Carbon is widely distributed in our

C H A P T E R WATER SOFTENING Carbon is widely distributed in our ecosystem through five major spheres, namely the lithosphere, hydrosphere, atmosphere, pedosphere, and biosphere (see Table -). Sediments and ocean water contain the greatest amounts of inorganic carbon. The carbonate in natural waters may have originated from mineral dissolution processes such as that of calcite or albite. For example, albite breaks down to form kaolinite, the most important alkaline species of the carbonate system. NaAlSi 3 O 6 (s) + CO + / H O = Na + + HCO 3 + H 4 SiO 4 + /Al Si O 5 (OH) 4 (s) Due to the dissolution process, certain dications such as Ca +, Mg +, Si +, Mn +, etc., often cause hardness in water. The major anions associated with them are HCO 3, SO 4, SiO 3, Cl, NO 3, etc. Boiler scale is a good example, resulting from the decomposition of bicarbonate in steam lines: 4

4 Chapter WATER SOFTENING Ca + + HCO 3 = CaCO 3 + H O + CO Table -. Carbon in Major Global Cycles Sediments Total on Earth (0 8 mol C) Carbonate 5,00 94,000 Organic carbon,000 8,500 Land Organic carbon 0..8 Ocean CO + H CO 3 0.09 0.35 HCO 3.9 54 CO 3 Total on Earth (units of atmospheric CO, A 0 ) 0.36 7 Dead organic 0.4 7 Living organic 0.0007 0.0 Atmospheric CO (A 0 ) 0.0540.0 Representative C r (M) a Alkalinity (eq lilter ) b Concentrations Seawater.3 0-3.5 0-3 River waters, average ~ 0-3 ~ 0-3 River waters, typical range 0-4 5 0-3 0-4 5 0-3 Groundwaters, typical range, 5 0-4 8 0-3 0-4 5 0-3 the United States Rainwater, typical range 0-5 5 0-5 0 4 0-5 c Atmospheric CO = 0.033% by volume in dry air: preindustrial 4 p CO = 3.3 0 atm 4 p CO =.9 0 atm ; a Cr = [CO (aq)] + [H CO 3 ] + [HCO 3 ] + [CO 3 ] (Oceanographers often use the symbol ΣCO ) b Alkalinity, the acid-neutralizing capacity, is expressed as mol protons per liter, or equivalents per liter. c Some rainwaters contain mineral acidity of up to 0-3 eq liter -. Source: Stumm and Morgan Atmospheric CO will interact and dissolve in almost any kind of water to establish equilibrium. In this chapter we will first review the carbonate system

. Carbonate System 43 and explain its common equilibrium constants. Then we will proceed to discuss examples of carbonate species in open and closed systems. Next, we will discuss the system points and equivalent points in relation to the different endpoints of acidic or alkaline titrations. One of the most economic processes used for water softening is the addition of lime and soda ash, which we have illustrated with typical examples. It is important to understand the carbonate system in gas-solution-solid equilibrium situations. For example, in the acid titration of carbonates, equilibrium can be either closed or open to the atmosphere, solution or gassolution type. In a natural water system, the hypolimnion of a lake is considered to be a solution, a solid equilibrium, since the top layer, the epilimnion, can act as a barrier to the atmosphere. In a manmade system, such as the oxidation pond of an activated sludge plant, the pond is considered open to the atmosphere but closed to the solid, because aeration occurs so vigorously within the pond by air or oxygen that any CO from carbohydrate degradation and the accompanying nitrification process will escape from the surface. On the other hand, the lime-soda ash softening process is treated as closed to the atmosphere but open to the solid precipitation. Because the ratio of the surface area to the volume of the treatment basin is small, the solid both added and formed is abundant and the contact time is short. In the lime-soda ash process, if excess soda ash is used, recarbonation will be necessary after the first stage; split treatment will be required if the excess lime is consumed by CO or the remaining portion of raw water from the split stream. Recarbonation is also often used for the removal of H S from stagnant water. The primary function of CO is to lower the ph, and consequently the dissolved sulfur species will be greatly decreased (e.g., at ph 5 the total undissolved species of sulfide is only %).. CARBONATE SYSTEM Carbonate occurs in natural waters due to the dissolution of atmospheric carbon dioxide as well as lithospheric mineral deposits of carbonate salts. In aquatic chemistry, [ ] is used to indicate concentration, and { } is used to indicate chemical activity. In this book, we use ( ) to indicate concentration, except in a few cases. In general, the governing equilibrium reactions and their constants are as follows:

44 Chapter WATER SOFTENING. Henry s Law Constant pk H =.5 CO (g) = CO (aq) [-] Equation [-] can have an equilibrium constant given by K D = ( CO) aq ( ) g CO [-] Using Dalton s law, PCO ( ) g = CO RT [-3] Thus, combining the above two expressions (Eqs. [-] and [-3]), K RT D ( CO ) aq = PCO = K H PCO [-4] where K H, the Henry s law constant, has a dimension measured as mole/l-atm. ( ) K D CO aq K H = = [-5] RT P K H is also related to the Bunsen absorption coefficient, α B, by CO (.44) α B = K H [-6] which has a dimension of atm. The variable P CO is the partial pressure of CO, which is related to the volume fraction X CO, the total pressure P T, and the water vapor pressure W by Usually the value is: and usually ( P W ) PCO = X T [-7] CO P = 0 5 atm CO 3..5 3.5 5 ( ) K P = 0 0 = 0 M CO = H CO aq [-8]

. Carbonate System 45 More discussion of Henry s law can be found in Chapter of Aeration and Stripping.. Hydration Constant Thus, pk h =.8 CO (aq) + H O = H CO 3 [-9] K h = ( H CO3) ( ) CO aq =.6 0 which means that (H CO 3 ) is only 0.6% of (CO ) aq. When (CO ) aq is in water, then (CO ) aq >> (H CO 3 ) and the majority of the species is the CO (aq) form. It is customary to combine the two forms into a new form: Since the (H CO 3 ) content is very low, 3. Pseudo First Ionization Constant 3 (H CO 3 ) + (CO ) aq = (H CO 3 * ) [-0] (H CO 3 * ) (CO ) aq [-0a] pk = 3.5 H CO 3 = H + + HCO 3 + ( H )( HCO ) ( ) H CO 3 3 = K [-] 4. First Ionization Constant pk = 6.3 H CO 3 * = H + + HCO 3

46 Chapter WATER SOFTENING + ( H )( HCO 3 ) = K * ( H CO ) 3 [-] It can be easily seen that To verify this, K can be written as 3.5.8 6.3 K K K h = 0 0 = 0 [-3] ( H )( HCO ) + ( H )( HCO ) 3 + 3 ( H CO3) K = = 3 CO aq + K h K = [-3a] ( H CO ) + ( ) + K h Since K h = /0.8 = 63, K h >>, then K K K h, and Equation [-3] is valid. 5. Second Ionization Constant pk = 0.3 HCO 3 = H + + CO 3 + ( H )( CO 3 ) = K ( HCO ) 3 [-4] 6. Solubility Product pk s = 8.3 CaCO 3(s) = Ca + +CO 3 K s = (Ca + )(CO 3 ) [-5] Let us define the total concentration of dissolved carbonate species as: C T = (H CO 3 * ) + (HCO 3 ) + (CO 3 ) [-6] We can also write the interrelation between any two carbonate species of the three species from Equation [-]:

. Carbonate System 47 (HCO 3 ) = K (H + ) (H CO * 3 ) [-] and from Equation [-4]: (CO 3 ) = K (H + ) (HCO 3 )= K K (H + ) (H CO * 3 ) [-7] If we define a distribution coefficient, α, for the three carbonate species as α 0 = (H CO 3 * )/C T α = (HCO 3 - )/C T α = (CO 3 )/C T [-8] and α 0 + α + α = [-9] then α 0 = [ + K (H + ) + K K (H + ) ] [-9a] α = [K (H + ) + + K (H + ) ] [-9b] α = [(K K ) (H + ) + K (H + ) + ] [-9c] A diagram can be constructed (Fig. -) to have the same general format as the log C-pH diagram used widely in aquatic chemistry. N.B. But C T (H + ) + K (H + ) + K K. Another way to express this is: α 0 = (H + ) /[(H + ) + k (H + ) + k k ] α = K (H + ) /[(H + ) + k (H + ) + k k ] α = K K /[(H + ) + k (H + ) + k k ] [-9d] [-9e] [-9f]

48 Chapter WATER SOFTENING Figure -. Ionization fractions of the carbonate species as a function of ph: α 0 = (H CO 3 )/CT; α = (HCO 3 )/CT; α = (CO 3 )/CT. [Example -] Construct the diagram of log α versus ph for the carbonate species as Figure -. Identify the slopes of each curve: α 0, α, and α. Let us determine the following three regions: Region ph < pk < pk Region pk < ph < pk Region 3 pk < pk < ph

. Carbonate System 49 for each α curve. We have nine zones to work out; assumptions and approximations can be made (see Table - for calculation details). Table -. Estimation of Ionization Fraction of Carbonate Species at Three Different Regions of ph Region ph < pk < pk Region pk < ph < pk Region 3 pk < pk < ph α 0 / (K /(H + )) (H + )/K (H + ) /K K Slope 0 α / ((H + )/K ) K /(H + ) (H + )/K Slope + 0 α / ((H + ) /K K ) K K /(H + ) /((H + )/ K ) K /(H + ) slope + + 0 Table -3 lists the essential carbonate equilibrium and solubility in terms of various constants, K, and always pk = log K. It is useful to recall that reactions add and that constants multiply. Consequently, when reactions subtract, then constants divide. These constants usually follow Arrhenius-type temperature dependence (for equilibrium constants at various temperatures, see Physical Chemistry, Environmental Chemistry in this series). The salinity corrections for the activity coefficients are also listed in Table -3. Since the K constants are computed based on thermodynamic principles, the operational K constants (usually referred to as c K), based on nonideal cases, must correlate with the activity coefficients. For example, pc = pa + log γ [-0] Since a = γ C and log C = log a + logγ, the C is related to operational constants, and a is related to thermodynamic principles. Therefore, aμ p c K = pk [-] +bμ

50 Chapter WATER SOFTENING Table -3. CaCO 3 Solubility and Carbonate Equlibria in Terms of Dissociation Constants, pk = ( logk ) Temperature, C Reaction a 5 0 5 0 5 40 60 CO(S) + HO = HCO3*; K.0.7.34.4.47.64.8 HCO3 = HCO3 - + H + ; K 6.5 6.46 6.4 6.38 6.35 6.30 6.30 HCO3 - = CO3 - + H + ; K 0.56 0.49 0.43 0.38 0.33 0. 0.4 CaCO3(S) = Ca + + CO3 - ; Ks 8.09 8.5 8. 8.8 8.34 8.5 8.74 CaCO3(S) + H + = Ca ++ + HCO3 - ; (Ks/K) -.47 -.34 -. -.0 -.99 -.7 -.40 CaCO3(S) + HCO3* = Ca + + HCO3 - ; (KsK/K) 4.05 4. 4. 4.8 4.36 4.59 4.90 HO = H + + OH - ; Kw 4.73 4.54 4.35 4.7 4.00 3.54 3.0 Salinity Corrections for Equilibrium Constants b pk = pk (µ).05 /[+.4(µ).05 ] pk = pk (µ).05 /[+.4(µ).05 ] PKs = pks 4 (µ).05 /[+ 3.9(µ).05 ] Log Ks /K = log Ks/K +.5(µ).05 /[+ 5.3(µ).05 + 5.5(µ).05 ] Estimation of µ µ.5 x 0-5 Sd where Sd is the total dissolved solids content of the water, or µ 4H T, where H = total hardness in moles/l; and T = alkalinity in eq/l (0 3 me/l) a H CO3 refers to the sum of dissolved CO and HCO3. Because [HCO3] << [CO], [CO] is essentially equal to [HCO3*]. b K and K are thermodynamic and operational constants, respectively. The operational constants can be used with mass law expressions containing concentration terms, with the exception of H +, which is always expressed in activities.

. Carbonate Concentration Diagrams 5. CARBONATE CONCENTRATION DIAGRAMS In the following section, we will discuss reversed carbonate equilibrium as well as carbonate concentration diagrams... A Closed System of Dissolved Carbonates In an aqueous system, pk w = 4 H O = H + + OH or For material balance, again, K w = (H + )(OH ) (OH ) = K w (H + ) [-] C T = (H CO 3 * ) + (HCO 3 ) + (CO 3 ) [-3] The electroneutrality of a carbonate solution is: (H + ) = (OH ) + (HCO 3 ) + (CO 3 ) [-4] In the presence of a base or an acid, this will become: (H + ) + (C B ) = (C A ) + (OH ) + (HCO 3 ) + (CO 3 ) [-5] Usually the three equilibrium concentrations of a carbonate system can be calculated by simultaneously solving three equations involving K constants. [Example -] Determine the operational component concentration of the carbonate species of a 3 0 3 molar carbonate solution where the ph is 7.5, based on thermodynamic considerations.

5 Chapter WATER SOFTENING Thermodynamically derived equilibrium constants can be correlated with the operationally concentration-based constants by an activity coefficient: K = γ c K [-6] K and K can be derived based on free energy. For example, H CO 3(aq) = H + (aq) + HCO 3 (aq) 49 0 40.3 ΔG 0 98 (cal/mol) ΔG 0 = 40.3 + 0 ( 49.0) = 8.69 kcal ln K = ΔG/RT = 4.65 K = 4.36 0 7 and pk = 6.36 Similarly, K = 5.0 0 and pk = 0.30. Since the individual concentration is unknown, ionic strength is difficult to evaluate (although approximates can be made), and we can calculate the concentrations by assessing as follows: (Case ), γ = K = (H + )(HCO 3 )/(H CO 3 * ) = 4.35 0 7 K = (H + )(CO 3 )/(HCO 3 ) = 5.0 0 K w = (0 7.5 )(OH ) = 0 4 C T = (H CO 3 * ) + (HCO 3 ) + (CO 3 ) = 3 0 3

. Carbonate Concentration Diagrams 53 Solving, (HCO 3 ) =.80 0 3 M (H CO 3 * ) =.97 0 4 M (CO 3 ) = 4.4 0 6 M (Case ) μ =.4 0 3 (OH ) = 3.6 0 7 M Based on HCO 3 and two other species and from the Davies equation, the activity coefficients can be determined. γ = HCO 3 0.96 γ = CO 3 0.85 γ = 0.96 OH We can now use c K for the evaluation of operational constants and a more accurate component concentration: + ( H )( HCO3 )( 0.96) * ( H CO ) K = = 4.35 0 + ( H )( CO )( 0.85) 3 3 ( HCO )( 0.96) K = = 5.0 0 K w = 3 7 7.5 4 ( 0 )( OH )( 0.96) = 0

54 Chapter WATER SOFTENING The C T equation is the same as for Case. The solution becomes: (HCO 3 ) =.8 0 3 M (H CO 3 * ) =.89 0 4 M (CO 3 ) = 5.04 0 6 M The values for both Case and Case are very close... Dissolved Carbonates in an Open System of Atmospheric Carbon Dioxide This system is exemplified by Figure -, in which six lines are formed when pc versus ph is plotted. We will show how each line is constructed. Here Henry's law becomes important: (CO ) aq = K H P CO = 0.5 0 3.5 = 0 5 (H CO * 3 ) = C T α 0 (from Eq. [-8]) For the above relation, the normal aqueous carbonate system open to the atmosphere contains P CO = 0 3.5 atm. Therefore, This is line in Figure - with a slope = 0. log (H CO 3 * ) = 5 [-7]

. Carbonate Concentration Diagrams 55 Figure -. Total concentration of carbonates, C T.CO3, versus ph for an open system at 5 C. (Modified from Ref. -.) Also, log (CO ) aq = 5 from the hydration constant, Thus, ( CO) aq ( ) H CO 3 = K h log (CO ) aq log (H CO 3 ) = pk h or log (H CO 3 ) = 5 +.8 = 7.8 [-8] which is line in Figure -.

56 Chapter WATER SOFTENING For (HCO 3 ) species (using Eqs. [-8], [-0a], [-8]), Therefore, α + ( HCO 3 ) = K H PCO = K ( ) H K H PCO α 0 log (HCO 3 ) = pk log K H log P CO ph =.3 ph [-9] This is line 3 for (HCO 3 ) with a slope of +. Now for the (H + ) line, line 4: log (H + ) = ph The intersection of line 3 and line 4 or the equivalent concentration point (point P) is: log (H + ) = log (HCO 3 ) = 5.65 [-30] which can be obtained by the simultaneous solution of y = x y =.3 x where y = log (H + ) and x = ph. The (OH - ) line, line 5, is then log (OH ) = pk w + ph = 4 + ph [-3] which also has a slope of +. Finally, the (CO 3 ) can be expressed by: α + ( CO 3 ) = K H PCO = K ( ) K H K H PCO α 0

. Carbonate Concentration Diagrams 57 Thus, log (CO 3 ) =.6 ph [-3] which has a slope of + and is represented by line 6...3 Dissolved Carbonates in Contact with Solid Calcium Carbonates (Open to Solid Minerals) The dissolution of CaCO 3 (s) in water assumes that the Ca + dissolved must be equal in concentration to the dissolved carbonate species. In other words, (Ca + ) = C T [-33] Furthermore, the electroneutrality condition must be: (Ca + ) + (H + ) = (HCO 3 ) + (CO 3 ) + (OH ) [-34] Our problem is to calculate the concentrations of all possible species in the water, which include Ca +, H CO 3 *, HCO 3, CO 3, H +, and OH. This will require the solution of six equations for the six unknowns. We already have two equations (Eqs. [-33] and [-34]). We can add the first and second ionization constants for the carbonic acid, the ion product of water, and the solubility product of calcium carbonate (K, K, K w, and K s ). In solving the equations, it is easy to start with the solubility product. (Ca + ) = K s (CO 3 ) Using the relations of ionization fractions, (CO 3 ) = α C T then (HCO 3 ) = α C T (Ca + ) = K s /α C T = C T = (K s /α ) / [-35]

58 Chapter WATER SOFTENING After substitution into the neutrality equation (Eq. [-34]), (K s /α) / ( α α ) + (H + ) K w (H + ) = 0 [-36] This equation can be solved by trial and error. The results are ph = 9.95, (Ca + ) =.7 0 4 M, (HCO 3 ) = 8.9 0 5 M, (CO 3 ) = 3.8 0 5 M, (H CO 3 * ) =.5 0 8 M, alkalinity =.54 0 4 eq/l, poh = 4.05. If the ph is adjusted with a strong acid or base, the neutrality condition becomes similar to Equation [-4] (C B ) + (Ca + ) + (H + ) = (C A ) + (HCO 3 ) + (CO 3 ) + (OH ) [-37] Here, for example, if the strong base were NaOH, (C B ) = (Na + ) and if the strong acid were HCl, then (C A ) = (Cl ). A logarithmic concentration versus ph plot can be constructed as shown in Figure -3. Some of the pertinent elements and principles are listed below. In order to see how these log C versus ph diagrams are constructed (e.g., the Ca + line or the C T line), the ph again can be separated into three regions. st region: ph < pk < pk nd region: pk < ph < pk 3rd region: pk < pk < ph From Equation [-35], (Ca + ) = (K s /α ) / log (Ca + ) = / log K s / log α = / log K s + / log [ + K (H + ) + K K (H + ) ] Using Table - as for Example -, log (Ca + ) / log K s + / log [K K (H + ) ] / pk s ph + / pk + / pk 4.5 ph [-38]

. Carbonate Concentration Diagrams 59 and for the slope, d log dph + ( Ca ) = 0.5= [-39] Figure -3. Logarithmic concentration diagram for carbonate solution in contact with solid calcium carbonate. (Adapted from Ref..4.) In the nd region, log (Ca + ) / log K s + / log K (H + ) / pk s / ph + / pk / ph [-40] d log Ca dph + ( ) and = 0. 05 [-4]

60 Chapter WATER SOFTENING and For the 3rd region, log (Ca + ) / log K s + / log / pk s 4.5 [-4] + d log( Ca ) 0 dph [-43] Lines connecting the three regions are shown in Figure -3 as the Ca + line or C T line. log Similarly, for the (HCO 3 ) line, ( HCO ) 3 = α C For the first region, ( HCO ) T s = α K α + ( KK ) ( H ) + K ( H ) K = s + + ( H ) + + K ( H ) + + ( K K ) ( H ) + K ( H ) K + K s 3 log log K log s + log H K K + ( H ) + ( ) ( HCO ) + ph pk pk ph + pk. log 3 pk s [-44] and the slope of this line is zero. Others values can be similarly obtained and the diagrams completed. K.3 SYSTEM POINTS AND EQUIVALENCE POINTS In general, alkalinity is a measure of the compounds that when added to water will neutralize a strong acid. In characterizing natural water systems, alkalinity is referred to as acid-neutralizing capacity. Most natural waters contain dissolved

.3 System Points and Equivalence Points 6 carbonate species, and therefore, it is best to look more closely at the carbonate system. Figure -4. Logarithmic concentration diagram for 0-3 molar carbonate solution. (Adapted from Ref...) Part A, B are system points; Part C, D and E are equivalence points. In the logarithmic concentration of a 0 3 molar carbonate solution, such as in Figure -4, we first introduce the system points, A and B. At point A, At point B, H CO 3 * = HCO 3 + H + ph 6.5 [-45] HCO 3 = CO 3 + H +, ph 0 [-46] In this instance, at point A (H CO 3 * ) = (HCO 3 ), and ph = pk, C T = (H CO 3 * ) + (HCO 3 ). At point A, C = C T, but (H CO 3 * ) = (HCO 3 ), so C = C T / or C T /C = Therefore, the actual concentration of point A is different from the pk value by log = 0.3 or 0.3 unit less in log scale. Similarly, the system point B can be determined. Points C, D, and E are termed equivalence points, at which a strong acid or base is needed to achieve a reference equivalence point.

6 Chapter WATER SOFTENING At point C, H CO 3 * predominates: (H + ) = (HCO 3 ) + (CO 3 ) + (OH ) We consider here only the K equation: ( + H )( HCO3) = K ( ) H CO Assuming that (H + ) = (HCO 3 ) and that in this case C T = (H CO 3 * ) as an approximation, and * 3 (H + ) = (C T K ) / ph = / log C T + / pk = 4.5 [-47] Point C is often referred to as the total alkalinity (and mineral acidity) equivalence point. Next, at point D, the proton condition is: At ph >7, (H + ) < (OH ), and Substituting K, K w, and K, (H + ) + (H CO 3 * ) = (OH ) + (CO 3 ) (H CO 3 * ) = (OH ) + (CO 3 ) + ( )( HCO ) w + ( H ) ( HCO3 ) + ( H ) H 3 K K In this case, C T = (HCO 3 ), and therefore, K = + + K wk ( H ) = + ( K K ) For C T 0 3 M, the first term is dropped, and C T

.3 System Points and Equivalence Points 63 ph pk + pk = 8.3 [-48] Point D is usually referred to as the carbonate alkalinity or carbon dioxide acidity equivalence point. Of course, at point D, (H CO 3 * ) = (CO 3 ). Thus, or + ( )( HCO ) ( HCO3 ) = K + ( H ) H 3 K (H + ) = (K K ) / Similarly, point E has a proton situation as follows: (H CO * 3 ) + (HCO 3 ) + (H + ) = (OH ) At ph >> 7, (HCO * 3 ) and (H + ) are negligible. Therefore, and (HCO 3 ) = (OH ) C T = (CO 3 ) and (H + ) = (K K w /C T ) / ph = / pk + / pk w / log C T = 0.6 Point E is called the caustic alkalinity, total acidity equivalence point, or endpoint. In practice titration using phenolphthalein with an endpoint at ph 8.3 is considered for the equivalence of carbonate alkalinity. Further titration using methyl orange with an endpoint at ph 4.5 is considered for the total alkalinity. The titration of water samples containing various forms of alkalinity can be demonstrated as in Figure -5 (e.g., the case of hydroxide only, carbonate only, mixed hydroxide-carbonate, mixed carbonate-bicarbonate, and bicarbonate only).

64 Chapter WATER SOFTENING Figure -5. Relationship of the pc vs. ph diagram to the titration curve for the carbonate system at 5 C. [Example -3] Locate all of the equivalence points (x,y) of a carbonate system if the carbonate solution is 0 3 M. For point C, in the graph of pc versus ph (see Fig. -4),

.3 System Points and Equivalence Points 65 C T = (H CO 3 * ) (HCO 3 ) = α C T K = + ( αct )( H ) + = ( H ) α CT log α = pk + ph = 6.5 + 4.5 = log C = 3 = 5 C is (4.5, 5) For point D, C T = (HCO 3 ) K ( ) ( HCO3 ) CO =α CT = + ( H ) 3 = K CT H + ( ) log α = ph pk = 8.3 0.5 =. log C = 3. = 5. For point E, D is (8.3, 5.) C T = (CO 3 ) = K (HCO 3 )(H + ) = K α C T (H + )

66 Chapter WATER SOFTENING (H + ) = K α log α = pk ph = 0.5 0.6 = 0. log C = 3 0. = 3. E is (0.6, 3.) [Example -4] A recarbonated water having ph 9.5 is chosen for phenolphthalein and methyl orange titrations, where in both titrations /50 N of H SO 4 is used for a 00-mL water sample. The first titration consumes.0 ml, and subsequently, 4.0 ml of acid is consumed. Find the total carbonate and the bicarbonate alkalinities of the water sample. Carbonate alkalinity = (meq of acid/l to reach phenolphthalein endpoint) = (.0 ml 0.0 eq/l 000 meq/eq /00 ml sample) = 0.40 meq/l Total alkalinity = meq of acid/l to reach methyl orange endpoint = (4.0 +.0) ml 0.0 eq/l 000 meq/eq /00 ml sample =.6 meq/l Bicarbonate alkalinity = total alkalinity carbonate alkalinity =.6 0.40 =. meq/l

.3 System Points and Equivalence Points 67 Figure -6. The Deffeyes diagram for 5 C. (From K. S. Deffeyes, Limnol. Ocean. 0: 4, 965.)

68 Chapter WATER SOFTENING A well-known Deffeyes diagram is used for the correlation among alkalinity, C T, and ph. Using the diagram, if two of the three properties are known, then the third property can be evaluated (Fig. -6). Also the removal of CO can cause a horizontal shift; the addition or removal of bicarbonate causes a 45 movement. Finally, the carbonate addition or removal causes a 60 movement from the horizontal. It can also be seen from the figure that dilution will cause a 45 downward movement..4 LIME-SODA ASH PROCESS The definition of alkalinity has been discussed in Chapter 3, especially Example 3- (see Eq. [3-0]), which illustrates how the alkalinity of solution species is expressed as a CaCO 3 equivalent. Like alkalinity, hardness can be expressed in terms of mg/l of CaCO 3. Therefore, the total hardness, TH, is: TH = [(Ca + ) + (Mg + )] 00 [-49] The carbonate hardness, CH, is the portion of the total hardness that is offset by bicarbonate and carbonate ions in the water: CH = [/ (HCO 3 ) + CO 3 ] 00 [-50] The portion of hardness in excess of the original carbonate hardness is termed noncarbonate hardness, or NCH, NCH = TH CH [-5] The NCH actually results from the Ca + and Mg + associated, for example, with chloride, sulfate anions, or other anions. The steps of the lime soda ash water softening process are as follows: (a) raise ph of water solution by adding lime to allow for precipitation for removal of CH H CO 3 * + Ca(OH) = CaCO 3 + H O (a) (b) use Le Chetelier s principle to remove CH and Ca + Ca + + HCO 3 + Ca(OH) = CaCO 3 + H O (b)

.4 Lime-Soda Ash Process 69 (c) remove Mg + (d) use soda ash Mg + + HCO 3 + Ca(OH) = Mg(OH) + CaCO 3 + H O (c) Ca + + Na CO 3 = CaCO 3 + Na + (d ) Mg + + Ca(OH) = Mg(OH) + Ca + Mg + + Ca(OH) + Na CO 3 = Mg(OH) + CaCO 3 + Na + (d) In practice, a stoichiometric model is used where a decision is made about (HCO 3 ) and (Ca + ). If (HCO 3 ) > (Ca + ), the remaining (HCO 3 ) goes to Equation (c) and all the (Ca + ) goes to Equation (b), then the remaining Mg + all goes to Equation (d). On the other hand, if (Ca + ) > (HCO 3 ), then put all (HCO 3 ) in Equation (b); do not use Equation (c); enter the left-over (Ca + ) to Equation (d ) and the left-over (Mg + ) to Equation (d) (see Fig. -7). The analytes can be expressed either in mm or mg/l of CaCO 3 using a bar diagram. The stoichiometric principle is based on: That is, (Ca + ) final = (C T ) final [-5] (Ca + ) 0 + (Ca + ) a = (C T ) 0 + (C T ) a [-53] Here, the subscript 0 is original, the subscript a is added, the (Ca + ) a pertains to added lime, and the (C T ) a pertains to added soda ash. Example -5 illustrates the principles of Equations [-5] and [-53].

70 Chapter WATER SOFTENING Figure -7. Logical decision structure of the stoichiometric model for the lime soda water softening process. (From Ref..3)

.4 Lime-Soda Ash Process 7 [Example -5] For the following two raw waters, how much lime and soda ash are required? Water Species mg/l mm Total hardness 5 (as CaCO 3 ).5 Mg + 5.8 0.65 Na + 8 0.35 SO 4 8.6 0.3 Cl 0 0.85 Alkalinity 85 3.7 CO 5.8 (as CO ) 0.59 ph = 6.7 Water Ca + 5 Mg + HCO 3 4 * H CO 3 4 SO 4 Cl 0.8 ph = 6.3 6 More information can be added from the analytes. That is, Water Ca + (from TH)(.5 0.65 =.5 mm HCO 3 (= alk) = 3.7 mm

7 Chapter WATER SOFTENING Figure -8. Cation-anion balance of waters to be softened by the lime soda water process. (From Ref..4.)

.4 Lime-Soda Ash Process 73 CO (H CO * 3 ) HCO 3 Ca + Mg + Lime Soda Eq(a) water 0.59 0.59 water 4 4 Eq(b) water 3.5.5 water 4 Eq(c) water 3.7-3 0.7/ 0.7 water Eq(d ) water water 5-3 Eq(d) water 0.65-0.7/ 0.3 0.3 water Σ (lime) = (Ca + ) a water 3.09 water 8 Σ (soda) = (C T ) a water 0.3 water 5 (Ca + ) 0 water.5 water 5 (C T ) 0 water 0.59 +3.7 water 4+4 Underlined values indicate the stoichiometric requirement from each reaction from (a) to (d) and (d ).

74 Chapter WATER SOFTENING Using Equations (a) to (d) for both waters (see also bar diagram in Fig. -8), then: (Ca + ) 0 + (Ca + ) a = (C T ) 0 + (C T ) a water 4.59 water 3 Thus, the requirements are in Σ (lime) and Σ (soda) as above in mm, but also can be expressed in other units.

REFERENCES 75 REFERENCES -. V. L. Snoeyink and D. Jenkins, Water Chemistry, Wiley, New York, 980. -. D. W. Sundstrom and W. E. Klee, Wastewater Treatment, Prentice-Hall, Englewood Cliffs, New Jersey, 979. -3. J. N. Butler, Carbon Dioxide Equilibria and Their Applications, Addison- Wesley, Reading, Massachusetts, 98. -4. G. M. Fair, J. C. Geyer, and D. A. Okun, Water and Wastewater Engineering, Vol., Water Purification and Wastewater Treatment and Disposal, Wiley, New York, 968. -5. R. E. Lowenthal and G.V.R. Marais, Carbonate Chemistry of Aquatic Systems: Theory and Application, Ann Arbor Science, Ann Arbor, Michigan, 976. -6. J. W. Clark, W. Viessman, and M. J. Hammer, Water Supply and Pollution Control, Harper and Row, New York, 977.

76 Chapter WATER SOFTENING PROBLEM SET. The total alkalinity of a sample of water with a ph of.0 is 00 mg/l. Find the distribution of the alkalinity as (a) bicarbonate, (b) carbonate, and (c) hydroxide alkalinity in mg/l as CaCO 3. Assume a water temperature of 5 C and that the effect of salinity can be discounted.. A water that is to be softened by the lime-soda process has the following carbon dioxide, calcium, magnesium, and bicarbonate concentrations: CO = 8 meq/l Mg + = 3 meq/l Ca + = 65 meq/l HCO 3 = 60 meq/l Determine the calcium oxide and soda ash required per million gallons if the purities are 85 and 95%, respectively. Use meq/l as the basis of computation for both alkalinity and hardness instead of mg/l as CaCO 3. 3. Compute the ph of a solution of limewater containing 00 mg/l of Ca(OH) at 5 C. 4. (a) Compute the pounds of lime (CaO = 56) and soda ash (Na CO 3 = 06) required to soften million gallons of water with analysis as shown: ph 7. Total hardness 300 ppm as CaCO 3 Total alkalinity 50 ppm as CaCO 3 Calcium 00 ppm as CaCO 3 Magnesium 00 ppm as CaCO 3 (b) If the solubility products K s of CaCO 3 and Mg(OH) are, respectively, 0.87 0 8 and 5. 0 for moles per liter, and the final effluent from the softening plant has a ph of 0.5 and a normal carbonate (CO 3 ) alkalinity of 300 ppm as CaCO 3, calculate the theoretical residual of calcium and magnesium hardness, both as CaCO 3 (00).