2.75)/(10-7.04) = 100.54 ? = 100.54/100.06 = 3.02 So siderite should precipitate!

    64. 64 BUFFER INTENSITY IN PRESENCE OF CaCO3 1) Constant pCO2 case. Presence of calcite (or any carbonate) can greatly enhance the buffer capacity!

    65. 65

    66. 66 Thus, the presence of calcite in excess of that required for saturation greatly increases the buffer intensity of a natural water. An extremely narrow range of pH is allowed as long as calcite is present. 2) Closed system with CaCO3(s) but no gas phase. Under these conditions, the variable D = 2(CT - [Ca2+]) is constant when calcite dissolves or precipitates. If calcite is present but is not in equilibrium with a constant CO2 reservoir, buffer intensity not as high and pH range not as narrow as in case (1).

    67. 67

    68. 68

    69. 69 RELATIVE STABILITIES OF HYDROXIDES VS. CARBONATES We may need to answer the question: Which phase controls solubility? General rule is that the least soluble phase is the most stable one, and should control solubility. Example: Fe(II) at 1 atm, 25C and I = 6x10-3 M Ks0(Fe(OH)2) = 10-14.7 mol3 L-3 Ks0(FeCO3) = 10-10.4 mol2 L-2 We cannot compare these values directly. First, the units are different. Second, the relative solubility depends on pH, pCO2, etc. So Fe(OH)2 is not necessarily more soluble than FeCO3.

    70. 70 Need to calculate which phase is least soluble at a given set of P, T, pH, CT, etc. conditions. To illustrate, lets assume [Alk] = 10-4 eq L-1 and pH = 6.8. Solubility equilibrium with FeCO3(s) yields FeCO3(s) ? Fe2+ + CO32- log cKs0 = -10.4 H+ + CO32- ? HCO3- -log cK2 = 10.1 FeCO3(s) + H+ ? Fe2+ + HCO3- log c*Ks0 = -0.3 log [Fe2+] = log c*Ks0 - pH - log [HCO3-] At this pH [Fe2+] ? [Fe(II)]T and [HCO3-] ? [Alk] log [Fe(II)]T = -0.3 - 6.8 + 4 = -3.10

    71. 71 Solubility equilibrium with Fe(OH)2(s): Fe(OH)2(s) ? Fe2+ + 2OH- log cKs0 = -14.5 2H+ + 2OH- ? 2H2O(l) -2log cKw = 27.8 Fe(OH)2(s) + 2H+ ? Fe2+ + 2H2O(l) log c*Ks0 = 13.3 log [Fe2+] = log c*Ks0 - 2pH log [Fe(II)]T = 13.3 - 2(6.8) = -0.30 Thus, even though FeCO3(s) has a greater solubility product than Fe(OH)2(s), under these specific conditions, FeCO3(s) is less soluble than Fe(OH)2(s), so FeCO3(s) is the most stable phase.

    72. 72 PREDOMINANCE DIAGRAMS A predominance diagram is any diagram with two compositional variables as axes and that shows the stable phases and solubilities in a mineral-solution system. pCO2-pH diagrams Plots with log pCO2 vs. pH as variables. An assumed ?Fe is employed. We will assume ?Fe = 10-4 mol L-1. 1) Fe(OH)2(s)/Fe2+ boundary Fe(OH)2(s) + 2H+ ? Fe2+ + 2H2O(l)

    73. 73 log *Ks0Fe(OH)2(s) -2pH + pFe = 0 12.85 - 2pH + 4 = 0 pH = 8.43 2) FeCO3(s)/Fe2+ boundary FeCO3(s) + 2H+ ? Fe2+ + CO2(g) + H2O(l) log pCO2 = log *Kps0FeCO3(s) - 2pH + pFe = 7.40 - 2pH + 4 = 11.90 - 2pH

    74. 74 3) FeCO3(s)/Fe(OH)2(s) boundary FeCO3(s) + H2O(l) ? Fe(OH)2(s) + CO2(g)

    75. 75

    76. 76 STABILITY OF CARBONATES IN THE SYSTEM Mg-CO2-H2O Brucite: Mg(OH)2(s) ?fGo = -200.0 kcal mol-1 Magnesite: MgCO3(s) ?fGo = -245.3 kcal mol-1 Nesquehonite: MgCO33H2O ?fGo = -411.7 kcal mol-1 Hydromagnesite: Mg4(CO3)3(OH)23H2O ?fGo = -1100.1 kcal mol-1 Mg2+ ?fGo = -109.0 kcal mol-1 CO32- ?fGo = -126.2 kcal mol-1 OH- ?fGo = -37.6 kcal mol-1 H2O(l) ?fGo = -56.69 kcal mol-1

    77. 77 Mg(OH)2(s) ? Mg2+ + 2OH- pKs0 = 11.6 MgCO3(s) ? Mg2+ + CO32- pKs0 = 7.5 MgCO33H2O ? Mg2+ + CO32- + 3H2O pKs0 = 4.7 Mg4(CO3)3(OH)23H2O ? 4Mg2+ + 3CO32- + 2OH- + 3H2O pKs0 = 29.5 We will now construct a pCO2-pH diagram to depict the phase relations in this system. First, note that the only difference between brucite and nesquehonite is the presence of water, so only one of these phases will appear on the phase diagram.

    78. 78 MgCO3(s) ? Mg2+ + CO32- log Ks0 = -7.5 Mg2+ + CO32- + 3H2O ? MgCO33H2O log Ks0 = 4.7 MgCO3(s) + 3H2O ? MgCO33H2O log K = -2.8 log K = -3log aH2O = -2.8 log aH2O = 0.933 or aH2O = 8.58 But as aH2O can never be greater than 1, nesquehonite can never be the stable phase, magnesite is always the stable carbonate. Magnesite/hydromagnesite boundary 4MgCO3(s) ? 4Mg2+ + 4CO32- log Ks0 = -30.0 4Mg2+ + 3CO32- + 2OH- + 3H2O ? Mg4(CO3)3(OH)23H2O log Ks0 = 29.5 4MgCO3(s) + 2OH- + 3H2O ? Mg4(CO3)3(OH)23H2O + CO32- log K = -0.5

    79. 79 4MgCO3(s) + 2OH- + 3H2O ? Mg4(CO3)3(OH)23H2O + CO32- log K = -0.5 2H2O ? 2OH- + 2H+ log K = -28.0 CO32- + H+ ? HCO3- log K = 10.3 HCO3- + H+ ? H2CO3* log K = 6.3 H2CO3* ? CO2 + H2O log K = 1.5 4MgCO3(s) + 4H2O ? Mg4(CO3)3(OH)23H2O + CO2 log K = -10.4 K = pCO2/(aH2O)4 log pCO2 = log K = -10.4 Thus, magnesite is stable relative to hydromagnesite down to very low pCO2.

    80. 80 Magnesite/brucite boundary MgCO3 ? Mg2+ + CO32- log K = -7.5 Mg2+ + 2OH- ? Mg(OH)2 log K = 11.6 2H2O ? 2OH- + 2H+ log K = -28.0 CO32- + H+ ? HCO3- log K = 10.3 HCO3- + H+ ? H2CO3* log K = 6.3 H2CO3* ? CO2 + H2O log K = 1.5 MgCO3 + H2O ? Mg(OH)2 + CO2 log K = -5.80 log pCO2 = log K = -5.80

    81. 81 Magnesite/Mg2+ boundary MgCO3 ? Mg2+ + CO32- log K = -7.5 CO32- + 2H+ ? CO2 + H2O log K = 18.1 MgCO3 + 2H+ ? Mg2+ + CO2 + H2O log K = 10.6 log K = 10.6 = log [Mg2+] + log pCO2 + 2pH we choose [Mg2+] = 10-3 mol L-1 10.6 = -3 + log pCO2 + 2pH log pCO2 = 13.6 - 2pH

    82. 82 Brucite/Mg2+ boundary Mg(OH)2(s) ? Mg2+ + 2OH- log K = -11.6 2OH- + 2H+ ? 2H2O log K = 28.0 Mg(OH)2(s) + 2H+ ? Mg2+ + 2H2O log K = 16.4 log K = 16.4 = log [Mg2+] + 2pH but [Mg2+] = 10-3 mol L-1 16.4 = -3 + 2pH pH = 9.70

    83. 83

    84. 84 SOLUBILITY OF DOLOMITE Conditions of dolomite precipitation are poorly understood. Precipitation of dolomite does not control compositions of natural waters. Dissolution of dolomite may exercise some control. Published values Ks0 range from 10-16.5 to 10-19.5.

    85. 85 CAN WE OBTAIN DOLOMITE Ks0 FROM FIELD DATA? CaMg(CO3)2 ? Ca2+ + Mg2+ + 2CO32- log Ks0,dolomite 2Ca2+ + 2CO32- ? 2CaCO3 -2log Ks0,calcite CaMg(CO3)2 + Ca2+ ? 2CaCO3 + Mg2+ log Kexchange

    86. 86

    87. 87 SOLUBILITY OF SULFIDES Estimate the solubility of ?-CdS at pH = 4.5 and pH2S = 1 atm (25C, I = 1 mol L-1). We have the following thermodynamic data: log c*Kpso(CdS) = -5.8 log cKH(H2S) = -1.05 log cK1(H2S) = -6.90 log cK2(HS-) = -14.0 For now we assume that no hydroxide, sulfide or carbonate complexes are formed.

    88. 88 The constant c*Kps0 refers to the reaction: ?-CdS(s) + 2H+ ? Cd2+ + H2S(g)

    89. 89 PREDOMINANCE DIAGRAM FOR SULFIDES Construct a predominance diagram in the system Cd2+-H2S-CO2-H2O at pCO2 = 10-3.5 atm. We need the following additional thermodynamic data: log c*Kps0(CdCO3) = 6.44 log cK1(H2CO3*) = -6.04 log cK2(HCO3-) = -9.57 log cKH(CO2) = -1.51 Our diagram will plot log (pCO2/pH2S) vs. pH.

    90. 90 CdCO3/CdS boundary CdCO3(s) + H2S(g) ? CdS(s) + H2O(l) + CO2(g) log K = log (c*Kps0(CdCO3)/c*Kps0(CdS)) = 6.44 - (-5.8) = 12.24 = log (pCO2/pH2S) CdS/Cd2+ boundary CdS(s) + 2H+ ? Cd2+ + H2S(g)

    91. 91 However, to plot this on the log (pCO2/pH2S) vs. pH diagram, we must subtract log pCO2 from both sides of the equation to get: -1.8 - log pCO2 = log pH2S - log pCO2 + 2pH but log pCO2 = -3.5 so 1.7 = log (pH2S/pCO2) + 2pH log (pCO2/pH2S) = 2pH - 1.7

    92. 92 CdCO3/Cd2+ boundary CdCO3(s) + 2H+ ? Cd2+ + CO2(g) + H2O(l)

    93. 93

    94. 94 PHOSPHATES - RELATIVE STABILITIES OF CALCITE AND APATITE To calculate a predominance diagram we must be concerned about the pH-dependent distribution of phosphate and carbonate. We write the reactions in terms of the predominant species at the pH of interest. Choose log PT (where PT is the total phosphate concentration) and pH as axes for our diagram. Assume CT = 10-4 mol L-1 and Ca is conserved in the solid phases.

    95. 95 pH < pK1(phosphate) < pK1(carbonate) 10CaCO3(s) + 6H3PO40 + 2H2O ? Ca10(PO4)6(OH)2(s) + 10H2CO30 log K = 39.5 = 10log [H2CO30] - 6log [H3PO40] 39.5 = 10(-4) - 6log PT log PT = -13.25 pK1(phosphate) < pH < pK1(carbonate) 10CaCO3(s) + 6H2PO4- + 6H+ + 2H2O ? Ca10(PO4)6(OH)2(s) + 10H2CO30 log K = 52.1 = 10log [H2CO30] - 6log [H2PO4-] + 6pH 52.1 = 10(-4) -6log PT + 6pH log PT = pH - 15.35 pK1(carbonate) < pH < pK2(phosphate) 10CaCO3(s) + 6H2PO4- + 2H2O ? Ca10(PO4)6(OH)2(s) + 10HCO3- + 4H+ -10.9 = 10log [HCO3-] - 6log [H2PO4-] - 4pH log PT = -4/6pH - 4.85

    96. 96 pK2(phosphate) < pH < pK2(carbonate) 10CaCO3(s) + 6HPO42- + 2H+ + 2H2O ? Ca10(PO4)6(OH)2(s) + 10HCO3- 32.3 = 10log [HCO3-] - 6log [HPO42-] + 2pH log PT = 1/3pH - 12.05 pK2(carbonate) < pH < pK3(phosphate) 10CaCO3(s) + 6HPO42- + 2H2O ? Ca10(PO4)6(OH)2(s) + 10CO32- + 8H+ -70.2 = 10log [CO32] - 6log [HPO42-] - 8pH log PT = -4/3pH + 5.03 pK3(phosphate) < pH 10CaCO3(s) + 6PO43- + 2H2O ? Ca10(PO4)6(OH)2(s) + 10CO32- + 2H+ 1.80 = 10log [CO32] - 6log [PO43-] - 2pH log PT = -1/3pH - 6.97

    97. 97

    98. 98 ACTIVITY OF SOLID PHASE OR WHAT TO DO WHEN SOLID NOT PURE Up to now we have assumed that asolids ? 1 in all our solubility calculations. This may not always be a valid assumption, especially where there is solid solution! Example: Two-component solid-solution between AgCl(s) and AgBr(s): AgCl(s) + Br- ? AgBr(s) + Cl-

    99. 99 Thus, the distribution coefficient is simply equal to the ratio of the solubility products of the two end member phases. aAgBr = XAgBr?AgBr and aAgCl = XAgCl?AgCl

    100. 100 The amount of dissolution of Br- into solid AgCl thus depends on: 1) the solubility product ratio 2) solution composition (aBr-/aCl-) 3) effects of non-ideal mixing (activity coefficients) To a first approximation, ?AgBr/ ?AgCl ? 1 and ?Br-/?Cl- ? 1 so

    101. 101 QUANTITATIVE IMPORTANCE OF SOLID SOLUTION Consider AgBr0.1Cl0.9(s) in equilibrium with solution. Solid: XAgBr = 0.1; XAgCl = 0.9 Solution: [Cl-] = 10-4.9 mol L-1; [Br-] = 10-8.4 mol L-1; [Ag+] = 10-4.9 mol L-1. D = 10-9.7/10-12.3 = 391 ? (10-4.9/10-8.4)(0.1/0.9) = 351 Br- is enriched in the solid phase relative to solution, i.e., Br- partitions preferentially into solid.

    102. 102 If we did not consider solid solution formation, we would erroneously conclude that the solution was undersaturated with respect to AgBr(s) because IAP = 10-4.910-8.4 = 10-13.3 < Ks0 = 10-12.3 However, the solutions is saturated with Br- because it is in equilibrium with a solid solution. IAP = 10-4.910-8.4/0.1 = 10-12.3 = Ks0 On the other hand, for AgCl(s) IAP = 10-4.9 10-4.9 = 10-9.8 ? Ks0 = 10-9.7 So the solubility of the minor component is greatly affected by solid solution, but the major component is less affected.

    103. 103 Example: Let us examine whether [Sr2+] in the ocean is controlled by solid solution in calcite and estimate the XSrCO3. The following is known: pcKs0(CaCO3) = 6.1; pcKs0(SrCO3) = 6.8 [CO32-] = 10-3.6 mol L-1; [Sr2+] ? 10-4 mol L-1

    104. 104 So XSrCO3 ?? 0.004 and XCaCO3 ? 0.996 Because D < 1, Sr is preferentially partitioned into the solution phase. A mole fraction of Sr of 0.004 is reasonable for marine calcite. If the solution were in equilibrium with pure strontianite (SrCO3), then [Sr2+] ? Ks0(SrCO3)/[CO32-] = 10-6.8/10-3.6 = 10-3.2 mol L-1 Which is much higher. Again, the solubility of the minor component is greatly reduced.

    105. 105 CALCIUM SELENATE CaSO42H2O ? Ca2+ + SO42- + 2H2O log Ks0 = -4.6 CaSeO42H2O ? Ca2+ + SeO42- + 2H2O log Ks0 = -3.09 Assume we have a water with [Ca2+] = 10-1 mol L-1 and [SeO42-] = 100 ppb = 1.27x10-6 mol L-1. If we assume aCaSeO4 = XCaSeO4 = 1, then IAP = 10-1(1.27x10-6) = 1.27x10-7 << 10-3.09 and solution is undersaturated in selenate. If XCaSeO4 = 10-5 (13 ppb), then IAP = 10-1(1.27x10-6)/10-5 = 1.27x10-2 > 10-3.09, and solution is saturated with selenate.

    106. 106 STOICHIOMETRIC SATURATION Stoichiometric saturation - refers to a metastable equilibrium between an aqueous phase and a multi-component solid, where, for kinetic reasons, the composition of the solid does not change. For example, magnesian calcite Ca(1-x)MgxCO3 ? (1-x)Ca2+ + xMg2+ + CO32- If this reaction was in complete equilibrium, there should be a differential partitioning of Ca and Mg between calcite and solution, e.g.,

    107. 107 A value of D = 0.02 indicates that Mg prefers the solution to the solid. However, for kinetic reasons when a mineral is dissolving, this equilibrium may not be attained fully. If the solid composition does not change throughout the dissolution process, then it can be treated thermodynamically as a one-component phase with an activity of 1.

    108. 108 NON-IDEALITY IN AQUEOUS SOLUTION Consider the dissolution of fluorite CaF2(s) ? Ca2+ + 2F-

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