Solving Ionic Equilibria Problems Objectives and scope

Transcription

1 7 Solving Ionic Equilibria Problems Objectives and scope In this chapter you will learn how to use the three main types of tools environmental scientists and engineers use to solve problems involving ionic equilibria in water: graphical, algebraic, and numerical (computer-based) techniques. The types of equations applicable to all three approaches are defined and described, along with the important concept of as the fraction of the total (analytical) concentration of a substance present in a particular chemical form. Construction and interpretation of single mastervariable diagrams (e.g., pc-ph diagrams) are explained, and their use to solve acid-base equilibrium problems is illustrated through examples. Construction of twomaster-variable (predominance-area) diagrams (e.g., ph-log P CO2 diagrams) and their use to illustrate mineral/phase equilibria also are described. The basic concepts and techniques used to solve ionic equilibria by algebraic techniques are developed through simple examples, with an emphasis on understanding the conditions under which it is appropriate to make simplifying assumptions. Because many real-world problems involving ionic equilibria in aquatic systems are too complicated to solve conveniently through graphical or manual algebraic techniques, emphasis is placed on the use of computer-based solutions for equilibrium problems. After a brief description of the history of computer solutions for ionic equilibria and the fundamental principles by which such programs operate, example-based instructions are given for use of the program MINEQL+. 220

2 SOLVING IONIC EQUILIBRIA PROBLEMS 221 Key terms and concepts Mass-action (equilibrium) expressions, mass-balance equations: X T and TOTX, charge-balance (electroneutrality) equations, proton-balance equations i, fraction of X T present as species i Master variables, pc = log C, pc-ph diagrams, system points Predominance-area diagrams Eight-step approach for manual algebraic solution of ionic equilibria Components and various types of species in computer-based equilibrium programs 7.1 Introduction A large part of aquatic chemistry involves the solution of ionic equilibrium problems. Such problems are important, for example, in water treatment chemistry, in describing the chemical composition of natural waters, and in analytical aspects of aquatic chemistry. Although the term ionic equilibria implies an exclusive focus on ions, as used here, the term is not meant to be so restrictive. Some ionic equilibria involve dissolved gases (e.g., CO 2 and O 2 ) and air-water equilibria of gases. Some important acid-base and complexation equilibria for aquatic systems involve organic compounds, particularly organic acids and bases. Finally, solubility equilibria involve solid (mineral) phases that are not ionic. This chapter describes three basic approaches graphical, algebraic, and numerical (computer-based) methods to solve and/or display chemical equilibria involving the kinds of species and equilibrium processes italicized above: gas transfer, acidbase, complexation, and mineral solubility reactions. All three approaches rely on the same set of mathematical expressions that define chemical equilibrium and associated constraints. Solutions for a fifth process, oxidation-reduction (redox) reactions, rely on similar types of expressions and similar graphical, algebraic, and numerical approaches. However, solving redox problems requires some additional understanding about redox fundamentals and additional equations and terms, and so we defer discussion of solving redox equilibria until Chapter Equations used to define and constrain ionic equilibria Five types of equations or relationships are used to define chemical equilibria: massaction (or equilibrium-constant, the term we use to refer to mass action) expressions, two kinds of mass-balance expressions (X T and TOTX), charge-balance expressions based on the electroneutrality constraint, and proton balances, which are analogous to charge balances but are based on balancing the products resulting from the release of protons and the products of consumption (or acceptance) of protons. The goal is to use these equations to obtain an equal number of equations and unknowns and thus have a solvable problem. To illustrate the development of these equations, we will use the carbonic acidbicarbonate-carbonate system (hereafter referred to simply as the carbonate system ) because of its importance in defining the acid-base chemistry and chemical composition of natural waters. Figure 7.1 is a schematic of important processes in the carbonate system; Chapter 8 discusses the carbonate system in more detail (see Section 8.5).

3 222 WATER CHEMISTRY CO 2(g) K H = M atm 1 = Δ K h = K a0 = K a2 = = H 2 O + CO 2(aq) H 2 CO H + + HCO H + + CO 2 Ca 2+ K a1 = = K s0 = H 2 CO * CaCO (s) Figure 7.1 Schematic representation of the main equilibria involved in the CO 2 -carbonate system. H 2 CO is the sum of CO 2(aq) and H 2 CO (true carbonic acid). K a0 is the acid dissociation constant for true carbonic acid, and K a1 is the acid dissociation constant for the composite species H 2 CO. K values are for 25 C Equilibrium constant or mass-action expressions These are based on stoichiometric relationships defined by chemical reactions or phasetransfer processes. For a general reaction, aa + bb cc + dd, where the lowercase letters are stoichiometric coefficients, the equilibrium expression is K eq = {C}c {D} d {A} a {B} b, (7.1) where {i} stands for the activity of species i. Recall that the equilibrium constant is related to G of the reaction ( G = RT lnk). As described in Chapter 4, under some circumstances we can approximate the activity terms using molar concentrations: c K eq = [C]c [D] d (7.2) [A] a [B] b From the general relationship between activity and concentration, {i} = i [i], we can see that the concentration-based equilibrium constant c K eq for the general reaction aa + bb cc + dd is related to the thermodynamic (activity-based) constant as follows: c K eq = K eq A a B b Cc D d (7.)

4 SOLVING IONIC EQUILIBRIA PROBLEMS 22 Important equilibrium constant expressions for the carbonate system (Figure 7.1) include acid dissociation constants (K a1 and K a2 ), air-water equilibria (K H for CO 2 ), and the solubility product, K s0, for calcium carbonate. Three important issues need to be discussed regarding the use of equilibrium constant expressions in solving ionic equilibrium problems. First, it frequently is useful to write these expressions in log form. The general example (Eq. 7.2) then becomes c log K eq = c log[c] + d log[d] a log[a] b log[b]. (7.4) As we shall see, such logarithmic expressions are particularly useful for graphical solutions, which usually involve logarithmic plots of variables. Second, if equilibrium constants are expressed as thermodynamic values (i.e., in terms of activities), there is an inconsistency between them and the other kinds of equations available to solve equilibrium problems, which are written in molar or equivalent concentrations. Several ways around this problem are described in Section 7.4 on algebraic solutions. Third, in solving ionic equilibria, one frequently encounters equilibrium expressions for which the value of K is unknown. Sometimes, K can be computed from tabulated values of G f (free energy of formation) of the products and reactants, as described in Chapter (see Section.11). In other cases, K for the reaction of interest can be obtained by adding or subtracting other reactions for which K values are known. This approach is used in several examples later in this chapter. A simple example here illustrates the basic idea. Consider the following expression for the dissolution of calcium carbonate: CaCO (s) + H + Ca 2+ + HCO (7.5) This expression is used in water treatment chemistry and is the basis for the Langelier index of calcium carbonate saturation (see Sections 10.2 and 10.4). Most tabulations, however, of equilibrium constants for the carbonate system do not list this specific reaction. Inspection of the reaction suggests that it can be derived by adding two wellknown reactions for which K values are available: the solubility product for calcium carbonate, and reverse reaction for the second acid dissociation of carbonic acid. The equilibrium constant for the net reaction is obtained by multiplying the K values for the two reactions: CaCO (s) Ca 2+ + CO 2 K s0 = CO 2 + H + HCO K 1 a2 = Net: CaCO (s) + H + Ca 2+ + HCO K net = K s0 K 1 a2 = = It is mathematically equivalent to write the second reaction as the dissociation of HCO and subtract it from the solubility expression for CaCO. In that case, we divide K s0 by K a2 (where K a2 = ). The above example can be formulated as a general rule: When two reactions are added, the equilibrium constant for the net reaction is the product of the

5 224 WATER CHEMISTRY K values for the individual reactions; when a reaction is subtracted from another, K for the net reaction is obtained by dividing K for the reaction being subtracted into K for the other reaction. The basis for this rule is the fact that equilibrium constants are exponentially related to free energies of reaction. Multiplying K values is equivalent to adding log K values, and the latter is equivalent to adding free energies of reaction ( G r) Mass-balance (X T ) expressions In many equilibrium problems, information is given about the total concentration of an acid-base system or some other substance that can exist as multiple species in solution, for which concentrations (or activities) of the individual species are not known. A massbalance equation (sometimes called a mole-balance equation ) then can be written as a constraint on the system. In general, we can write X T = i [X] i, (7.6) where X is the species/element of interest, concentrations are in molar units, subscript T stands for total dissolved, and the summation is over all chemical species of the substance in solution. For a simple carbonic acid-carbonate system, X is carbon, and thus C T is C T =[H 2 CO ]+[HCO ]+[CO2 ], (7.7) where H 2 CO is the sum of dissolved CO 2 and true carbonic acid, as shown in Figure 7.1: [H 2 CO ] = [CO 2] aq + [H 2 CO ] (7.8) Routine analytical methods do not distinguish between true carbonic acid and dissolved CO 2, so it is convenient to lump the two species together into the composite species H 2 CO. The expression in Eq. 7.7 assumes that from a mass-balance perspective there are no significant soluble complexes involving bicarbonate or carbonate ions, which usually is the case in freshwater systems. C T for the carbonate system can be measured by several analytical techniques and thus is a common item of information in problems involving this system TOT X expressions For systems in which both dissolved and gaseous or precipitated species affect the mass balance, the total amount present is X T plus any material in the solid or (occasionally) gaseous phases. For example, in a closed system containing calcium carbonate, and TOTC =[CaCO (s) ]+[H 2 CO ]+[HCO ]+[CO2 ], (7.9) TOTCa =[CaCO (s) ]+[Ca 2+ ], (7.10)

6 SOLVING IONIC EQUILIBRIA PROBLEMS 225 Although these equations are useful in determining the total mass in the system or for completing total mass balances (i.e., fraction of metal precipitated), they are not often used as one of the equations in solving equilibrium problems The last equation At this point, the equilibrium expressions (including that for the dissociation of water) and the mass-balance equations for the relevant species provide us with n unknowns but generally only n 1 equations. Thus, one more equation usually is needed, and there are three choices for this final expression: the charge-balance, the proton-balance, and the TOTH equation. In the end, each is a different way of expressing the same equation/concept and the choice depends on which makes most sense to the solver of the problem. That said, understanding of the TOT H equation is necessary to understand and use most computer-based solution programs Charge-balance (electroneutrality) expressions All solutions are electrically neutral. Thus, the sum of the cation concentrations must equal the sum of the anion concentrations when both are expressed on a charge equivalents basis that is, as the molar concentration times the integer value of the charge on each ion (ignoring the sign of the charge). In general, this constraint can be expressed as cation equivalents = anion equivalents, [ i M i + ] = i j j [ A j ]. (7.11) The concentrations of cations (M) and anions (A) are expressed in moles per liter, and i and j are the absolute values of the integer charges on each cation and anion, respectively. Because one mole of a substance with a ±2 charge contributes two moles of charge, it is important to remember to multiply by the factor of 2. For a solution of sodium bicarbonate (NaHCO ) in pure water, the charge-balance equation is [H + ] + [Na + ] = [HCO ] + 2[CO2 ]+[OH ], (7.12) and for a solution of carbon dioxide in pure water, the charge balance is [H + ] = [HCO ] + 2[CO2 ] + [OH ]. (7.1) The starting substance (CO 2(aq) ) does not appear in the charge balance because it is not an ion; only the ionic products of its reaction with water and ionic products of the dissociation of water appear in the expression.

7 226 WATER CHEMISTRY Proton-balance expressions These equations represent a special case of the electroneutrality condition, and they apply primarily to problems dealing with acid-base equilibria. For any acid that dissociates and releases a proton, another substance (a base) must accept the proton. The sum of the concentrations of all the products of proton release in an acid-base system thus must equal the sum of the concentrations of the products of proton consumption (or proton acceptance). This is because free protons do not exist in aqueous solution. As discussed in Chapter 8, when acids dissociate in water, the proton is accepted by a base; if no other base is available, water itself acts as a base, accepting the proton to become H O +. Thus we can write the general proton-balance expression as PPC = PPR, (7.14) where PPC stands for products of proton consumption (or proton acceptance) and PPR stands for products of proton release. The simplest proton balance is for water itself. When it dissociates, the product of proton release is hydroxide ion, OH ; the product of proton consumption is the hydronium ion, H O +. Therefore, the proton balance for water is PPC = PPR, [H O + ]=[OH ]. (7.15) For simplicity we usually ignore the fact that H + is attached to H 2 O and write it just as H +. The proton balance thus becomes PPC = PPR [H + ]=[OH ]. (7.16) Equation 7.16 is the starting point for all proton-balance expressions of aqueous solutions; that is, [H + ] is always a product of proton consumption (because it represents H O + ), and OH is always a product of proton release in such proton balances. Proton balances for some carbonic acid-carbonate systems are given in the following examples. In Example 7.2, note that it is always possible to convert a charge balance to a proton balance (and vice versa). EXAMPLE 7.1 Dissolution of CO 2 in water: As shown in Figure 7.1, dissolved CO 2 becomes hydrated to form true carbonic acid, H 2 CO, which then dissociates to form a bicarbonate ion, HCO, and one H+ (again, really H O + ). In turn, HCO can dissociate to form a carbonate ion, CO 2, and another H+. However, CO 2(aq) and its hydration product H 2 CO (the sum of which, as we noted above, is H 2 CO ) cannot accept another proton. The proton-balance expression thus is (1) PPC = PPR, [H + ] = [HCO ] + 2[CO2 ] + [OH ].

8 SOLVING IONIC EQUILIBRIA PROBLEMS 227 Note that the reactant (CO 2 or H 2 CO ) is not a part of the proton balance. This is a general rule for proton balances; only the products of reaction are a part of the expression. Note also that the only product of proton consumption is H + itself, from the dissociation of water, carbonic acid, and bicarbonate. The factor of 2 multiplying the CO 2 concentration derives from the stoichiometric fact that for every CO 2 formed from H 2 CO, two H + are produced. Finally, it is pertinent to note that the proton-balance equation in this case is the same as the charge-balance equation. EXAMPLE 7.2 Dissolution of sodium bicarbonate (NaHCO ) in water: Sodium is a strong base cation and has no significant tendency to react with water or its dissociation products. Therefore, it does not contribute to the proton balance. In contrast, bicarbonate is both a weak acid and a weak base. It can accept a proton to form carbonic acid and can release a proton to form carbonate ion. The proton-balance expression for NaHCO and water thus is (2) PPC = PPR, [H + ] + [H 2 CO ] = [CO2 ] + [OH ]. Recall that C T =[H 2 CO ] + [HCO ] + [CO2 ]. Rearranging, we obtain [H 2CO ] = C T [HCO ] [CO2 ]. Also note that because for each unit of bicarbonate added to the solution, one unit of sodium is added, so C T = [Na + ]. Substituting [H 2 CO ] into the proton balance, [H + ] + [Na + ] [HCO ] [CO2 ] = [CO2 ] + [OH ]. Combining the anions on the right hand side gives the charge balance: [H + ] + [Na + ] = [HCO ] + 2[CO2 ]+[OH ]. Thus, either the charge balance or the proton balance (but not both) can be used in solving a problem TOTH expressions In development of the TOT H expression, the acid/base groups must be categorized. For each group, one must be chosen as the reference species. For the example of adding sodium bicarbonate to water, the groups are (1) H +,OH,H 2 O; (2) H 2 CO, HCO, and CO 2 ; and () Na+. The choice of the reference species is arbitrary, but it is convenient if the reference species is chosen as the one that will be dominant in the system. Here we will choose H 2 O and HCO. All species are now related to the reference species: 1H 2 O+0H + 1H 2 O 0H 2 O+1H + 1H + 1H 2 O 1H + 1OH 1Na + +0H + 1Na + 1 HCO +1H+ H 2 CO 1 HCO +0H+ 1 HCO 1 HCO 1H+ 1CO 2

9 228 WATER CHEMISTRY The coefficients on the protons (n i ) indicate the excesses or deficiencies of H + in the product species relative to the reference species. For example, the proton (or H O + ) contains one more proton that water so n = 1. Hydroxide contains one less proton than water, and thus n for hydroxide is 1. The TOTH at equilibrium (TOTH eq )is n i [X i ] i for all X: TOTH eq =[H 2 O] 0 +[H + ] 1 +[OH ] ( 1) +[Na + ] 0 +[H 2 CO ] 1 +[HCO ] 0 +[CO2 ] ( 1) (7.17a) TOTH eq =[H + ] [OH ]+[H 2 CO ] [CO2 ] (7.17b) The expression that arises is dependent on the choice of reference species. If H + and CO 2 are chosen as the reference species, one will not obtain the same equation as shown in Eq. 7.17b. To complete the TOTH expression, the proton concentration added (TOTH in ) relative to the reference species needs to be determined. If Na 2 CO is added, the contribution to TOTH in is the negative of the concentration of the sodium carbonate added (i.e., if 1 10 M is added, TOTH in = 1 10 ), because CO 2 is proton deficient (n = 1) compared to the reference species HCO. If NaOH is added, the contribution is again the negative of the concentration added (n = 1) because hydroxide is proton deficient relative to water. If NaHCO is added in any amount, the contribution to TOTH in is 0 (because HCO is the reference). The equation is completed by setting TOTH in = TOTH eq, which is the general form of the TOTH expression. If we add only NaHCO to the solution, TOTH in = 0, and thus substituting in this value and the expression for TOTH eq (Eq. 7.17b) into TOTH in = TOTH eq gives This equation can be rearranged to give 0 =[H + ] [OH ]+[H 2 CO ] [CO2 ]. (7.18a) [H + ] + [H 2 CO ] = [CO2 ] + [OH ], (7.18b) which, of course, is the proton balance shown in Example 7.2, and we already know that this is equivalent to the charge balance. Thus, the charge-balance, the proton-balance, and the TOTH equations are three different ways of looking at the same equation i as the fraction of X T present as species i The alpha ( ) notational system is widely used in ionic equilibria to express the fraction of the total analytical concentration of an acid-base system present as a particular ionic species. In general, i = [i] X T, (7.19) where [i] is the molar concentration of the ith species and X T is the sum of the molar concentrations of all species in the system. For the carbonate system illustrated in

10 Figure 7.1 and its C T, as represented in Eq. 7.7, we can write 0 = 1 = 2 = SOLVING IONIC EQUILIBRIA PROBLEMS 229 [H 2 CO ] [H 2 CO ] + [HCO ] + [CO2 ] = [H 2CO ], (7.20a) C T [HCO ] [H 2 CO ] + [HCO ] + [CO2 ] = [HCO ], (7.20b) C T [CO 2 ] [H 2 CO ] + [HCO ] + [CO2 ] = [CO 2 ]. (7.20c) C T By convention, for all acid-base systems 0 is the fraction of C T present as the completely protonated acid; 1 is the fraction of C T present as the first dissociation product of the acid; 2 is the fraction of C T present as the second dissociation product of the acid, and n is the fraction of C T present as the completely deprotonated form of an n-protic acid. By simple rearrangement of Eqs , we have [H 2 CO ]= 0C T, (7.21) [HCO ]= 1C T, (7.22) [CO 2 ]= 2C T. (7.2) The usefulness of the notation lies in the fact that the values can be expressed in terms solely of equilibrium constants and {H + }. Values of for a given species thus can be calculated and tabulated as a function of ph, independent of the concentration of individual acid-base species or of C T for the system of interest. For any diprotic acid, the expressions for 0, 1, and 2 can be derived readily from the corresponding expressions for C T, K a1, and K a2 (see Example 7.): 0 = 1 = 2 = {H + } 2 {H + } 2 +{H + }K a1 + K a1 K a2 (7.24) {H + }K a1 {H + } 2 +{H + }K a1 + K a1 K a2 (7.25) K a1 K a2 {H + } 2 +{H + }K a1 + K a1 K a2 (7.26) Equations can be generalized for a n-protic acid as follows: i = DT i {H + } n +{H + } n 1 K a1 +{H + } n 2 K a1 K a {H + }K a1...k an 1 + K a1...k an, (7.27) where DT i stands for the corresponding denominator term for species i. Note that {H + } n is the denominator term for the fully protonated species H n B, and that K a1... K an is the

11 20 WATER CHEMISTRY Table 7.1 Alpha ( ) expressions for triprotic acids 0 = 1 = 2 = = {H + } {H + } +{H + } 2 K 1 +{H + }K 1 K 2 + K 1 K 2 K {H + } 2 K 1 {H + } +{H + } 2 K 1 +{H + }K 1 K 2 + K 1 K 2 K {H + } 2 K 1 K 2 {H + } +{H + } 2 K 1 +{H + }K 1 K 2 + K 1 K 2 K K 1 K 2 K {H + } +{H + } 2 K 1 +{H + }K 1 K 2 + K 1 K 2 K term for the fully deprotonated species B n. For a monoprotic acid, HB, Eq thus yields and 0 = [HB] C T = 1 = [B ] C T = Expressions for 0 to for triprotic acids are given in Table 7.1. {H + } {H + }+K a1 (7.28) K a1 {H + }+K a1. (7.29) EXAMPLE 7. Derivation of α expressions for diprotic acids: Derivations of expressions for any acid-base system starts with writing the basic expression as in Eqs and then solving the acid dissociation constants to express all the components of the denominator in terms of the numerator term. For example, for 0, (1) 0 = [H 2B] [H = 2 B] C T [H 2 B] + [HB ] + [B 2 ]. The acid dissociation (equilibrium) constants are (2) K 1 = {H+ }[HB ] [H 2 B], () K 2 = {H+ }[B 2 ] [HB. ] Solving K 1 for [HB ] and solving K 2 for [B 2 ] yields (4) [HB ]= K 1[H 2 B] {H + }, (5) [B 2 ]= K 2[HB ] {H +, } For consistency here we write K a1 and K a2 as mixed constants with H + expressed as activity and other terms in molar concentrations (because ph is a measure of the log of H + activity and the other constituents are expressed in molar concentration in X T ). Values of the equilibrium constants thus are not the true thermodynamic values but apply only to a particular ionic strength. See Eq. 7. for the relationship between these Ks and the thermodynamic values.

12 SOLVING IONIC EQUILIBRIA PROBLEMS 21 which is not quite what we need. However, if we substitute Eq. (4) for [HB ] into Eq. (5), we obtain an expression for [B 2 ] in terms of [H 2 B]: (6) [B 2 ]= K 1K 2 [H 2 B] {H + } 2 Substituting Eqs. (4) and (6) into Eq. (1) yields [H (7) 0 = 2 B] [H 2 B] + K 1 [H 2 B]/{H + }+K 1 K 2 [H 2 B]/{H + }. 2 We can eliminate the term [H 2 B] from Eq. (7) because it appears in every term in the numerator and denominator. Furthermore, we can simplify the result by multiplying the numerator and denominator by {H + } 2 to obtain {H + } 2 (8) 0 = {H + } 2 + K 1 {H +. }+K 1 K 2 For 1, we solve K 1 and K 2 for [H 2 B] and [B 2 ], respectively, in terms of [HB ] and substitute the results into the denominator of the 1 expression: (9) [H 2 B]= {H+ }[HB ], (10) [B 2 ]= K 2[HB ] K 1 {H +, } [HB ] (11) 1 = [H 2 B]+[HB ]+[B 2 ] = [HB ] {H + }[HB ]/K 1 +[HB ]+K 2 [HB ]/{H + }. Again, we can eliminate the term [HB ] from the right side of Eq. (11) and simplify by multiplying the numerator and denominator by the term K 1 {H + }toget K 1 {H + } (12) 1 = {H + } 2 + K 1 {H +. }+K 1 K 2 Finally, for 2 we solve K 2 for [HB ] in terms of [B 2 ] and substitute that into K 1 to get [H 2 B] in terms of [B 2 ] and then substitute both results into the denominator of the 2 expression: (1) [HB ]= {H+ }[B 2 ], K 2 (14) [H 2 B]= {H+ }[HB ] = {H+ } 2 [B 2 ], K 1 K 1 K 2 and [B 2 ] (15) 2 = [H 2 B]+HB ]+[B 2 ] = [B 2 ] {H + } 2 [B 2 ]/K 1 K 2 +{H + }[B 2 ]/K 2 +[B 2 ], which upon simplification yields K 1 K 2 (16) 2 = {H + } 2. + K 1 {H + }+K 1 K 2 Finally, we note that the concept of values is not limited to acids and bases but also is used for metal ions and their complexes with various ligands. For example, for metal M

13 22 WATER CHEMISTRY and ligand L we can write M = [Mm+ ], ML = [ML+m 1 ],... MLn = [ML+m n n ], (7.0) M TM M TM M TM where M TM = [M m+ ] + [ML +m 1 ] [ML n +m n ], and the equilibrium constants used in the expressions are stepwise complex dissociation constants. 7. Graphical methods of displaying ionic equilibria and solving equilibrium problems 7..1 Introduction Many types of graphs are used to display ionic equilibrium, but two types are by far the most common and are the focus of this chapter. The first involves plots of the log of concentration of various ionic species versus some independent variable, most commonly ph. Because the negative log of concentration is denoted pc, and because ionic concentrations in natural waters generally have negative log values, such plots often are called pc-ph diagrams. The independent variable sometimes is referred to as a master variable, and the plots also are called single master-variable diagrams. The second type involves plots of master variables on both axes, e.g., log P CO2 versus ph or redox potential, E h versus ph. These are often called predominance-area diagrams because the lines on such plots delimit regions in which a given species (ionic form or mineral phase) is the predominant species in the system pc-ph diagrams Equilibrium expressions for acid-base reactions have multiplicative terms, and concentrations of reactants and products vary over many orders of magnitude over the ph range of 0 14, which itself is a range of 14 orders of magnitude for {H + }. Consequently, it is not feasible to graph concentrations of acid-base species as a linear function of ph. A more useful way is to plot concentrations on a log scale ( log C or pc) versus ph ( log{h + }). This results in generally simple, straight line relationships. The following paragraphs and equations describe how such diagrams are developed from acid-base equilibrium and mass-balance relationships. Consider the monoprotic acid, acetic acid (HAc): K a = {H+ }{Ac } {HAc} = ; pk a = 4.76 (7.1) For ease in developing the equations for pc-ph diagrams, we will ignore differences between activity and concentration, and from here forward will use brackets, [i], to denote either. We can write the mass-balance expression: Ac T =[HAc]+[Ac ] (7.2)

14 Solving Eq. 7.1 in terms of [Ac ] and [HAc], we obtain and SOLVING IONIC EQUILIBRIA PROBLEMS 2 [HAc] = [H+ ][Ac ] K a, (7.) [Ac ]= K a[hac] [H +. (7.4) ] Substituting Eq. 7. and 7.4 into Eq. 7.2 and solving for [HAc] and [Ac ], we obtain Ac T =[HAc]+K a [HAc][H + ] (7.5a) or [HAc] = Ac T[H + ] [H + ]+K a, (7.5b) and Ac T =[Ac ]+[H + ][Ac ]/K a (7.6a) or [Ac ]= Ac TK a [H + ]+K a. (7.6b) Furthermore, we know that K w =[H + ][OH ]. (7.7) For the system HAc and water, we have four unknowns: [HAc], [Ac ], [H + ], and [OH ]. We wish to plot the unknowns versus ph, or more precisely, plot log X versus ph, where X is one of the unknown concentrations. From equations 7.5b, 7.6b, and 7.7, we can do this plotting log X on the ordinate versus ph on the abscissa. It is apparent from Eqs. 7.5b and 7.6b that we need to know the values of Ac T and K a to draw the diagram. Figure 7.2 shows the pc-ph diagram for acetic acid (pk a = 4.76) at Ac T = M (pac T = 2.0). Instructions on how the lines are developed are as follows. (1) [H + ]. By definition, ph = log[h + ] since we are ignoring differences between activity and concentration. This defines a line with a slope of 1 and an intercept of ph = 0: At ph = 0, pc = 0; that is, [H + ] = 1.0. pc = log[h + ]=ph; (7.8)

15 24 WATER CHEMISTRY ph [HAc] system point [Ac ] 4 [OH ] pc = logc 6 8 [Ac ] (A) [H + ] (B) [HAc] 10 Figure 7.2 pc-ph diagram for M acetic acid (2) [OH ]. From Eq. 7.7, log K w = 14.0 = log[h + ] + log[oh ], or pc = log[oh ] = 14 + log[h + ], or pc=14 ph, (7.9) which defines a line for log[oh ] with a slope of +1 and an intercept of log [OH ] = 0 when ph = 14. () [HAc]. When ph < pk a, that is, when [H + ] >> K a, the denominator in Eq. 7.20b approaches [H + ] and [HAc] approaches Ac T : or [HAc] =Ac T [H + ]/{[H + ]+K a } Ac T [H + ]/[H + ] Ac T p[hac] = log[hac] = pac T when ph < pk a (7.40) Thus, for ph < pk a, log[hac] is a straight line parallel to the ph axis at pc = log[hac] =2.0. If ph > pk a,[h + ] << K a, and Eq. 7.5b becomes or [HAc] =Ac T [H + ]/{[H + ]+K a } Ac T [H + ]/K a, p[hac] = log[hac] = logac T log[h + ]+logk a, p[hac] = log[hac] = pac T pk a +ph. (7.41)

16 SOLVING IONIC EQUILIBRIA PROBLEMS 25 Equation 7.41 defines a line with a slope of 1 for log[hac] in the region ph > pk a. It can be constructed from the slope by assuming that it intersects the line log[hac] = pac T = 2.0 (Eq. 7.40) at ph = pk a. (4) [Ac ]. Similarly, Eq. 7.6b can be represented by two asymptotes. If ph > pk a, then K a >> [H + ], and or [Ac ] Ac T K a /K a = Ac T, p[ac ] = log[ac ] = pac T ; (7.42) that is, the line is parallel to the ph axis at pc = 2.0. If ph < pk a, then [H + ] >> K a, and or [Ac ] Ac T K a /[H + ], or p[ac ]= log[ac ]= logac T logk a + log[h + ], p[ac ] = log[ac ] = pac T +pk a ph, (7.4) which has a slope of +1 and intersects the line p[ac ] = pac T (Eq. 7.42) at ph = pk a. Finally, it is clear from the nature of Eqs. 7.5b and 7.6b that both denominator terms must be taken into account when the ph is close to pk a and that the lines are curved in this region.at ph = pk a,p[hac]=p[ac ]=p(ac T /2) = pac T log 2 = pac T 0.. The exact curves for [HAc] and [Ac ] thus intersect at this point. The pc-ph diagram thus should be completed by hand drawing curved lines through this point and connecting asymptotically to the horizontal lines for p[hac] and p[ac ] = pac T and the sloping lines for p[hac] and p[ac ]. The size of the curved region for each species should be approximately one log unit above and below ph = pk a. This can be seen from the Henderson-Hasselbach equation: ph = pk a + log [X ] [HX] (7.44) When the ph is one unit greater than pk a (i.e., at ph 5.76 for acetic acid), log[ac ]/[HAc] = 1.0. Thus, the ratio [Ac ]/[HAc] = 10, and [Ac ] approaches Ac T. Conversely, when the ph is one unit less than pk a (i.e., ph.76 for acetic acid), log[ac ]/[HAc] = 1.0. Thus, the ratio [Ac ]/[HAc] = 0.10, and [HAc] approaches Ac T. 7.. Simple approach to drawing pc-ph diagrams Although it is useful to derive equations at least once to understand the basis for the diagrams, it is not necessary to go through all the algebraic manipulations described in the preceding section to draw a pc-ph diagram for any acid-base system. Once you have this understanding, the only information you need to draw a pc-ph diagram is

17 26 WATER CHEMISTRY Table 7.2 Step-by-step instructions for drawing ph-pc diagrams. 1. Draw and label the axes as a fourth quadrant graph with x-axis at top of graph labeled ph and the y-axis extended downward labeled log C or pc. 2. Draw identity lines for ph = log[h + ] and poh = log[oh ]. For H +, draw straight line with slope 1 intercepting pc = 0atpH= 0; for OH, draw a straight line with slope +1 intercepting pc = 0 at ph = 14.. Draw (lightly) a horizontal line parallel to and across the range of the ph axis at pc = log X T. 4. Insert a system point at pc = px T and ph = pk a. 5. Insert another point 0. log units below the system point. 6. For HX, at ph < pk a, p[hx] = px T ; this is the line drawn in #; label it [HX]. At ph > pk a, p[hx] is a straight line with slope = 1 that intersects the system point. Draw the line lightly and label it. 7. For X,k at ph > pk a, p[x ] = px T ; this is the line drawn in #; label it [X ]. At ph < pk a, p[x ] is a straight line with slope =+1 that intersects the system point. Draw this line lightly and label it. 8. For ph = pk a ±1, lines for p[hx] and p[x ] are curved. They intersect at ph = pk a and the point 0. log units below the system point. The line for p[hx] approaches px T at ph pk a 1.0. The line for p[x ] approaches px T at ph pk a Sketch these curved lines to connect asymptotically with the horizontal and sloping lines for p[hx] and p[x ] and darken all lines. 9. For polyprotic acids, repeat steps 4 8 for pk 2,.... Slopes for the acid-base species change across the ph range as follows. For diprotic acids: p[h 2 X] has a slope of 2 in the ph region above pk 2, and p[x 2 ] has a slope of +2 in the ph region below pk 1. For triprotic acids, p[h X] has a slope of for ph > pk, and p[x ] has a slope of + for ph < pk 1. Similar changes in slope apply to higher order polyprotic acids. (1) the pk a of the acid (or the pk a values for polyprotic acids), (2) the total analytical concentration (X T ) of the acid, and () the simple instructions given in Table 7.2. Alternatively, it is easy to use a spreadsheet program to generate the pc-ph diagram. This is accomplished by first making a column of ph values (e.g., from 0 to 14). In the next column of cells, generate the corresponding [H + ] concentrations (e.g., [H + ] = 10 ph or in spreadsheet notation, = 10ˆ-A1, where A1 is cell you are referencing). Subsequent columns for [HAc], [Ac-], and [OH ] are generated using Eqs. 7.5b, 7.6b, and 7.7. In spreadsheet notation with Ac T = 10 M, [HAc] would be =((10ˆ- ) B1)/(B1+ 10ˆ-4.767), where cell B1 contains the [H + ] concentration based on the ph in cell A1. Dragging these formulas down for all ph values will generate the concentrations at each ph value. The logarithms of these values are needed to generated the pc-ph diagram, and these can be obtained using the log function in the spreadsheet program Use of pc-ph diagrams to solve acid-base equilibrium problems pc-ph diagrams like that in Figure 7.2 are useful not only for gaining a quick visual perspective of the equilibrium concentrations of acid and base species as a function of ph but also for quantitatively solving simple to moderately complicated

18 SOLVING IONIC EQUILIBRIA PROBLEMS 27 acid-base equilibrium problems. We illustrate this problem-solving capability with Examples , the first two of which rely on the acetic acid/acetate diagram in Figure 7.2. In some cases (e.g., Examples 7.4 and 7.5), equilibrium conditions can be inferred directly from the stoichiometry of the reaction. In all the examples, we write and analyze the proton-balance expression. In most cases (and in all cases we shall analyze), the proton-balance expression can be simplified to only two important species (one on each side of the PPC = PPR expression, and this tells us that the equilibrium condition (e.g., the equilibrium ph) can be found on the pc-ph diagram where the lines for these two species intersect (i.e., where their concentrations are equal). EXAMPLE 7.4 Dissolution of M acetic acid in pure water: The equilibrium reaction for this case is HAc + H 2 O H O + + Ac or HAc H + + Ac. From stoichiometric considerations, we see that [H + ] must equal [Ac ] at equilibrium, and therefore the equilibrium ph must occur where the lines for [H + ] and [Ac ] intersect. The proton-balance expression is PPC = PPR [H + ]=[Ac ]+[OH ]. From the pc-ph diagram in Figure 7.2, we see that [OH ] is negligibly small compared with [Ac ] except at high ph (>> 7). Because we are adding a weak acid to pure water, we know that the equilibrium ph must be < 7.0. Consequently, we can ignore the term for [OH ] in the proton balance, and we find the equilibrium condition at [H + ] [Ac ], which agrees with the conclusion from the stoichiometry of the reaction expression. This condition occurs at the intersection of the lines for [H + ] and [Ac ], which occurs at ph =.4 (point (A) in Figure 7.2). EXAMPLE 7.5 Dissolution of M sodium acetate in pure water: The reaction is Ac + H 2 O HAc + OH. equilibrium Thus, from stoichiometry, we see that at equilibrium [HAc] = [OH ] (assuming that the above reaction is the only significant source of both HAc and OH ). The proton-balance expression is PPC = PPR, [H + ]+[HAc] =[OH ]. From Figure 7.2, we see that the line for [H + ] is much lower than the line for [HAc], except at low ph, which we would not expect to be the equilibrium situation since we are adding a weak base to pure water. Thus, [H + ] can be assumed to be negligible in the proton balance, and the equilibrium condition is defined by [HAc] =[OH ], that is,

19 28 WATER CHEMISTRY the intersection of these lines on the diagram. This occurs at ph 8.4 (point (B) in Figure 7.2). EXAMPLE 7.6 Dissolution of M sodium bicarbonate (NaHCO ) in pure water: To solve this problem we drew the pc-ph diagram for the carbonic acid (H 2 CO )- bicarbonate-carbonate system (pk 1 = 6.5, pk 2 = 10.) for C T = M (Figure 7.) using the procedures described in Table 7.1. Note the change in slope for p[h 2 CO ]atph> pk a2 and p[co 2 ]atph< pk a1. The proton-balance expression is (see Example 7.2) PPC = PPR, [H 2 CO ]+[H+ ]=[CO 2 ]+[OH ]. From the pc-ph diagram (Figure 7.), we see that [H 2 CO ] >> [H+ ] and [CO 2 ] > [OH ] in the circumneutral ph range where we expect the equilibrium condition to occur. The reader may ask, How can we make this inference? The answer is that we know that we have added to pure water a substance (HCO ) that acts as both a very weak acid and a weak base; that is, it can release a proton becoming CO 2 and it also can accept a proton becoming H 2 CO. Thus, we do not expect the solution to be highly acidic or highly basic, and the equilibrium ph should not be very high or very low. Given the above inequalities, we conclude that the equilibrium condition is defined approximately as [H 2 CO ] = [CO2 ], and the ph is defined by the intersection of these lines (ph = 8.; point (A) in Figure 7.). We note that [OH ]is not totally negligible compared with [CO 2 ]; careful analysis of the figure will show that it is about one-fifth the concentration of [CO 2 ]. Consequently, use of the term is defined approximately is appropriate in the previous sentence. 0 ph [H + ] [OH ] 2 [H 2 CO *] [HCO ] [CO 2 ] 4 pc = logc Figure 7. pc-ph diagram for the carbonate system at C T = M. Point (A) gives the ph if NaHCO is added to water 12 and point (B) gives the ph if Na 2 CO is added to water as described in Examples 7.6 and [HCO ] [CO 2 ] (A) (B) [H 2 CO *]

20 SOLVING IONIC EQUILIBRIA PROBLEMS 29 EXAMPLE 7.7 Dissolution of M sodium carbonate (Na 2 CO ) in pure water: We can use the same diagram (Figure 7.) to solve this problem. Here the proton balance is defined as PPC = PPR, 2[H 2 CO ]+[H+ ]+[HCO ]=[OH ]. Note that the factor of 2 multiplying the concentration of [H 2 CO ] arises because for every molecule of H 2 CO formed from the starting species (CO2 ), two H+ are consumed. From the pc-ph diagram we see that both [H 2 CO ] and [H+ ] are small compared with [HCO ]. Therefore, the proton balance simplifies to [HCO ] [OH ], and the equilibrium ph is defined by the intersection of the lines for these species: ph 10.8 (point (B) in Figure 7.). EXAMPLE 7.8 Dissolution of equimolar NH 4 Cl and NaHCO (both at M ) in water: Because Na + and Cl constitute a neutral salt, the additions effectively represent addition of NH 4 HCO (ammonium bicarbonate) to water. To solve this diagram we need to superimpose the pc-ph diagrams for the carbonate system (as in Figure 7.) and the ammonium-ammonia system, where both C T and N T = M, on the same graph, as shown in Figure 7.4. The proton balance is the sum of the proton balances for the two acid-base substances (NH + 4 and HCO ): PPC = PPR, [H 2 CO ]+[H+ ]=[CO 2 ]+[NH ]+[OH ]. From the combined pc-ph diagram, we see that [H 2 CO ] >> [H+ ], and [NH ] >> [CO 2 ] >> [OH ] for almost the entire ph range shown in the figure. Therefore, 0 ph [NH 4 + ] [NH ] [HCO ] pc = logc 4 6 A [CO 2 ] [NH + 4 ] [H 2 CO ] [OH ] [H + ] Figure 7.4 pc-ph diagram for ammonium bicarbonate at C T = M. Point (A) gives the ph if equimolar NH 4 Cl and NaHCO are added to water as described in Example 7.8.

21 240 WATER CHEMISTRY [H 2 CO ] [NH ], and the equilibrium ph occurs where the lines for these species intersect (ph = 7.8, point A in Figure 7.4) pc-ph diagrams for metal hydroxide solubility In addition to displaying conventional acid-base equilibria, pc-ph diagrams can be used to display the effects of ph on some metal hydroxides like Al(OH) and Fe(OH). Construction of such diagrams, although similar to the approach used for the diagrams described in the previous section, differs in several important respects. A brief summary of the general procedure for drawing such diagrams is as follows. (1) First, identify all the soluble forms of the metal ion that can exist in equilibrium with the solid metal hydroxide. These include the free metal ion and various metal-hydroxide (M-OH) complexes. (2) Write an equilibrium equation relating (as a function of ph) the concentration (or activity) of each soluble metal form to the solid phase metal hydroxide, and draw the lines representing these equations on a pc-ph diagram. () Sum up the concentrations of the various soluble species at a given ph (X T,M for M, hereafter referred to as M T ), and draw the line representing M T versus ph. Figure 7.5 shows the pc-ph solubility diagram for amorphous ferric hydroxide (Fe(OH) am, ferrihydrite) for an assumed maximum M T = 1 M. Example 7.9 lists the equilibrium expressions for the soluble Fe III species and describes how they are used to define lines for the species. Step sounds like a formidable task, but it usually is easy to do because only one metal ion species is important in the mass-balance expression for M (M T ) over wide 0 0 ph A Figure 7.5 Solubility diagram (pc-ph) for ferrihydrite (amorphous ferric hydroxide) in water. The shaded area is the region of supersaturation, where solid Fe(OH) will precipitate spontaneously. See Example 7.9. pc = logc [Fe(OH) aq ] [Fe + ] Fe T aq [Fe(OH) 4 ] [FeOH 2+ ] [Fe(OH) 2 + ]

22 SOLVING IONIC EQUILIBRIA PROBLEMS 241 ranges of ph, and it is unusual for more than two species to contribute significantly to M T. When two species are important, the largest impact on M T occurs where the lines intersect. At this point, M T is 0. log unit greater than the value of either species at that ph; that is, pm T is 0. units less than the value of pm at the intersection point. When the ph is more than one unit higher or lower than the ph at the intersection, the line for pm T approaches the pm line for the more soluble species at that ph. Where three species are approximately coequal in concentration, as in point A in Figure 7.5, where p[fe + ] p[feoh 2+ ] p[fe(oh) + 2 ], the line for pfe T is three times higher than the point of intersection, or 0.5 log units higher. Consequently, the line for pm T can be sketched on the diagram easily once the lines for the individual species are drawn. EXAMPLE 7.9 Derivation of the lines for Fe species in Figure 7.5: We need equilibrium relationships for the following Fe III species with solid Fe(OH) am (ferrihydrite): Fe +, FeOH 2+, Fe(OH) + 2, Fe(OH)0, and Fe(OH) 4. There are several other soluble FeIII species, such as Fe 2 (OH) 4+ 2, but they generally do not contribute significantly to Fe T. In the interests of simplicity, we will not consider them here in drawing the solubility diagram. Also, for simplicity we will not consider effects of ionic strength on activity and we equate activity to molar concentration. For Fe +, the simple solubility product is Thus, or or Fe(OH) am = Fe + + OH ; K s0 = ; log K s0 = K s0 =[Fe + ][OH ], log K s0 = 8.81 = log[fe + ]+ log[oh ]=log[fe + ] poh, pfe + =8.81 (14.00 ph) =.19 + ph. The line of pfe + thus has a slope of and crosses the ph-axis (where pfe + = 0) at ph =.19/, or ph For FeOH 2+ we cannot find a direct equilibrium expression with Fe(OH) am, but we can obtain it from K s0 and K 1, the first acidity constant of Fe + : Fe(OH) = Fe + + OH ; K s0 = Fe + + H 2 O = FeOH 2+ + H + ; K 1 = H + + OH = H 2 O; K 1 w = Net: Fe(OH) am = FeOH OH ; K = K s0 K 1 K 1 w = =

23 242 WATER CHEMISTRY Thus, we can write log[feoh 2+ ]+2log[OH ]=log K = 27.00, or pfeoh 2+ = (pK w ph) = pH, or pfeoh 2+ =2pH 1. The line for pfeoh 2+ thus has a slope of 2 and crosses the ph-axis at ph = 0.5 Similarly, for Fe(OH) + 2, we can obtain an equilibrium expression with Fe(OH) am as follows: Fe(OH) = Fe + + OH ; K s0 = Fe + + 2H 2 O = Fe(OH) H+ ; K 2 = H + + 2OH = 2H 2 O; K 2 w = Net: Fe(OH) am = Fe(OH) OH ; K = K s0 K 2 K 2 w = = We can write log[fe(oh) + 2 ]+log[oh ]=log K = or pfe(oh) + 2 =15.4 (pk w ph) = 1.4+pH. The line for pfe(oh) + 2 ph = 1.4. thus has a slope of 1 and crosses the ph-axis at

24 SOLVING IONIC EQUILIBRIA PROBLEMS 24 For the un-ionized soluble species Fe(OH) 0, a similar exercise yields the following relationship: Fe(OH) = Fe + + OH ; K s0 = Fe + + H 2 O = Fe(OH) 0 + H+ ; K = H + + OH = H 2 O; K w = Net: Fe(OH) am = Fe(OH) 0 ; K = K s0k K w = = We thus can write pfe(oh) 0 =pk = 9.7. Note that the line for this species is not ph dependent; instead, it is a horizontal line at pc = 9.7. Finally, for Fe(OH) 4, a similar exercise leads to Fe(OH) = Fe + + OH ; K s0 = Fe + + 4H 2 O = Fe(OH) 4 + 4H+ ; K 4 = H + + 4OH = 4H 2 O; K 4 w = Net: Fe(OH) am + OH = Fe(OH) 4 ; K = K s0k 4 K 4 w = = We can write log[fe(oh) 4 log[oh ]=logk = 4.4. or pfe(oh) 4 = (pk w ph) = 18.4 ph. The line for pfe(oh) 4 at pc = 4.4. has a slope of 1 and intersects with the right y-axis (at ph 14)

25 244 WATER CHEMISTRY and In summary, the equations for all the Fe lines in the diagram are pfe + =.19 + ph, pfeoh 2+ =2pH 1, pfe(oh) + 2 =1.4+pH, pfe(oh) 0 = 9.7, pfe(oh) 4 =18.4 ph. Finally, we note that the lines have meaning only where solid Fe(OH) am exists. Because we limited Fe T to 1 M (maximum pfe T = 0), the lines at ph < 1.2 (where we start to exceed the mass-balance constraint) are not meaningful. Diagrams like Figure 7.5 are useful in illustrating how the solubility of a metal hydroxide varies with ph, but they generally are not used to solve problems as are the pc-ph diagrams in Figures Interpretation of solubility diagrams is straightforward. Any combination of ph and pm T values that falls within the shaded envelope in Figure 7.5 represents supersaturated conditions with respect to the solubility of Fe(OH) am, and the precipitation reaction will proceed spontaneously (although the figure implies nothing about the rate at which that may occur). Any combination of ph and pm T values that falls outside the shaded envelope in Figure 7.5 represents undersaturated conditions with respect to the solubility of Fe(OH) am, and if solid phase Fe(OH) am is present, it will tend to dissolve. The gray line denoted Fe T in Figure 7.5 represents the equilibrium condition the total solubility of Fe(OH) am expressed in terms of Fe as a function of ph Predominance-area diagrams These diagrams involve master variables on both axes. Examples include plots of (1) log P CO2 versus ph for carbonate and hydroxide mineral solubility and (2) redox potential, E h versus ph to describe the effects of those variables on the speciation of an element. They are called predominance-area diagrams because the lines on such plots delimit regions in which a given species (ionic form or mineral phase) is predominant in the system. The lines themselves represent one of three possibilities the locations (combination of values of the master variables) at which (1) two mineral phases coexist at equilibrium; (2) a mineral (solid) phase coexists in equilibrium with some (arbitrarily defined) concentration of a free ion component of the solid; or () two different soluble species of the element coexist at the same concentration (or a specified ratio of concentrations). Construction of E h -ph diagrams is covered in Chapter 11; here we describe how to construct and interpret predominance-area diagrams for mineral solubility as controlled