THE GEOCHEMISTRY OF NATURAL WATERS

1. 1 THE GEOCHEMISTRY OF NATURAL WATERS MINERAL WEATHERING AND MINERAL SURFACE PROCESSES - II INCONGRUENT DISSOLUTION AND ACTIVITY DIAGRAMS CHAPTER 4 - Kehew (2001) Activity diagrams - 1

2. 2 LEARNING OBJECTIVES Learn about incongruent dissolution of silicates. Learn to calculate and use activity diagrams. Learn about the use of mass-balance calculations to infer weathering reactions. Apply the knowledge gained to rationalize compositions of natural waters in igneous and metamorphic rocks. Explore the implications of acid rain in igneous and metamorphic terrains. In Lectures 4 and 5, we dealt with congruent dissolution of oxides and carbonates, and incongruent dissolution of carbonates. In this lecture we concentrate on methods for dealing with incongruent dissolution of silicates. This will involve learning to construct and use a type of phase diagram called an activity diagram. We will also learn how to do mass-balance calculations to infer the weathering reaction that control the composition of waters during weathering of silicate minerals. In the process, we will apply what we learn to understanding how incongruent dissolution affects the compositions of natural waters in igneous and metamorphic rocks, and we will also briefly discuss the effect of acid rain on natural waters occurring in such rocks. In Lectures 4 and 5, we dealt with congruent dissolution of oxides and carbonates, and incongruent dissolution of carbonates. In this lecture we concentrate on methods for dealing with incongruent dissolution of silicates. This will involve learning to construct and use a type of phase diagram called an activity diagram. We will also learn how to do mass-balance calculations to infer the weathering reaction that control the composition of waters during weathering of silicate minerals. In the process, we will apply what we learn to understanding how incongruent dissolution affects the compositions of natural waters in igneous and metamorphic rocks, and we will also briefly discuss the effect of acid rain on natural waters occurring in such rocks.

3. 3 EVALUATION OF THE SATURATION STATE OF A SILICATE MINERAL To determine whether or not a water is saturated with an aluminosilicate such as K-feldspar, we could write a dissolution reaction such as: KAlSi3O8 + 4H+ + 4H2O ? K+ + Al3+ + 3H4SiO40 We could then determine the equilibrium constant: from Gibbs free energies of formation. The IAP could then be determined from a water analysis, and the saturation index calculated. In order to determine whether or not solubility of an aluminosilicate mineral is controlling the composition of a natural water (in other words, to determine if a natural water is saturated with an aluminosilicate mineral), we could proceed as in previous work by calculating an IAP and then a saturation index. We would write a dissolution reaction assuming that the mineral dissolves congruently (i.e., only dissolved reaction products) and proceed accordingly. This would require that we measure the activities of silicic acid, Al and any other ions involved in the dissolution reaction. In order to determine whether or not solubility of an aluminosilicate mineral is controlling the composition of a natural water (in other words, to determine if a natural water is saturated with an aluminosilicate mineral), we could proceed as in previous work by calculating an IAP and then a saturation index. We would write a dissolution reaction assuming that the mineral dissolves congruently (i.e., only dissolved reaction products) and proceed accordingly. This would require that we measure the activities of silicic acid, Al and any other ions involved in the dissolution reaction.

4. 4 WHAT’S WRONG WITH THIS APPROACH? Dissolved Al is not routinely determined in natural waters. When determined, Al is often below detection limit, or so low that accurate analysis is difficult. Much of the Al present is probably not actually dissolved, but in colloidal form. The dissolved Al needs to be converted to free Al3+ by calculation. However, there are a number of complications to this approach. First and foremost, Al is not routinely determined in the analysis of natural waters. This is because the concentrations of Al are typically very low owing to the low solubility of Al oxides, hydroxides and silicates (see Lecture 5). These low concentrations are often near or below the detection limits of many instruments used in the analysis of natural waters. Even when instruments are available with which Al can be determined, there are many problems in interpreting the results. Often, much of the Al in natural waters is not present in true solution, but in the form of very fine-grained suspended solids (colloidal material) that often cannot be efficiently separated from the solution. Because these colloids are actually suspended solids, their unintentional inclusion in the water analysis makes the dissolved Al concentration seem higher than it really is. Thus, extreme care is required to obtain accurate measurements of the actual concentration of truly dissolved Al. Finally, even if accurate determinations of the dissolved Al concentration can be obtained, we saw in Lecture 6 that most of the Al is probably not in the form of uncomplexed Al3+, the activity of which we need for calculation of a saturation index. Most of the Al3+ is probably present as hydroxide or other complexes. Thus, we would have to perform a calculation to correct for the amount of Al in complexes, so that we would be able to determine the free, uncomplexed Al3+ activity. However, there are a number of complications to this approach. First and foremost, Al is not routinely determined in the analysis of natural waters. This is because the concentrations of Al are typically very low owing to the low solubility of Al oxides, hydroxides and silicates (see Lecture 5). These low concentrations are often near or below the detection limits of many instruments used in the analysis of natural waters. Even when instruments are available with which Al can be determined, there are many problems in interpreting the results. Often, much of the Al in natural waters is not present in true solution, but in the form of very fine-grained suspended solids (colloidal material) that often cannot be efficiently separated from the solution. Because these colloids are actually suspended solids, their unintentional inclusion in the water analysis makes the dissolved Al concentration seem higher than it really is. Thus, extreme care is required to obtain accurate measurements of the actual concentration of truly dissolved Al. Finally, even if accurate determinations of the dissolved Al concentration can be obtained, we saw in Lecture 6 that most of the Al is probably not in the form of uncomplexed Al3+, the activity of which we need for calculation of a saturation index. Most of the Al3+ is probably present as hydroxide or other complexes. Thus, we would have to perform a calculation to correct for the amount of Al in complexes, so that we would be able to determine the free, uncomplexed Al3+ activity.

5. 5 INCONGRUENT DISSOLUTION Aluminosilicate minerals usually dissolve incongruently, e.g., 2KAlSi3O8 + 2H+ + 9H2O ? Al2Si2O5(OH)4 + 2K+ + 4H4SiO40 As a result of these factors, relations among solutions and aluminosilicate minerals are often depicted graphically on a type of mineral stability diagram called an activity diagram. In any event, writing a congruent dissolution reaction for aluminosilicates does not make a whole lot of sense, because at the pH values of most natural waters, these phases dissolve incongruently, producing clay minerals such as kaolinite, gibbsite or smectite. Because of all these problems, relations among solutions and aluminosilicate minerals are usually handled using a different approach. This involves the calculation of a type of phase diagram called an activity diagram. An activity diagram typically has activities of selected dissolved species as its coordinates, and it shows graphically the solution conditions under which selected phases are stable. In any event, writing a congruent dissolution reaction for aluminosilicates does not make a whole lot of sense, because at the pH values of most natural waters, these phases dissolve incongruently, producing clay minerals such as kaolinite, gibbsite or smectite. Because of all these problems, relations among solutions and aluminosilicate minerals are usually handled using a different approach. This involves the calculation of a type of phase diagram called an activity diagram. An activity diagram typically has activities of selected dissolved species as its coordinates, and it shows graphically the solution conditions under which selected phases are stable.

6. 6 ACTIVITY DIAGRAMS: THE K2O-Al2O3-SiO2-H2O SYSTEM We will now calculate an activity diagram for the following phases: gibbsite {Al(OH)3}, kaolinite {Al2Si2O5(OH)4}, pyrophyllite {Al2Si4O10(OH)2}, muscovite {KAl3Si3O10(OH)2}, and K-feldspar {KAlSi3O8}. The axes will be a K+/a H+ vs. a H4SiO40. The diagram is divided up into fields where only one of the above phases is stable, separated by straight line boundaries. We will illustrate the construction and interpretation of activity diagrams by plotting the fields of stability of the phases listed in this slide as a function of the coordinates aK+/aH+ vs. aH4SiO40, i.e., aK+/aH+ on the y-axis and aH4SiO40 on the x-axis. Activity diagrams consist of a series of straight lines which separate fields where only one of the aluminosilicate phases is stable. Along these straight lines, or phase boundaries, two of the aluminosilicate phases coexist in equilibrium. At this point it may be helpful to cheat a little and take a peak at the final diagram to see what the goal is. We will illustrate the construction and interpretation of activity diagrams by plotting the fields of stability of the phases listed in this slide as a function of the coordinates aK+/aH+ vs. aH4SiO40, i.e., aK+/aH+ on the y-axis and aH4SiO40 on the x-axis. Activity diagrams consist of a series of straight lines which separate fields where only one of the aluminosilicate phases is stable. Along these straight lines, or phase boundaries, two of the aluminosilicate phases coexist in equilibrium. At this point it may be helpful to cheat a little and take a peak at the final diagram to see what the goal is.

7. 7 This is what we will end up with after all the calculations and plotting are through. To calculate the positions of each of the boundaries shown above, we need to have thermodynamic data so we can calculate equilibrium constants for the reactions that occur at each of the boundaries. For example, along the boundary between the stability fields for the phases gibbsite and kaolinite, a reaction takes place involving these two phases. If we cross this boundary from the gibbsite field into the kaolinite field, the reaction is one in which gibbsite is converted to kaolinite. The thermodynamic data we require for this exercise are given in the following table. This is what we will end up with after all the calculations and plotting are through. To calculate the positions of each of the boundaries shown above, we need to have thermodynamic data so we can calculate equilibrium constants for the reactions that occur at each of the boundaries. For example, along the boundary between the stability fields for the phases gibbsite and kaolinite, a reaction takes place involving these two phases. If we cross this boundary from the gibbsite field into the kaolinite field, the reaction is one in which gibbsite is converted to kaolinite. The thermodynamic data we require for this exercise are given in the following table.

8. 8 THERMODYNAMIC DATA The thermodynamic data given here come from Drever, J.I. (1997) The Geochemistry of Natural Waters, 3rd. ed., Prentice-Hall (Table 10-2, p. 208). The thermodynamic data given here come from Drever, J.I. (1997) The Geochemistry of Natural Waters, 3rd. ed., Prentice-Hall (Table 10-2, p. 208).

9. 9 From looking at slide 7, you will realize that a number of different boundaries will need to be calculated to complete the entire diagram, and these boundaries will come together in a number of three-way intersections. At the start, if we have never seen the phase diagram for the system before, we do not know exactly how many boundaries we will have on our diagram, or how the phase boundaries will intersect one another. One way to approach the problem systematically and save yourself the trouble of calculating boundaries that end up not appearing on the diagram is to rank the phases in the order you expect them to appear on the diagram. For example, we note that gibbsite, kaolinite and pyrophyllite do not contain K. Therefore, we would expect boundaries between these phases to be independent of aK+/aH+ and plot as vertical lines. Moreover, these phases can be ranked in order of increasing silica content as gibbsite < kaolinite < pyrophyllite. As a result, we would expect gibbsite to plot at the lowest values of aH4SiO40, and pyrophyllite to plot at the highest values of aH4SiO40. Similarly, we can order the phases with respect to increasing K content per atom of Al according to: gibbsite = kaolinite = pyrophyllite < muscovite < K-feldspar. Thus, we would expect the stability field of K-feldspar to lie at the highest values of aK+/aH+ and the three K-free phases will lie at the lowest values. Finally, K-feldspar has more Si per Al atom than muscovite, so K-feldspar should be stable at higher values of aH4SiO40. We thus arrive at a rough mapping of the phases where we would expect them to occur on our final diagram. From looking at slide 7, you will realize that a number of different boundaries will need to be calculated to complete the entire diagram, and these boundaries will come together in a number of three-way intersections. At the start, if we have never seen the phase diagram for the system before, we do not know exactly how many boundaries we will have on our diagram, or how the phase boundaries will intersect one another. One way to approach the problem systematically and save yourself the trouble of calculating boundaries that end up not appearing on the diagram is to rank the phases in the order you expect them to appear on the diagram. For example, we note that gibbsite, kaolinite and pyrophyllite do not contain K. Therefore, we would expect boundaries between these phases to be independent of aK+/aH+ and plot as vertical lines. Moreover, these phases can be ranked in order of increasing silica content as gibbsite < kaolinite < pyrophyllite. As a result, we would expect gibbsite to plot at the lowest values of aH4SiO40, and pyrophyllite to plot at the highest values of aH4SiO40. Similarly, we can order the phases with respect to increasing K content per atom of Al according to: gibbsite = kaolinite = pyrophyllite < muscovite < K-feldspar. Thus, we would expect the stability field of K-feldspar to lie at the highest values of aK+/aH+ and the three K-free phases will lie at the lowest values. Finally, K-feldspar has more Si per Al atom than muscovite, so K-feldspar should be stable at higher values of aH4SiO40. We thus arrive at a rough mapping of the phases where we would expect them to occur on our final diagram.

10. 10 GIBBSITE/KAOLINITE BOUNDARY - I The reactions are always written to conserve Al in solid phases: 2Al(OH)3(s) + 2H4SiO40 ? Al2Si2O5(OH)4 + 5H2O ?rG° = (-3800) + 5(-237.13) - 2(-1151) - 2(-1316.6) = -50.45 kJ mol-1 The preliminary map we sketched in slide 9 suggests that the first boundary we might consider calculating is the vertical (independent of aK+/aH+) boundary between gibbsite and kaolinite. To calculate this phase boundary, we place gibbsite on one side of a chemical reaction and kaolinite on the other side. We write the reaction so as to conserve Al in the solid phases. In other words, the Al is balanced using the two solid phases only; no aqueous species are permitted. If we do not conserve Al, and include an aqueous Al species in addition to the two Al-bearing solids, the reaction can not be balanced in a single unique manner. Moreover, we would then have a third compositional variable to deal with. That would require a 3-D diagram, which is not so easy to construct or read. Once you have balanced the Al using only the two solid phases, you next balance Si using H4SiO40. We use the latter aqueous species because its activity is the coordinate on our x-axis. When the Si is balanced, it is likely that the H will not balance, so you are free to add H2O to balance H (never, ever use H2 or H+ here). If you balanced the Al, Si and H correctly, then the O will be balanced automatically. If the O does not balance, you made a mistake, and you need to try again. When your equation is balanced, you can then calculate the ?rG° value for the reaction, and then the log K value. The preliminary map we sketched in slide 9 suggests that the first boundary we might consider calculating is the vertical (independent of aK+/aH+) boundary between gibbsite and kaolinite. To calculate this phase boundary, we place gibbsite on one side of a chemical reaction and kaolinite on the other side. We write the reaction so as to conserve Al in the solid phases. In other words, the Al is balanced using the two solid phases only; no aqueous species are permitted. If we do not conserve Al, and include an aqueous Al species in addition to the two Al-bearing solids, the reaction can not be balanced in a single unique manner. Moreover, we would then have a third compositional variable to deal with. That would require a 3-D diagram, which is not so easy to construct or read. Once you have balanced the Al using only the two solid phases, you next balance Si using H4SiO40. We use the latter aqueous species because its activity is the coordinate on our x-axis. When the Si is balanced, it is likely that the H will not balance, so you are free to add H2O to balance H (never, ever use H2 or H+ here). If you balanced the Al, Si and H correctly, then the O will be balanced automatically. If the O does not balance, you made a mistake, and you need to try again. When your equation is balanced, you can then calculate the ?rG° value for the reaction, and then the log K value.

11. 11 GIBBSITE/KAOLINITE BOUNDARY - II But the equilibrium constant is written: So this plots as a vertical boundary, independent of aK+/aH+. The next step is to write the mass-action expression for the reaction. In this case, the only term in the mass-action expression that does not have an activity of unity is the activity of H4SiO40. Recall that pure solids and H2O have activities of unity. We take the logarithm of both sides of the equation and simplify so that we have an expression whereby the activity of H4SiO40 is a constant. We then plot a vertical line on our diagram that intersects the x-axis at -4.42. The next step is to write the mass-action expression for the reaction. In this case, the only term in the mass-action expression that does not have an activity of unity is the activity of H4SiO40. Recall that pure solids and H2O have activities of unity. We take the logarithm of both sides of the equation and simplify so that we have an expression whereby the activity of H4SiO40 is a constant. We then plot a vertical line on our diagram that intersects the x-axis at -4.42.

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13. 13 KAOLINITE/PYROPHYLLITE BOUNDARY - I Once again, Al is conserved: Al2Si2O5(OH)4 + 2H4SiO40 ? Al2Si4O10(OH)2 + 5H2O ?rG° = (-5275) + 5(-237.13) - (-3800) - 2(-1316.6) = -27.45 kJ mol-1 We will calculate the other vertical boundary, that between kaolinite and pyrophyllite, next. As before, we place one of the Al-bearing phases on either side of the diagram, conserving Al. We then balance the rest of the components using H4SiO40 and H2O, and calculate the log K as shown. We will calculate the other vertical boundary, that between kaolinite and pyrophyllite, next. As before, we place one of the Al-bearing phases on either side of the diagram, conserving Al. We then balance the rest of the components using H4SiO40 and H2O, and calculate the log K as shown.

14. 14 KAOLINITE/PYROPHYLLITE BOUNDARY - II But the equilibrium constant is written: So this also plots as a vertical boundary, independent of aK+/aH+. As before, we write the mass-action expression for the reaction taking place at the boundary and we ultimately obtain the equation of a vertical line intersecting the x-axis at -2.4. As before, we write the mass-action expression for the reaction taking place at the boundary and we ultimately obtain the equation of a vertical line intersecting the x-axis at -2.4.

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16. 16 GIBBSITE/MUSCOVITE BOUNDARY - I 3Al(OH)3(s) + 3H4SiO40 + K+ ? KAl3Si3O10(OH)2 + H+ + 9H2O ?rG° = (-5606) + (0) + 9(-237.13) - 3(-1151) - 3(-1316.6) - (-283.27) = -54.10 kJ mol-1 Referring back to our preliminary mapping in slide 9, the next logical choice of boundary to calculate is the gibbsite/muscovite boundary. This boundary should close off the stability field of gibbsite. We put gibbsite on one side of a reaction and muscovite on the other side. Once again we conserve Al in the solid phases. Then, we must balance the K in muscovite by putting a K+ ion on the left-hand side of the reaction. We use K+ because the activity of this ion is one of the components of the coordinates of our y-axis (aK+/aH+). Now introduction of K+ on the left-hand side causes a charge-imbalance that is rectified by placing an appropriate number (in this case 1) of H+ ions on the right-hand side. Thus, H+, which we can not use to balance H atoms, is reserved for balancing charge. The requirement of charge-balance is the reason for plotting a ratio of activities (i.e., aK+/aH+) instead of simply aK+ on the y-axis. Next, we proceed as before, balancing Si with H4SiO40 and H with H2O, and as before the O should automatically balance. Referring back to our preliminary mapping in slide 9, the next logical choice of boundary to calculate is the gibbsite/muscovite boundary. This boundary should close off the stability field of gibbsite. We put gibbsite on one side of a reaction and muscovite on the other side. Once again we conserve Al in the solid phases. Then, we must balance the K in muscovite by putting a K+ ion on the left-hand side of the reaction. We use K+ because the activity of this ion is one of the components of the coordinates of our y-axis (aK+/aH+). Now introduction of K+ on the left-hand side causes a charge-imbalance that is rectified by placing an appropriate number (in this case 1) of H+ ions on the right-hand side. Thus, H+, which we can not use to balance H atoms, is reserved for balancing charge. The requirement of charge-balance is the reason for plotting a ratio of activities (i.e., aK+/aH+) instead of simply aK+ on the y-axis. Next, we proceed as before, balancing Si with H4SiO40 and H with H2O, and as before the O should automatically balance.

17. 17 GIBBSITE/MUSCOVITE BOUNDARY - II But the equilibrium constant is written: So this plots as a straight line with a slope of -3. Now our mass-action expression contains terms other than the activity of silicic acid. We take the logarithm of both sides of this reaction and manipulate it until we have an equation of the form y = mx + b, i.e., an equation of a straight line, where y = aK+/aH+ and x = aH4SiO40. In this case our line has a slope of -3 and a y-intercept of -9.48. Now our mass-action expression contains terms other than the activity of silicic acid. We take the logarithm of both sides of this reaction and manipulate it until we have an equation of the form y = mx + b, i.e., an equation of a straight line, where y = aK+/aH+ and x = aH4SiO40. In this case our line has a slope of -3 and a y-intercept of -9.48.

18. 18 This is the first instance in which we encounter two boundaries that intersect one another. When two phase boundaries intersect, they divide one another into four line segments. For each boundary, usually only one of the segments is stable. Here we have chosen the two solid line segments, and will eventually erase the two dashed line segments (these are metastable). Why do we make this choice instead of some other combination of line segments? Recall that the goal is to enclose each solid phase in a well-defined stability field. We already established in our preliminary mapping that the stability field of gibbsite will plot in the lower left-hand side of the diagram. We introduce here a rule, called the Morey-Schreinemaker rule, which states that two lines bounding a stability field cannot intersect at an angle greater than 180°. These considerations dictate that the solid line segments, and not the dashed line segments, are the stable portions of the two intersecting phase boundaries. The next slide shows in detail why the choice we have made is the only logical one. This is the first instance in which we encounter two boundaries that intersect one another. When two phase boundaries intersect, they divide one another into four line segments. For each boundary, usually only one of the segments is stable. Here we have chosen the two solid line segments, and will eventually erase the two dashed line segments (these are metastable). Why do we make this choice instead of some other combination of line segments? Recall that the goal is to enclose each solid phase in a well-defined stability field. We already established in our preliminary mapping that the stability field of gibbsite will plot in the lower left-hand side of the diagram. We introduce here a rule, called the Morey-Schreinemaker rule, which states that two lines bounding a stability field cannot intersect at an angle greater than 180°. These considerations dictate that the solid line segments, and not the dashed line segments, are the stable portions of the two intersecting phase boundaries. The next slide shows in detail why the choice we have made is the only logical one.

19. 19 Note that each of the line segments have been labeled with A, B, C or D in the above diagram. Let us suppose that we choose segments A and B to be the stable segments of each of the boundaries. If start at the beginning of the arrow and move counter clockwise, we see that, muscovite reacts to gibbsite when we cross segment B. Gibbsite is stable between B and A, and then gibbsite reacts to kaolinite as we cross A. During this process, we run into no logical inconsistencies, nor do we violate the Morey-Schreinemaker rule, because the line segments limiting the gibbsite stability field intersect at an angle less than 180°. Now let us suppose that we had chosen segments C and D to be the stable segments (and we erased segments A and B). In this case we see that the gibbsite stability field would be bounded by segments intersecting at an angle >180°, in violation of the Morey-Schreinemaker rule; can’t do it! Let’s try segments A and D. Once again, we start at the beginning of the arrow and move counterclockwise. When we encounter segment A, gibbsite will react to kaolinite. Gibbsite is no longer stable. Kaolinite is the stable phase and gibbsite is gone. However, if we continue moving around the intersection in a counterclockwise manner, we find gibbsite wanting to react to form muscovite as we cross segment D, but gibbsite can’t react because it disappeared when we crossed line segment A. Hence, with the choice of segments A and D, we run into an impossible, logically inconsistent situation. We get a similar logical inconsistency if we choose segments B and C (the inconsistency is best revealed by moving along the arrow in a clockwise direction). Thus, only if we choose segments A and B do we find ourselves with a logically consistent situation that does not break any rules. Moreover, it is apparent that the muscovite/kaolinite boundary should plot between line segments C and D, so that the muscovite and kaolinite stability fields will be bounded by line segments intersecting at less than 180°. Note that each of the line segments have been labeled with A, B, C or D in the above diagram. Let us suppose that we choose segments A and B to be the stable segments of each of the boundaries. If start at the beginning of the arrow and move counter clockwise, we see that, muscovite reacts to gibbsite when we cross segment B. Gibbsite is stable between B and A, and then gibbsite reacts to kaolinite as we cross A. During this process, we run into no logical inconsistencies, nor do we violate the Morey-Schreinemaker rule, because the line segments limiting the gibbsite stability field intersect at an angle less than 180°. Now let us suppose that we had chosen segments C and D to be the stable segments (and we erased segments A and B). In this case we see that the gibbsite stability field would be bounded by segments intersecting at an angle >180°, in violation of the Morey-Schreinemaker rule; can’t do it! Let’s try segments A and D. Once again, we start at the beginning of the arrow and move counterclockwise. When we encounter segment A, gibbsite will react to kaolinite. Gibbsite is no longer stable. Kaolinite is the stable phase and gibbsite is gone. However, if we continue moving around the intersection in a counterclockwise manner, we find gibbsite wanting to react to form muscovite as we cross segment D, but gibbsite can’t react because it disappeared when we crossed line segment A. Hence, with the choice of segments A and D, we run into an impossible, logically inconsistent situation. We get a similar logical inconsistency if we choose segments B and C (the inconsistency is best revealed by moving along the arrow in a clockwise direction). Thus, only if we choose segments A and B do we find ourselves with a logically consistent situation that does not break any rules. Moreover, it is apparent that the muscovite/kaolinite boundary should plot between line segments C and D, so that the muscovite and kaolinite stability fields will be bounded by line segments intersecting at less than 180°.

20. 20 KAOLINITE/MUSCOVITE BOUNDARY - I 3Al2Si2O5(OH)4 + 2K+ ? 2KAl3Si3O10(OH)2 + 2H+ + 3H2O ?rG° = 2(-5606) + 2(0) + 3(-237.13) - 3(-3800) - 2(-283.27) = 43.15 kJ mol-1 Next, we calculate the kaolinite/muscovite boundary according to the rules we have already established: 1) put kaolinite and muscovite on opposite sides of the reaction; 2) balance Al in the solids only; 3) balance K using the K+ ion; 4) balance the charge using H+; 4) balance Si using H4SiO40 (but in this case Si is already balanced after step 2); 5) balance H with H2O; and 6) check that the O balances, as it should automatically. Next calculate the Gibbs free energy of reaction and the equilibrium constant. Next, we calculate the kaolinite/muscovite boundary according to the rules we have already established: 1) put kaolinite and muscovite on opposite sides of the reaction; 2) balance Al in the solids only; 3) balance K using the K+ ion; 4) balance the charge using H+; 4) balance Si using H4SiO40 (but in this case Si is already balanced after step 2); 5) balance H with H2O; and 6) check that the O balances, as it should automatically. Next calculate the Gibbs free energy of reaction and the equilibrium constant.

21. 21 KAOLINITE/MUSCOVITE BOUNDARY - II This boundary plots as a horizontal line, independent of silicic acid activity. Again, we manipulate the mass-action expression into the y = mx + b form of a straight line. However, in this case, H4SiO40 does not appear on either side of the reaction, so this boundary plots as a horizontal line which intersects the y-axis at aK+/aH+ = 3.78. The boundary is independent of the activity of silicic acid.Again, we manipulate the mass-action expression into the y = mx + b form of a straight line. However, in this case, H4SiO40 does not appear on either side of the reaction, so this boundary plots as a horizontal line which intersects the y-axis at aK+/aH+ = 3.78. The boundary is independent of the activity of silicic acid.

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23. 23 K-FELDSPAR/MUSCOVITE BOUNDARY - I 3KAlSi3O8 + 2H+ + 12H2O ? KAl3Si3O10(OH)2 + 2K+ + 6H4SiO40 ?rG° = (-5606) + 2(-283.27) + 6(-1316.6) - 3(-3767) - 2(0) - 12(-237.13) = 74.42 kJ mol-1 To answer the question posed in the previous slide, we next calculate the K-feldspar/muscovite boundary. By now, the steps in the calculation are familiar so we won’t repeat what’s been said before. To answer the question posed in the previous slide, we next calculate the K-feldspar/muscovite boundary. By now, the steps in the calculation are familiar so we won’t repeat what’s been said before.

24. 24 K-FELDSPAR/MUSCOVITE BOUNDARY - II So this plots as a straight line with a slope of -3. The K-feldspar/muscovite boundary line has a slope of -3 and an intercept of -6.52. This boundary is parallel to the gibbsite/muscovite boundary which also was shown to have a slope of -3. The K-feldspar/muscovite boundary line has a slope of -3 and an intercept of -6.52. This boundary is parallel to the gibbsite/muscovite boundary which also was shown to have a slope of -3.

25. 25 We draw the stable part of the K-feldspar/muscovite boundary above its intersection with the muscovite/kaolinite boundary because this results in a muscovite stability field enclosed by boundaries making less than 180° angles with one another. If we chose the portion of the K-feldspar/muscovite boundary below the muscovite/kaolinite boundary, we would end up with a logical inconsistency as discussed in slide 19. We draw the stable part of the K-feldspar/muscovite boundary above its intersection with the muscovite/kaolinite boundary because this results in a muscovite stability field enclosed by boundaries making less than 180° angles with one another. If we chose the portion of the K-feldspar/muscovite boundary below the muscovite/kaolinite boundary, we would end up with a logical inconsistency as discussed in slide 19.

26. 26 K-FELDSPAR/KAOLINITE BOUNDARY - I 2KAlSi3O8 + 2H+ + 9H2O ? Al2Si2O5(OH)4 + 2K+ + 4H4SiO40 ?rG° = (-3800) + 2(-283.27) + 4(-1316.6) - 2(-3767) - 2(0) - 9(-237.13) = 35.23 kJ mol-1 Looking at slide 25, we see that the next logical boundary to calculate is the K-feldspar/kaolinite boundary. Looking at slide 25, we see that the next logical boundary to calculate is the K-feldspar/kaolinite boundary.

27. 27 K-FELDSPAR/KAOLINITE BOUNDARY - II So this plots as a straight line with a slope of -2. This boundary has a slope of -2 and a y-intercept of -3.09. This boundary has a slope of -2 and a y-intercept of -3.09.

28. 28 The kaolinite/K-feldspar boundary splits the >180° sector between the muscovite/kaolinite and muscovite/K-feldspar boundaries into two sectors <180°. The kaolinite/K-feldspar boundary extends downward until it intersects the kaolinite/pyrophyllite boundary. Now the kaolinite stability field is enclosed by boundaries intersecting at <180°. The last boundary we need to draw is the K-feldspar/pyrophyllite boundary. The kaolinite/K-feldspar boundary splits the >180° sector between the muscovite/kaolinite and muscovite/K-feldspar boundaries into two sectors <180°. The kaolinite/K-feldspar boundary extends downward until it intersects the kaolinite/pyrophyllite boundary. Now the kaolinite stability field is enclosed by boundaries intersecting at <180°. The last boundary we need to draw is the K-feldspar/pyrophyllite boundary.

29. 29 K-FELDSPAR/PYROPHYLLITE BOUNDARY - I 2KAlSi3O8 + 2H+ + 4H2O ? Al2Si4O10(OH)2 + 2K+ + 2H4SiO40 ?rG° = (-5275) + 2(-283.27) + 2(-1316.6) - 2(-3767) - 2(0) - 4(-237.13) = 7.78 kJ mol-1

30. 30 K-FELDSPAR/PYROPHYLLITE BOUNDARY - II So this plots as a straight line with a slope of -1.

31. 31 This is our activity diagram plotted to correct scale. The dashed lines represent the solubilities of quartz and amorphous silica; we will see how to plot these in the next slide. Note that no phase boundaries intersect at angles greater than 180°. If we now have analyses of a natural water, we can convert the measured concentrations to activities and plot them on the diagram to predict which aluminosilicate phase we would expect to be in equilibrium with the water. Note that, because the diagram does not involve the activity of Al, we cannot be certain, if a water plots in the K-feldspar stability field, for example, that the water is actually saturated with respect to K-feldspar. To know this for sure, we would have to accurately determine the aluminum activity and calculate the ion activity product, and then the saturation index. However, because the solubility of Al is low, many natural waters probably are saturated with some aluminosilicate, and the activity diagram can tell us which aluminosilicate is the stable one. Another point to make is that, we can only make conclusions about minerals we have chosen to plot on the diagram. If a water plots in the muscovite field, we can say that muscovite is the most stable of the phases we chose to include on our diagram. However, there may be a more stable phase that we did not include on our diagram. Of course, we cannot plot a phase unless we have thermodynamic data for it. This is our activity diagram plotted to correct scale. The dashed lines represent the solubilities of quartz and amorphous silica; we will see how to plot these in the next slide. Note that no phase boundaries intersect at angles greater than 180°. If we now have analyses of a natural water, we can convert the measured concentrations to activities and plot them on the diagram to predict which aluminosilicate phase we would expect to be in equilibrium with the water. Note that, because the diagram does not involve the activity of Al, we cannot be certain, if a water plots in the K-feldspar stability field, for example, that the water is actually saturated with respect to K-feldspar. To know this for sure, we would have to accurately determine the aluminum activity and calculate the ion activity product, and then the saturation index. However, because the solubility of Al is low, many natural waters probably are saturated with some aluminosilicate, and the activity diagram can tell us which aluminosilicate is the stable one. Another point to make is that, we can only make conclusions about minerals we have chosen to plot on the diagram. If a water plots in the muscovite field, we can say that muscovite is the most stable of the phases we chose to include on our diagram. However, there may be a more stable phase that we did not include on our diagram. Of course, we cannot plot a phase unless we have thermodynamic data for it.

32. 32 SILICA SATURATION Lines representing conditions of silica saturation can also be added to the diagram. For quartz, we start with the mass-action expression for: SiO2(quartz) + 2H2O(l) ? H4SiO40 or Thus, the quartz saturation line is a vertical line. Similarly for amorphous silica we have: The lines representing saturation with respect to quartz and amorphous silica are almost trivial to plot. Clearly, the two reactions: SiO2(quartz) + 2H2O ? H4SiO40 SiO2(amorphous) + 2H2O ? H4SiO40 involve the activity of silicic acid only, and not the K+/H+ activity ratio, so they plot as vertical lines. We noted in Lecture 5 that supersaturation with respect to amorphous silica is almost never observed in natural waters, so we would not expect to find water compositions that plot to the right of the dashed line label “Amorphous silica” in slide 31. This means that we should not expect to find pyrophyllite as a stable phase in equilibrium with water at 25ºC because the entire pyrophyllite field lies at silicic acid activities that exceed saturation with amorphous silica. Another point to make about the activity diagram in the previous slide is that there are boundaries that we did not calculate. For example, a reaction can be written between gibbsite and pyrophyllite and a boundary between these two fields can be plotted. However, the boundary plots in the middle of the kaolinite field. Thus, kaolinite is more stable that gibbsite + pyrophyllite. We say that the gibbsite/pyrophyllite boundary is metastable. Similarly, the K-feldspar/gibbsite and muscovite/pyrophyllite boundaries are also metastable and are not plotted. The systematic approach we applied to the problem allowed us to avoid the effort of calculating metastable boundaries that will ultimately not appear on the diagram. The lines representing saturation with respect to quartz and amorphous silica are almost trivial to plot. Clearly, the two reactions: SiO2(quartz) + 2H2O ? H4SiO40 SiO2(amorphous) + 2H2O ? H4SiO40 involve the activity of silicic acid only, and not the K+/H+ activity ratio, so they plot as vertical lines. We noted in Lecture 5 that supersaturation with respect to amorphous silica is almost never observed in natural waters, so we would not expect to find water compositions that plot to the right of the dashed line label “Amorphous silica” in slide 31. This means that we should not expect to find pyrophyllite as a stable phase in equilibrium with water at 25ºC because the entire pyrophyllite field lies at silicic acid activities that exceed saturation with amorphous silica. Another point to make about the activity diagram in the previous slide is that there are boundaries that we did not calculate. For example, a reaction can be written between gibbsite and pyrophyllite and a boundary between these two fields can be plotted. However, the boundary plots in the middle of the kaolinite field. Thus, kaolinite is more stable that gibbsite + pyrophyllite. We say that the gibbsite/pyrophyllite boundary is metastable. Similarly, the K-feldspar/gibbsite and muscovite/pyrophyllite boundaries are also metastable and are not plotted. The systematic approach we applied to the problem allowed us to avoid the effort of calculating metastable boundaries that will ultimately not appear on the diagram.