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**THE GEOCHEMISTRY OF NATURAL WATERS**

THE CARBONATE SYSTEM CHAPTER 3 - Kehew (2001) Alkalinity

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**LEARNING OBJECTIVES Understand sources of CO2 in natural waters.**

Define and understand alkalinity. Learn to calculate the solubility of carbonate minerals such as calcite. Understand the common-ion effect. Become familiar with the concept of incongruent dissolution. Apply these concepts to some case studies. In this lecture we will pick up where we left off in Lecture 3, and continue with Chapter 3 in Kehew (2001). First we will consider sources of CO2, other than atmospheric, in natural waters. We will then define the concept of alkalinity and attempt to understand its relevance. The bulk of this lecture is devoted to understanding solubility equilibria of carbonate minerals, such as calcite. This includes being able to calculate the solubility of carbonates under specified conditions of pCO2 and pH, considering the common-ion effect and its relevance to natural waters, and familiarizing ourselves with the concept of incongruent dissolution. With these tools under your belt, you are then directed to the latter half of Chapter 3 in Kehew (2001), where he applies these concepts in a discussion of some interesting case studies. Because Kehew (2001) does a very good job describing these case studies, and we have limited time together, I will not be reviewing the case studies in class. However, I direct you to the Lecture 4 page on the course web site for some study questions to help guide your reading in the text.

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BRIEF REVIEW We saw in Lecture 3 that pH, pCO2 and bicarbonate ion concentrations are all interrelated. Rearrangement of the equations we have worked with previously yields: Thus, if we measure pH and bicarbonate ion concentration, we can calculate pCO2. As we saw in Lecture 3, pH, pCO2, and the concentration of HCO3- are all interrelated through the Henry’s Law constant and the first dissociation constant. Rearrangement of an expression from Lecture 3 (slide 35) yields the one shown in this slide. From this relationship, we see that if we measure pH and bicarbonate ion concentration, we can calculate the pCO2 with which the system is apparently in equilibrium. Note that we can also work this expression in other ways, i.e., if we know any two of the variables, pH, pCO2, and the concentration of HCO3-, we can calculate the third.

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**SOURCES OF CO2 IN NATURAL WATERS**

When the equation in the previous slide is applied to natural waters, particularly ground waters and soil solutions, pCO2 values greater than atmospheric are commonly obtained. Why? A system closed to atmospheric CO2 is implied. Respiration by plant roots and microbes consumes organic matter and produces CO2: CH2O + O2 CO2 + H2O Amount of CO2 production depends on temperature, soil moisture content, and the amount of organic matter. When we apply the equation in slide 3 to chemical analyses of ground waters and soil solutions, we often calculate pCO2 > atm, or in other words, values greater than we would expect for equilibrium with the Earth’s atmosphere. Why are these pCO2 values so different? Clearly, these systems must be closed to atmospheric CO2; CO2 is not free to escape or be buffered by the atmosphere, so something else is controlling its partial pressure. The process that produces additional CO2 is the decomposition of organic matter, primary through the respiration of plant roots and microbes. A schematic reaction for the process is given in this slide, using formaldehyde (CH2O) as a proxy for the organic matter. Commonly, simple organic molecules such as formaldehyde are employed to illustrate the general trends expected in reactions involving organic matter. The actual organic matter present in a soil or aquifer will, of course, be more complex and probably have a slightly different overall chemical composition. However, this does not change the fact that oxidation of such organic matter leads to the production of CO2, and if the system is not open to the atmosphere, the partial pressure of CO2 can build up to values much higher than atmospheric. Many factors control the amount of CO2 produced, including temperature, the humidity of the soil, and the composition and amount of organic matter. These are all reflections of the particular climate in the area of interest.

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ALKALINITY - I In aqueous solutions, positive and negative charges must balance. In a pure CO2-H2O system, the charge-balance condition is: This equation shows that, as H2CO3* dissociates to form HCO3-, the concentration of H+ also increases to maintain charge balance. Often, because CO32- and OH- are negligible, the charge-balance expression can be approximated as: One of the most fundamental laws of nature is that aqueous solutions must be electrically neutral. This means that, the total sum of all the positive charges must equal the total sum of all negative charges. We make use of this fact to judge the validity of chemical analyses of natural waters using the charge-balance error (Lecture 1). In a pure CO2-H2O system, the only possible species besides H2O are H+, OH-, H2CO3*, HCO3- and CO32-, and so the charge-balance constraint is written as the first equation in this slide. Note that, each mole of CO32- provides two equivalents of negative charge in solution, so the concentration of carbonate must be multiplied by 2. This charge-balance constraint tells us that, as we pump more CO2 into a system, and the concentration of H2CO3* increases, and H2CO3* dissociates to form HCO3- and CO32-, an equivalent amount of H+ must also be formed to balance the charge. This is why increased partial pressure of CO2 leads to decreased pH (increased H+ activity) in solution. In a pure CO2-H2O solution, the pH will always be less than 7, and therefore bicarbonate will be much more concentrated than carbonate, and H+ will be present in much higher concentrations than OH-. This leads to the simplified charge-balance at the bottom of the slide.

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ALKALINITY - II Reactions with minerals can affect this relationship. For example, dissolution of calcite would result in: If this solution is removed from contact with calcite, and strong acid is added, the concentration of H+ and all the carbonate species would change, but the concentration of Ca2+ would not change. Thus, Ca2+ is a conservative ion, and HCO3-, CO32-, H+ and OH- are non-conservative. Grouping these ions accordingly we get: If there are additional components in the system, resulting, e.g., from the dissolution of calcite, then any other charged components must be added to the charge-balance expression. In this case, the concentration of Ca2+ is added (note the multiplier of 2) to the charge-balance constraint. Let’s imagine that we have allowed a CO2-H2O solution to equilibrate with CaCO3(s), and that we then remove the solution from the solid. If we then add strong acid, e.g., HCl, to the system, the concentration of H+ would change, and because they are a function of pH, the concentrations of OH- and all the carbonate species would change. However, because the solution is no longer in contact with the calcite, the Ca2+ concentration cannot change. We call ions, the concentrations of which do not change under such circumstances, conservative. Ions, the concentrations of which do change, are called non-conservative. The last line of this slide shows the charge-balance constraint rewritten such that all the conservative ions are on the left, and all the non-conservative ions are on the right. Note that, if the left-hand side of the equation is conservative, then the sum of all the non-conservative species on the right-hand side of the equation must also be conservative. Thus, although none of the individual species on the right is conservative, their mathematical combination as shown is conservative. Therefore, the quantity on the right-hand side of the equation is a special combination, and we call this quantity the total alkalinity.

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**ALKALINITY - III The quantity: is called the total alkalinity.**

Another definition of total alkalinity: the equivalent sum of bases titratable with a strong acid. Total alkalinity is the neutralizing capacity of a solution; the greater the total alkalinity, the more acid the solution can neutralize. For a general natural water, the charge-balance can be written: In a pure CO2-H2O solution, there is no Ca2+, and so from the last expression in slide 7, we see that the total alkalinity must be zero for such a solution. However, if some additional base, such as CaCO3 and/or NaOH is added to the system, then the total alkalinity will not be equal to zero, because the conservative ions Ca2+ and/or Na+ would appear on the left-hand side of the equation. Another way of thinking about alkalinity is that it is the equivalent sum of all bases in solution stronger than H2CO3*, that can be titrated in the laboratory using a strong acid. Thus, total alkalinity represents the ability of a natural water to neutralize acid. It is thus an important characteristic to consider when dealing with the possible effects of acidification of natural waters. For example, acid-mine drainage, acid-rain or an acidic industrial effluent added to a river, lake or an aquifer would tend to decrease the pH, which can lead to undesirable environmental consequences. Natural waters with low alkalinity are unable to resist a decrease in pH when strong acid is added, whereas waters with high alkalinity will be less affected by addition of acid. It is important not to confuse alkalinity and pH. Alkalinity is the total equivalent concentration of bases in a water, whereas pH is a measure of the H+ concentration only. It is possible for a slightly acidic water (e.g., pH = 6.0) to have a non-zero value of total alkalinity. All that is required is that the water contain some bicarbonate. Finally, in general, all charged species must show up in the charge-balance expression, so in a complex natural water, the left-hand side of the equation may contain a large number of conservative cations and anions. A common symbol for total alkalinity is [alk].

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ALKALINITY - IV Because all the terms on the left-hand side of the previous charge-balance expression are conservative, then the alkalinity must also be conservative. The only way to change alkalinity is to either add strong acid or base, or for solids to dissolve or precipitate. This is why it is important to measure alkalinity in the field before precipitation can occur. Alkalinity is measured by titration with strong acid. A known volume of sample is titrated (usually with H2SO4) until an endpoint. Thus, for alkalinity of a water to change, we must add strong acid or base, or have solids dissolve or precipitate. Addition of CO2, in the absence of any of these other events, has no effect whatsoever on alkalinity. However, changes in CO2 partial pressure, combined with dissolution or precipitation of a solid phase, can affect alkalinity. Because precipitation/dissolution reactions can change alkalinity, it is important to measure this quantity immediately in the field. This is because changes in temperature or partial pressure of CO2 once the sample is removed from its environment may induce the precipitation (or less commonly dissolution) of solid phases which will cause the alkalinity value to change. Waiting to measure alkalinity in the laboratory is one of the most common reasons for unacceptable charge-balance errors in chemical analyses of natural waters. Alkalinity is determined by titration with a strong acid to an endpoint (often signaled by a change in color of an indicator). Knowing the volume of the water sample (V), the volume of acid solution required to reach the endpoint (V0) and the normality of the acid (Nacid), one can calculate the alkalinity in eq L-1 according to: V·[alk] = V0·Nacid

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Titration curve for a 510-3 m Na2CO3 solution, together with a Bjerrum plot for the same solution. A is the beginning of the titration, B is the carbonate endpoint, C is the region of strong carbonate buffering, and D is the bicarbonate endpoint. This diagram shows a titration curve (mL acid added vs. pH) for the titration of a 510-3 molal Na2CO3 solution. The titration curve for a natural water containing carbonate as the only significant base would look similar. Superimposed on the titration curve is a Bjerrum diagram for the carbonate system with CT = 510-3 molal. The titration curve starts at point A, where almost all the carbonate is in the form of CO32-. The flat portions of the titration curve, where pH changes rapidly for small changes in added acid are the endpoints. Point B is the endpoint where essentially all CO32- present has been titrated to HCO3-, and point D is the endpoint where essentially all HCO3- has been titrated to H2CO3*. In between the two endpoints is the carbonate buffering region (point C). Note that, in this region, large additions of acid are required to change the pH; in other words, the system strongly resists pH change (the system is well buffered). In contrast, at the endpoints B and D, the system is very poorly buffered. The alkalinity is defined as the total number of equivalents of acid per liter required to bring the system from point A, to the bicarbonate endpoint (D). There is also a quantity called the p-alkalinity, which is the total number of equivalents of acid per liter required to bring the system from point A, to the carbonate endpoint (B). Note that, not all systems will have a pH such that the titration begins at point A. In fact, most natural systems will have a pH substantially lower than that at point A. Many natural waters may have essentially no p-alkalinity at all (i.e., their initial pH falls below that at point B. Some natural waters, particularly those impacted by acid-mine drainage, may have zero or negative alkalinity. This occurs when the pH of the water is lower than the pH at the bicarbonate endpoint (D). This means that the solution contains no strong bases, but contains excess strong acid.

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**(100 g mol-1)/(2 eq mol-1) = 50 g eq-1**

ALKALINITY - V Alkalinity is often expressed as the equivalent weight of calcium carbonate (mg L-1 CaCO3). Calculation of alkalinity from a titration is according to: The equivalent weight of CaCO3 is 50 g eq-1. Example: A 100 mL sample is titrated to the methyl orange end point with 2 mL of 0.5 N H2SO4. What is the total alkalinity in mg L-1 as CaCO3 and what is the concentration of HCO3- in mg L-1? Alkalinity can be expressed in many different ways. If you employ the formula given in the notes to slide 8, your alkalinity value will be in eq L-1, or normality. For thermodynamic calculations, these are the units you want. However, in published water analyses, the alkalinity is often expressed as mg L-1 CaCO3 equivalent. This does not mean that CaCO3 is necessarily present in the solution. What it means is that the solution has numerically the same alkalinity as a solution in which the same weight of CaCO3 per liter has been dissolved. The equivalent weight of CaCO3 is 50 g eq-1. This value results from the fact that the formula weight of calcium carbonate is approximately 100 g mol-1, and there are 2 eq mol-1 in CaCO3 (recall that an equivalent is essentially a mole of charge). Thus, we get: (100 g mol-1)/(2 eq mol-1) = 50 g eq-1 In the example given, the solution in question is titrated to the methyl orange endpoint. Methyl orange is a common indicator dye which changes color from yellow to red as pH decreases from about 4.5 to 3. This is the general location of the bicarbonate endpoint for most natural solutions. The problem states that it requires 2 mL of 0.5 N sulfuric acid to reach this endpoint. Recall that 0.5 N = 0.5 eq L-1. Note that there is an error in equation (3-27) in Kehew (2001). In this equation, the denominator is given as mLacid. It should be mLsample.

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**ALKALINITY - VI The total alkalinity in mg L-1 as CaCO3 is given by:**

In most natural waters, bicarbonate is the dominant contributor to the total alkalinity, so the concentration of HCO3- is given as: As the calculation shows, the alkalinity of our sample is equivalent to 500 mg L-1 of CaCO3. Another way to express alkalinity is as mg L-1 equivalent HCO3-. This is useful because in many natural waters, bicarbonate is by far and away the dominant contributor to alkalinity, that is, [alk] MHCO3-. Thus, we can directly obtain the concentration of bicarbonate as shown in the second formula in this slide. As pointed out by Kehew (2001), the ratio of equivalent weight of HCO3- to that of CaCO3 is 61/50 = Thus, alkalinity expressed as mg L-1 CaCO3 can be readily converted to alkalinity expressed as mg L-1 by multiplying the former by a factor of 1.22.

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**LEARNING OBJECTIVES Understand sources of CO2 in natural waters.**

Define and understand alkalinity. Learn to calculate the solubility of carbonate minerals such as calcite. Understand the common-ion effect. Become familiar with the concept of incongruent dissolution. Apply these concepts to some case studies. In this lecture we will pick up where we left off in Lecture 3, and continue with Chapter 3 in Kehew (2001). First we will consider sources of CO2, other than atmospheric, in natural waters. We will then define the concept of alkalinity and attempt to understand its relevance. The bulk of this lecture is devoted to understanding solubility equilibria of carbonate minerals, such as calcite. This includes being able to calculate the solubility of carbonates under specified conditions of pCO2 and pH, considering the common-ion effect and its relevance to natural waters, and familiarizing ourselves with the concept of incongruent dissolution. With these tools under your belt, you are then directed to the latter half of Chapter 3 in Kehew (2001), where he applies these concepts in a discussion of some interesting case studies. Because Kehew (2001) does a very good job describing these case studies, and we have limited time together, I will not be reviewing the case studies in class. However, I direct you to the Lecture 4 page on the course web site for some study questions to help guide your reading in the text.

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**CARBONATE MINERAL EQUILIBRIA**

The solubility of calcite at 25°C is governed by: For aragonite we have: Aragonite is more soluble (less stable) than calcite. The solubility of dolomite at 25°C is governed by: Now we are ready to add carbonate minerals to our system and investigate solubility equilibria of these minerals. Carbonate minerals are more abundant than more soluble minerals such as sulfates and chlorides, and much more soluble than silicate minerals, and so carbonates are the most important solid phases, the solubility of which controls the composition of natural waters. As we saw in Lecture 2, a fundamental control on the solubility of minerals, carbonates included, is the solubility product. In general, the larger the solubility product, the more soluble the mineral. But remember that we can only directly compare solubility products of minerals of the same stoichiometry. For example, we can directly compare the solubility products of calcite and aragonite (CaCO3), magnesite (MgCO3), rhodochrosite (MnCO3), smithsonite (ZnCO3), siderite (FeCO3) and cerussite (PbCO3), because these all have identical MCO3 stoichiometries. On the other hand, we cannot directly compare the solubility product of calcite and dolomite, because these two minerals have different stoichiometries. The slide tells us that aragonite has a somewhat larger solubility product than calcite, so aragonite is more soluble than calcite. The less soluble of two minerals will also always be the most stable of the two, because the less soluble mineral will tend to dissolve and reprecipitate as the less soluble phase. Thus, calcite is the more stable of the CaCO3 polymorphs. Both aragonite and calcite are supersaturated in sea water, but because aragonite is less stable, we should expect calcite to precipitate. However, in sea water, calcite precipitation is inhibited by adsorption of Mg to the surface of incipient crystals, so aragonite precipitates instead; it later alters to the more stable calcite. It also should be mentioned that Kdolomite is rather poorly known. Part of the reason for this is that the rate of precipitation of dolomite from solutions is painfully slow.

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**SOLUBILITY OF CALCITE IN AN OPEN SYSTEM - I**

We have six dissolved species: H+, OH-, H2CO3*, HCO3-, CO32- and Ca2+ whose concentrations are unknown. We need six independent equations to solve for these concentrations. Mass Action Expressions: In an open system, in addition to the solubility product (Kcalcite), other equilibrium constants that influence the solubility of calcite are the Henry’s Law constant (KCO2), the dissociation constants of carbonic acid (K1, K2) and the dissociation constant of water (Kw). We have six unknowns (the concentrations of Ca2+, H+, OH-, H2CO3*, HCO3- and CO32-) to solve for. There are five mass-action expressions that relate the activities (or concentrations if activity coefficients are assumed to be unity) of these species. We need one additional constraint. Can you guess what this constraint is?

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**SOLUBILITY OF CALCITE IN AN OPEN SYSTEM - II**

The sixth constraint is the charge-balance equation: This can be simplified to: At a constant value of pCO2, the logarithms of the concentrations of each of the species can be expressed as a straight-line function of pH. For example: The sixth constraint is the charge-balance constraint! Remember, this will almost always be one of the constraints we use to solve such problems. The charge-balance constraint should simplify as shown in this slide. Why can we expect that this simplification will be valid? Well, first of all, we know that, if the system were pure CO2-H2O, we would expect the pH to be less than 7, that is, acidic, and therefore carbonate should be negligible in comparison with bicarbonate, and hydroxide ion should be negligible compared to hydrogen ion. Now, in adding CaCO3 to the system, we are adding a base, which will tend to make the pH more basic. However, calcite has a moderately low solubility, which means the pH should not shift to such a high pH that carbonate or hydroxide become more concentrated than bicarbonate. Because the pH should now be slightly basic, we can expect that the concentration of hydrogen ion will be very low compared to that of Ca2+. By this reasoning, the only terms left in the charge-balance expression are Ca2+ and HCO3-. Even if the reasons for making this simplification are not entirely clear to you yet, remember that in these types of calculations, it is often useful to make an assumption, carry out a much simpler calculation, and then check to see if the answer you get makes sense. This approach makes the calculations much simpler, and also teaches you more about how the chemistry of the system works. The first step in the solution is to express the concentrations of each of the carbonate species in terms of pH and pCO2. Convince yourself you know how to do this; we did similar manipulations in Lecture 3.

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**SOLUBILITY OF CALCITE IN AN OPEN SYSTEM - III**

Bicarbonate can be calculated from: And carbonate from: Next, we write an expression for the activity of bicarbonate in terms of pH, and do the same for carbonate.

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**SOLUBILITY OF CALCITE IN AN OPEN SYSTEM - IV**

Calcium ion concentration is obtained from: Finally, we calculate the activity of Ca2+ as a function of pH. In this case we start out with the solubility product for calcite, and substitute the expression we have derived for the activity of CO32-. Now, we have expressions for the concentrations of Ca2+, H2CO3*, HCO3- and CO32- as a function of pH, and all of these are straight lines with various slopes and intercepts. If we plot these on a log activity vs. pH diagram, we might be able to find a graphical solution to the problem (in case you have forgotten, we are trying to calculate the solubility of calcite in an aqueous solution in equilibrium with atmospheric CO2).

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Log-log plot of concentrations of species in solution in equilibrium with calcite vs. pH at constant pCO2 = atm. Here is a plot of log concentration vs. pH for pCO2 = for a solution in equilibrium with calcite. The lines were plotted based on the equations we just derived, assuming activity coefficients can be taken to be unity. What we need to do to find our solution is find the pH at which our simplified charge-balance expression holds, that is, 2MCa2+ MHCO3-. Thus, we need to find the location on the graph where the concentration of bicarbonate is twice as high as that of calcium ion. On the log scale, we are looking for the pH where the line representing the bicarbonate ion concentration lies exactly (log 2) log units above the line representing the concentration of the calcium ion. This occurs at the pH indicated by the vertical red line on this diagram, i.e., at approximately pH = 8.3. Note that at this pH, the concentration of carbonate ion is almost two orders of magnitude less than that of bicarbonate, and the concentrations of all the other species are even lower. This means that our assumption that the charge-balance expression may be written as 2MCa2+ MHCO3- is valid. At a pH = 8.3, log MCa2+ -3.3, or MCa2+ 510-4 mol L-1. Another thing to note about this diagram: the lines representing the concentrations of the carbonate species, and H+ and OH- are exactly the same as the ones we plotted on the second-to-last diagram in Lecture 3. The only difference between the two plots is that here we have added the line representing the concentration of Ca2+, and we have left off the line representing the total carbonate concentration.

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**SOLUBILITY OF CALCITE IN AN OPEN SYSTEM - V**

In addition to the graphical solution, we have the numerical solution based on: Because of limitations in our ability to read the graph in the previous slide, we normally would like to obtain a numerical solution to the problem. This can be obtained by starting with our simplified charge-balance expression, and then make the appropriate substitutions for the concentrations of Ca2+ and HCO3-. We end up with an expression that is third-order in the activity of hydrogen ion. If we solve for this value, then we can back-substitute to get the calcium ion concentration.

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**SOLUBILITY OF CALCITE IN AN OPEN SYSTEM - VI**

Substituting the appropriate values for the K’s we get: pH = 8.28 Now based on the equation: we obtain This slide shows that, on substituting the appropriate equilibrium constants and CO2 partial pressure, we solve for a pH of 8.28, and a calcium concentration of about 510-4 mol L-1. Both of these values are close to our graphical solution.

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**SOLUBILITY OF CALCITE IN AN OPEN SYSTEM - VII**

If we take into account activity coefficients, then the following expressions can be derived (see pp in Kehew, 2001 for details): These equations may be used to derive the plots on the next two slides (assuming activity coefficients are unity). If we need to take into account activity coefficients, e.g., we are dealing with a water with high ionic strength, the expressions giving the solution to the calcite solubility problem shown in this slide can be used. We can, in turn, use these equations to make plots of pH vs. pCO2 and of MCa2+ vs. pCO2.

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The pH of pure water in equilibrium with calcite at 25°C as a function of the partial pressure of CO2. Note that pH decreases linearly with increasing CO2 partial pressure. This plot shows that the pH of a solution in equilibrium with calcite at a fixed partial pressure of carbon dioxide decreases linearly with increasing log pCO2. The higher the pCO2, the greater the quantity of acid that is being added to the solution, so the pH decreases.

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**Plot of calcium concentration vs**

Plot of calcium concentration vs. partial pressure of CO2 for a water in equilibrium with calcite at 25°C. Mixing of two saturated waters A and B can lead to undersaturation and calcite dissolution. This plot shows how Ca2+ concentration varies with pCO2 for a solution in equilibrium with calcite. The solubility curve is concave downward; conditions above the curve are supersaturated with respect to calcite, whereas below the curve the system is undersaturated. This concave-downward shape leads to a counterintuitive finding: two solutions saturated with respect to calcite can mix and result in an undersaturated solution. For example, if we take two solutions, A and B, that are saturated with calcite, and mix them in any proportion, the composition of the resultant solution will lie on a straight line connecting the two starting solutions. As the line falls below the solubility curve in the undersaturated region, any mixtures will be undersaturated and will have a tendency to dissolve calcite. Such mixing of two solutions saturated with calcite at two different values of pCO2 can occur in karst terrains. The dissolution that can result from such mixing has been employed as an explanation for how caves form.

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**LEARNING OBJECTIVES Understand sources of CO2 in natural waters.**

Define and understand alkalinity. Learn to calculate the solubility of carbonate minerals such as calcite. Understand the common-ion effect. Become familiar with the concept of incongruent dissolution. Apply these concepts to some case studies. In this lecture we will pick up where we left off in Lecture 3, and continue with Chapter 3 in Kehew (2001). First we will consider sources of CO2, other than atmospheric, in natural waters. We will then define the concept of alkalinity and attempt to understand its relevance. The bulk of this lecture is devoted to understanding solubility equilibria of carbonate minerals, such as calcite. This includes being able to calculate the solubility of carbonates under specified conditions of pCO2 and pH, considering the common-ion effect and its relevance to natural waters, and familiarizing ourselves with the concept of incongruent dissolution. With these tools under your belt, you are then directed to the latter half of Chapter 3 in Kehew (2001), where he applies these concepts in a discussion of some interesting case studies. Because Kehew (2001) does a very good job describing these case studies, and we have limited time together, I will not be reviewing the case studies in class. However, I direct you to the Lecture 4 page on the course web site for some study questions to help guide your reading in the text.

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**THE COMMON-ION EFFECT - I**

Calcite solubility is governed by the reaction: CaCO3(s) Ca2+ + CO32- (1) Suppose we added a second compound containing carbonate, and this compound is more soluble than calcite, e.g., Na2CO3. This compound will dissolve according to: Na2CO3(s) 2Na+ + CO32- (2) To the extent that reaction (2) proceeds to the right, by Le Chatlier’s principle, this will force reaction (1) to the left, precipitating calcite. According to Le Chatlier’s principle, any increase in the concentration of CO32- will shift the following reaction to the left and cause calcite to precipitate: CaCO3(s) Ca2+ + CO32- One way to increase the concentration of CO32- is to add a salt to the solution that both contains carbonate ion, and is more soluble than calcite. For example, Na2CO3 is much more soluble than calcite. Thus, when we add this salt to a solution saturated with calcite, its addition will cause CO32- to increase and therefore, calcite will precipitate. This is called the common-ion effect. The anion CO32- is the common-ion between CaCO3 and Na2CO3.

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**THE COMMON-ION EFFECT - II**

The effect of adding sodium carbonate to the solution can be demonstrated by adjusting the charge-balance expression to be: By repeating the derivation of the equations on a previous slide using this charge-balance expression we obtain: Increasing Na+ concentration leads to decreased Ca2+ concentration. If sodium carbonate is added to our system, we must modify our simplified charge-balance expression to include sodium ion. Manipulating the mass-action expressions as we did for the Na-free system in an earlier slide, we obtain the second equation shown above (convince yourself you can derive this equation). This equation allows us to calculate the effect of adding sodium carbonate on the solubility of calcite. However, it is important to note that the equation above only applies when Na+ is added in the form of Na2CO3 (or NaHCO3). If Na+ were added as NaCl, the above equation would not be valid. If it is assumed that activity coefficients can be neglected, then the above equation can be solved relatively easily for Ca concentration at a given Na concentration. However, if activity coefficients are taken into account, we have to do an iterative calculation, much like the ones we did in Lecture 2.

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Figure 3-14 from Kehew (2001). Curves showing Ca concentration in equilibrium with calcite as increasing amounts of NaHCO3 are added to solution. Addition of the common ion (HCO3-) in the form of sodium bicarbonate causes precipitation of calcite and a consequent decrease in the concentration of dissolved Ca. This diagram shows the effect of adding various amounts of NaHCO3 on the solubility of calcite as a function of pCO2. We can clearly see that at a given pCO2 addition of Na+ in the form of NaHCO3 causes a dramatic decrease in the solubility of calcite.

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**ANOTHER EXAMPLE OF THE COMMON-ION EFFECT**

Consider a groundwater just saturated with respect to calcite. This water encounters a rock formation containing gypsum. Gypsum is more soluble than calcite; it dissolves according to: CaSO4·2H2O Ca2+ + SO H2O(l) To the extent that this reaction goes to the right, it pushes the following reaction to the left: CaCO3(s) Ca2+ + CO32- causing calcite to precipitate. This example of the common-ion effect is probably more common in nature because gypsum is a much more common mineral than NaHCO3. However, the solubility of gypsum is closer to that of calcite than is NaHCO3, so the common-ion effect here is somewhat less dramatic. In fact, according to Appelo and Postma (1993) Geochemistry, groundwater and pollution. A.A. Balkema, the solubility of gypsum must be at least a factor of three greater than that of calcite for calcite precipitation to occur by the common-ion effect. For gypsum and calcite, this requirement is not fulfilled until the temperature rises above 25°C.

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**INCONGRUENT DISSOLUTION OF CALCITE AND DOLOMITE - I**

Incongruent dissolution - when one mineral dissolves simultaneously with the precipitation of another. Example: when calcite and dolomite are both encountered along a ground water flow path. How do we determine what will happen when both dolomite and calcite are present? Start by rearranging the KSP for dolomite: Thus far we have dealt primarily with congruent dissolution. Congruent dissolution is where the solid completely dissolves to form aqueous species, and no other solid forms. Incongruent dissolution is a situation in which one mineral dissolves, and another takes its place. An example of this is when both calcite and dolomite occur together along the same flow path, and the solution is close to saturation with respect to both minerals. If, for example, calcite is slightly more soluble than dolomite under the conditions of interest, then some calcite will dissolve. This will cause dolomite to precipitate by the common-ion effect, because the concentrations of Ca2+ and CO32-, which are constituents of dolomite, will be increased. This will increase the IAP for dolomite and cause it to precipitate. In order to predict exactly what will happen in this situation, we start with the mass-action expression for the dissolution of dolomite according to the reaction: CaMg(CO3)2(dolomite) Ca2+ + Mg2+ + 2CO32- We can rewrite the mass-action expression as shown in the slide above.

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**INCONGRUENT DISSOLUTION OF CALCITE AND DOLOMITE - II**

If a solution were in equilibrium with dolomite alone, then the activities of Ca2+ and Mg2+ would be equal so that: At 10°C we have Kdol½ = , which is exactly equal to Kcal = for this temperature. If dolomite had first reached equilibrium, then calcite would not be able to dissolve because IAP = Kcal! Let us now imagine that dolomite alone is dissolving (that is, calcite is not present). If dolomite were the only mineral dissolving, then by the stoichiometry of its dissolution reaction, the concentrations (and therefore the activities, if activity coefficients of Mg2+ and Ca2+ are similar) of Mg2+ and Ca2+ must be equal. Thus, we may make the substitution shown in the slide above. Taking the square root of both sides of the equation on the left, we obtain the equation on the right, which we see has the same form as the mass-action expression for the solubility of calcite. One way to determine what will happen when both calcite and dolomite are dissolving is to compare Kdol½ with Kcal. If Kdol½ = Kcal, then both calcite and dolomite should exist together in equilibrium, because their solubilities are equal. If Kdol½ > Kcal, then dolomite is more soluble than calcite; dolomite should dissolve and calcite should precipitate. If Kdol½ < Kcal, then dolomite is less soluble than calcite; calcite should dissolve and dolomite should precipitate. One way to think about the situation at 10°C, is to imagine that dolomite reached saturation first. In this case, saturation with respect to dolomite alone will fix IAPdol½ = Because IAPdol½ = (aCa2+)(aCO32-) = IAPcal, this means that IAPcal = Kcal, so the solution is also just saturated with calcite, and so calcite cannot dissolve. Both calcite and dolomite are saturated simultaneously.

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**INCONGRUENT DISSOLUTION OF CALCITE AND DOLOMITE - III**

However, at other temperatures, in general IAP would not be equal to Kcal. For example, at 30°C we have: Kdol½ = , and Kcal = In this case calcite would dissolve, because the ion activity product would be less than the solubility product for calcite. Dissolution of calcite would then cause dolomite to precipitate via the common-ion effect. We see in this slide that, at 30°C, Kdol½ < Kcal, making calcite more soluble than dolomite, and it should dissolve. One way to think about this is to imagine that at 30°C dolomite reached saturation first. Under these conditions, Kdol½ = But this also fixes the product (aCa2+)(aCO32-) at The latter product happens to be the IAP value for calcite. Because, IAP = < Kcal = , calcite will dissolve in an attempt to raise the IAP for calcite. Of course, this raises the IAP for dolomite as well, so dolomite must precipitate as calcite dissolves. The simultaneous precipitation of dolomite and dissolution of calcite is referred to as incongruent dissolution of calcite.

32
**INCONGRUENT DISSOLUTION OF CALCITE AND DOLOMITE - IV**

The latter process would be termed incongruent dissolution of calcite. At 0°C we have: Kdol½ = , Kcal = In this case, calcite would precipitate and dolomite would dissolve incongruently. We might also get incongruent dissolution because calcite dissolves more rapidly than dolomite. In this case, Ca2+ and CO32- concentrations increase more rapidly than Mg2+, so calcite may reach supersaturation while dolomite is still undersaturated. At 0°C, we have the situation that, Kdol½ > Kcal. Thus, according to what we have said thus far, we would expect calcite to precipitate and dolomite to dissolve. This is called incongruent dissolution of dolomite. So far we have only considered incongruent dissolution under equilibrium conditions. It is also possible to have a situation in which incongruent dissolution occurs owing to kinetic factors. For example, a water flowing into a formation containing both calcite and dolomite may dissolve calcite faster than dolomite. If the dissolution of calcite is considerably faster than that of dolomite, then calcite may attain saturation well before dolomite. However, because the concentration of Mg2+ is still low, dolomite may continue to be undersaturated and dissolve. Dissolution of dolomite may then induce calcite to precipitate, because as dolomite dissolves, increased calcium and carbonate ion concentrations result. This process may occur irrespective of the relative values of Kdol½ and Kcal, because it only depends on the fact that calcite dissolves more rapidly than dolomite.

33
**SOLUBILITY PRODUCTS FOR CALCITE AND DOLOMITE IN PURE WATER AT 1 BAR**

This table summarizes the relative values of pKcal and pKdol½. From this table we can see that equilibrium incongruent dissolution of dolomite can occur at all temperatures less than 10°C, and equilibrium incongruent dissolution of calcite can occur at all higher temperatures. Source: Freeze and Cherry (1979)

34
**LEARNING OBJECTIVES Understand sources of CO2 in natural waters.**

Define and understand alkalinity. Learn to calculate the solubility of carbonate minerals such as calcite. Understand the common-ion effect. Become familiar with the concept of incongruent dissolution. Apply these concepts to some case studies. In this lecture we will pick up where we left off in Lecture 3, and continue with Chapter 3 in Kehew (2001). First we will consider sources of CO2, other than atmospheric, in natural waters. We will then define the concept of alkalinity and attempt to understand its relevance. The bulk of this lecture is devoted to understanding solubility equilibria of carbonate minerals, such as calcite. This includes being able to calculate the solubility of carbonates under specified conditions of pCO2 and pH, considering the common-ion effect and its relevance to natural waters, and familiarizing ourselves with the concept of incongruent dissolution. With these tools under your belt, you are then directed to the latter half of Chapter 3 in Kehew (2001), where he applies these concepts in a discussion of some interesting case studies. Because Kehew (2001) does a very good job describing these case studies, and we have limited time together, I will not be reviewing the case studies in class. However, I direct you to the Lecture 4 page on the course web site for some study questions to help guide your reading in the text.

35
THE MADISON AQUIFER - I The Madison aquifer is located east of the Rocky Mountains. In this aquifer, we can see the effects of both the common-ion effect, and incongruent dissolution. The aquifer comprises Mississippian carbonates in which the primary minerals are calcite, dolomite and anhydrite. The Madison aquifer shows the effects of both the common-ion effect and incongruent dissolution. This aquifer is located in Montana, Wyoming, and the Dakotas. The aquifer is mostly confined, except where recharge occurs (along the flanks of the Rocky Mountains). The predominant minerals in the aquifer are anhydrite, dolomite and calcite.

36
**Predevelopment potentiometric surface of the Madison aquifer**

Predevelopment potentiometric surface of the Madison aquifer. Contours in meters above sea level (from Plummer et al., 1990, Water Resources Research, v. 26, pp ). This map shows the pre-development potentiometric surface for the Madison aquifer and is taken directly from your text.

37
**THE MADISON AQUIFER - II**

Along the flow path to the northeast, dissolution of anhydrite induces calcite precipitation via the common-ion effect. Precipitation of calcite results in decreased pH and Ca2+ concentration and increased pCO2 according to: Ca2+ + 2HCO3- CaCO3(s) + CO2(s) + H2O This results in dolomite becoming undersaturated, so it dissolves. In the northeast part of the aquifer, anhydrite dissolves. The further along the flow path, the more anhydrite dissolves, and so the concentration of SO42- for a particular water sample is an indication of how far along the flow path the water sample was taken. As anhydrite dissolves, the calcium ion concentration is temporarily increased, supersaturating the water with respect to calcite, which precipitates. Precipitation of calcite lowers the calcium ion concentration and pH, and increases the partial pressure of CO2, all of which tend to increase the solubility of dolomite. Thus, dolomite becomes undersaturated and dissolves. To summarize, dissolution of anhydrite causes the precipitation of calcite by the common-ion effect, and this results in the incongruent dissolution of dolomite.

38
**THE MADISON AQUIFER - III**

The dissolution of dolomite is a type of incongruent dissolution called dedolomitization. Sulfate concentration increases along the flow path until saturation with respect to anhydrite, thus sulfate serves as a measure of distance from the recharge area. Calcite is close to saturation throughout the aquifer, so very little anhydrite dissolution is required to precipitate calcite. As mentioned in the slide above, incongruent dissolution of dolomite can also be referred to as dedolomitization (incongruent dissolution of calcite to form dolomite would be dolomitization). As we will see, calculation of the saturation index for calcite shows that, throughout the aquifer, the water is very close to saturation with calcite. This means that only small amounts of calcium are required to cause calcite to precipitate, and therefore, relatively small amounts of anhydrite need to dissolve to drive calcite precipitation.

39
Saturation indices of gypsum as a function of SO42- concentration for the Madison aquifer. The relationship between these variables shows that dissolution of gypsum /anhydrite is the source of sulfate in these waters. (Data from Plummer et al., 1990, Water Resources Research, v. 26, pp ) This figure shows data from the Madison Aquifer. It is a plot of the calculated saturation index for gypsum vs. the measured sulfate concentration. It is not entirely clear why Kehew (and the original authors, Plummer et al. for that matter) chose to plot the indices for gypsum when the mineral in the aquifer is anhydrite. However, the difference in the solubility products of gypsum and anhydrite is slight, and a plot of the saturation indices for anhydrite would look essentially the same as the plot above. Waters with the lowest sulfate concentration, and therefore gypsum saturation index, are those that are closest to the recharge area of the aquifer. Those that have the highest sulfate concentrations are farthest from the recharge area, that is, farther along the flow path.

40
Saturation indices of calcite as a function of SO42- concentration for the Madison aquifer. The majority of the waters are saturated to slightly oversaturated. (Data from Plummer et al., 1990, Water Resources Research, v. 26, pp ) The majority of the waters, as shown in this plot, are at or above saturation with calcite. Thus, as mentioned previously, only small amounts of anhydrite dissolution are required to effect calcite precipitation. The only waters that are significantly undersaturated with calcite are those with very low sulfate concentrations. These are the waters in which little or no anhydrite dissolution has occurred, and they are all located near the recharge zone.

41
Saturation indices of dolomite as a function of SO42- concentration for the Madison aquifer. The SI for dolomite is more variable, with undersaturation present across the aquifer. (Data from Plummer et al., 1990, Water Resources Research, v. 26, pp ) According to this plot of dolomite saturation index vs. sulfate concentration, we can see that dolomite bounces back and forth from undersaturation to oversaturation all along the aquifer.

42
Precipitation of calcite by the common-ion effect as a function of anhydrite dissolution in the Madison aquifer. The dashed line shows the trend due to dedolomitization alone, and the solid arrow shows dedolomitization plus cation exchange reactions. This diagram shows the results of mass-balance calculations, in which the water chemistry is employed to calculate the amount of anhydrite dissolved and the amount of calcite precipitated. Most of the data plot along the dashed line; this trend can be explained by dedolomitization (triggered by anhydrite dissolution) alone. Along flow path 2, it appears that additional ion-exchange reactions are also required to explain the water chemistry. We will cover ion-exchange in a subsequent lecture.

43
Dissolution of dolomite by the common-ion effect as a function of anhydrite dissolution in the Madison aquifer. The dashed line shows the trend due to dedolomitization alone, and the solid arrow shows dedolomitization plus cation exchange reactions. This diagram shows more or less the same thing as the previous diagram, except that it plots the amount of dolomite dissolved vs. the amount of anhydrite dissolved. Once again, most of the data conform well to the trend expected for dedolomitization alone.

44
**Plot of calcite precipitated vs**

Plot of calcite precipitated vs. dolomite dissolved in the Madison aquifer. The remarkably linear relationship demonstrates the nature of the incongruent dissolution. Given the good correlations shown in the previous two plots, the correlation shown here, between the amounts of calcite precipitated and dolomite dissolved, should not be surprising. However, this excellent correlation is further evidence of the relationship between calcite and dolomite during dedolomitization.

45
**IONIC STRENGTH EFFECT ON SOLUBILITY - I**

If NaCl is added to a solution saturated with calcite, what will happen? NaCl contains no ions in common with calcite, so we would not expect solubility to decrease from a direct common-ion effect. On the other hand, NaCl will increase the ionic strength of the solution. What will this do? Consider the equation: Adding additional salts to a solution can have a number of different effects on the solubility of minerals. As we determined in Lecture 2, if we add a salt which contains a ligand that can form a complex with a component of the mineral, the solubility will increase. For example, if we added sodium acetate to a solution, the acetate anion might form a complex with calcium and increase the solubility of calcite. We have already seen that, if we add a salt that contains a common ion, e.g., NaHCO3, then the solubility of calcite will decrease by the common-ion effect. However, if we add a salt that is more or less inert, i.e., contains no complexing ligand or common ion, then what will its effect be on solubility? To answer this question, we need to expand the mass-action expression for the solubility of calcite in terms of ion concentrations and activity coefficients. This expression is written on the last line of this slide.

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**IONIC STRENGTH EFFECT ON SOLUBILITY - II**

Addition of NaCl will increase ionic strength, which in general decreases the activity coefficients. To keep the solubility product constant, if the activity coefficients decrease, the concentration terms must increase. Thus, addition of NaCl will generally increase calcite solubility. Minerals tend to be more soluble in concentrated solutions than in dilute ones (providing there is no common-ion effect). The effect of an inert salt like NaCl is to increase the overall ionic strength of the solution. We have seen previously (Lecture 2), that activity coefficients generally decrease with increasing ionic strength (at least until an ionic strength of about 0.5 mol L-1). According to the equation above, if the activity coefficients of Ca2+ and CO32- decrease below one, then to maintain the activity product constant and equal to Kcal, the concentrations of these ions must increase to offset the decrease in the activity coefficients. Thus, more calcite will dissolve in a high-ionic strength solution than in a more dilute one. Finally, to see some applications of the principles we have learned in this lecture, read carefully the examples on pages in Kehew (2001). I will not go over these examples, but you are responsible for understanding them. To help guide your reading, you will find some study guide questions on the Lecture 4 page of the course web site.

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