TX249_frame_C11.fm Page 513 Friday, June 14, 2002 2:27 PM 11 Water Stabilization As mentioned in Chapter 10 on water softening, as long as the concentrations of CaCO3 and Mg(OH)2 exceed their solubilities, the solids may continue to precipitate This condition can cause scale to form, a solid that deposits due to precipitation of ions in solution To prevent scale formation, the water must be stabilized A water is said to be stable when it neither dissolves nor deposits precipitates If the pH is high, stabilization may be accomplished using one of several acids or using CO2, a process called recarbonation If the pH is low, stabilization may be accomplished using lime or some other bases Because of the universal presence of carbon dioxide, any water body is affected by the reaction products of carbon dioxide and water The species produced from this reaction form the carbonate system equilibria As discussed later, the stability or instability of water can be gaged using these equilibria Thus, this chapter discusses this concept It also discusses criteria for stability and the recarbonation process after water softening 11.1 CARBONATE EQUILIBRIA The carbonate equilibria is a function of the ionic strength of water, activity coefficient, and the effective concentrations of the ionic species The equilibrium coefficients that are calculated from the species concentrations are a function of the temperature This functionality of the coefficients can, in turn, be calculated using the Van’t Hoff equation, to be addressed later One of the major cations that can form scales as a result of the instability of water is calcium Calcium plays an important role in the carbonate equilibria We will therefore express the carbonate equilibria in terms of the interaction of the calcium ion and the carbonate species which are the reaction products of carbon dioxide and water In addition, since the equilibria occur in water, the dissociation of the water molecule must also be involved Using calcium as the cation, the equilibrium equations of the equilibria along with the respective equilibrium constants at 25°C are as follows (Rich, 1963): K w = 10 – 14 + − = { H } { OH } + K = 10 © 2003 by A P Sincero and G A Sincero – 6.35 (11.1) − { H } { HCO } = -∗ { H CO } (11.2) TX249_frame_C11.fm Page 514 Friday, June 14, 2002 2:27 PM 514 Physical–Chemical Treatment of Water and Wastewater + K = 10 – 10.33 2− { H } { CO } = -− { HCO } –9 2+ (11.3) 2− K sp,CaCO = 4.8 ( 10 ) = { Ca } { CO } (11.4) The Ks are the values of the respective equilibrium constants K sp,CaCO is the equilibrium constant for the solubility of CaCO3 The pair of braces, { }, are read as “the activity of,” the meaning of which is explained in the Background Chemistry and Fluid Mechanics chapter in the Background Prerequisites section As shown, the equilibrium constants are calculated using the activity In simple language, activity is a measure of the effectiveness of a given species in its participation in a reaction It is proportional to concentration; it is an effective or active concentration and has units of concentrations Because activity bears a relationship to concentration, its value may be obtained using the value of the corresponding concentration This relationship is expressed as follows: { sp } = γ [ sp ] (11.5) 2+ 2− − where sp represents any species involved in the equilibria such as Ca , CO , HCO and so on The pair of brackets, [], is read as “the concentration of,” γ is the activity coefficient 11.1.1 IONIC STRENGTH As the particle ionizes, the number of particles increases Thus, it is not a surprise that activity coefficient is a function of the number of particles in solution The number of particles is characterized by the ionic strength µ This parameter was devised by Lewis and Randall (1980) to describe the electric field intensity of a solution: µ = ∑ [ sp i ]z i (11.6) i is the index for the particular species and z is its charge The concentrations are in gmmols/L In terms of the ionic strength, the activity coefficient is given by the DeBye-Huckel law as follows (Snoeyink and Jenkins, 1980; Rich, 1963): 0.5z i ( µ ) −log γ = + 1.14 ( µ ) (11.7) γ = 10 0.5z i ( µ ) – 1+1.14 ( µ ) (11.8) In 1936, Langelier presented an approximation to the ionic strength µ Letting TDS in mg/L represent the total dissolved solids, his approximation is µ = 2.5 ( 10 )TDS –5 © 2003 by A P Sincero and G A Sincero (11.9) TX249_frame_C11.fm Page 515 Friday, June 14, 2002 2:27 PM Water Stabilization 515 Also, in terms of the specific conductance, sp conduc (in mmho/cm), Russell, another researcher, presented yet another approximation as µ = 1.6 ( 10 )sp conduc –5 (11.10) Example 11.1 The pH of a solution is Calculate the hydrogen ion concentration? Solution: + pH = – log 10 { H } + + = – log 10 { H } –7 { H } = 10 gmols/L Ans Example 11.2 The concentration of carbonic acid was analyzed to be 0.2 mgmol/L If the pH of the solution is 7, what is the concentration of the bicarbonate ion if the temperature is 25°C? Solution: − + K = 10 – 6.35 − –7 10 { HCO } { H } { HCO } = = ∗ 0.2/1000 { H CO } − − –4 { HCO } = 8.93 ( 10 ) gmol/L = [ HCO ] Ans Example 11.3 A sample of water has the following composition: CO2 = 22.0 2+ 2+ + − mg/L, Ca = 80 mg/L, Mg = 12.0 mg/L, Na = 46.0 mg/L, HCO = 152.5 mg/L , 2− and SO = 216 mg/L What is the ionic strength of the sample? Solution: µ = ∑ [ sp i ]z i Ion 2+ Ca 2+ Mg + Na − HCO 2− SO mg/L Mol Mass gmols/L 80 12.0 46.0 152.5 216 40.1 24.3 23 61 96.1 0.001995 0.0004938 0.002 0.0025 0.0022 2 µ = [ 0.001995 ( ) + 0.0004938 ( ) + 0.002 ( ) + 0.0025 ( ) + 0.0022 ( ) ] = 0.023 Ans Example 11.4 In Example 11.3, calculate the activity coefficient and the activity in mg/L of the bicarbonate ion © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 516 Friday, June 14, 2002 2:27 PM 516 Physical–Chemical Treatment of Water and Wastewater Solution: 0.5z i ( µ ) – 1+1.14 ( µ ) 0.5 ( ) ( 0.023 ) -=0.86 – 1+1.14 ( 0.023 ) γ = 10 = 10 { sp } = γ [ sp ] = 0.86 ( 0.0025 ) = 0.00215 mg/L 11.1.2 EQUILIBRIUM CONSTANT AS A FUNCTION OF Ans TEMPERATURE The equilibrium constants given previously were at 25°C To find the values of the equilibrium constants at other temperatures, the Van’t Hoff equation is needed According to this equation, the equilibrium constant K (Ksp for the solubility product constants) is related to temperature according to a derivative as follows: o dlnK ∆H = -2 dT RT (11.11) ο T is the absolute temperature; ∆Η is the standard enthalpy change, where the standard enthalpy change has been adopted as the change at 25°C at one atmosphere of pressure; and R is the universal gas constant The value of R depends upon the unit used for the other variables Table 11.1 ο gives its various values and units, along with the units used for ∆Η and T By convention, the concentration units used in the calculation of K are in gmmols/L Enthalpy is heat released or absorbed in a chemical reaction at constant pressure Table 11.2 shows values of interest in water stabilization It is normally reported as enthalpy changes There is no such thing as An absolute value of an enthalpy does not exist, only a change in enthalpy Enthalpy is a heat exchange at constant pressure, so enthalpy changes are measured by allowing heat to transfer at constant pressure; the amount of heat measured during the process is the enthalpy change Also, the table indicates enthalpy of formation This means that the values in the table are the heat TABLE 11.1 Values and Units of R o R Units K Concentration Units Used ∆H Units 0.08205 L atm gmmol.K ° gmmols -L — 8.315 J gmmol.K ° gmmols -L J -gmmol °K 1.987 cal gmmol.K° gmmols -L cal -gmmol °K gmmols -L — R Value 82.05 atm.cm gmmol.K° From J M Montgomery Engineers, Pasadena, CA © 2003 by A P Sincero and G A Sincero T Units °K °K TX249_frame_C11.fm Page 517 Friday, June 14, 2002 2:27 PM Water Stabilization 517 TABLE 11.2 Enthalpies of Formation of Substances of Interest in Stabilization o Substance ∆H 298 , kcal/gmmol HOH(l) + H ( aq ) − OH ( aq ) ∗ H CO CO2(aq) 2− CO ( aq ) − HCO ( aq ) CaCO3(s) 2+ Ca ( aq ) Ca(OH)2(aq) Mg(OH)2(aq) 2+ Mg −68.317 −54.96 −167.0 −98.69 −161.63 −165.18 −288.45 −129.77 −239.2 −221.0 −110.41 measured when the particular substance was formed from its elements For example, when calcium carbonate solid was formed from its elements calcium, carbon and oxygen, −288.45 kcal of heat per gmmol of calcium carbonate was measured The negative sign means that the heat measured was released or liberated in the chemical reaction Also, the state of the substance when it was formed is also indicated in the table For example, the state when calcium carbonate is formed liberating heat in the amount of −288.45 kcal/gmmol is solid, indicated as s The symbol l means that the state is liquid and the symbol aq means that the substance is being formed in water solution Also, note the subscript and superscript They indicate that the values in the table were obtained at standard temperature and pressure and one unit of activity for the reactants and products The standard temperature is 25°C; thus the 298, which is the Kelvin equivalent of 25°C The standard pressure is atmosphere The supero script symbolizes unit activity of the substances This means that the elements from which the substances are formed were all at a unit of activity and the product substances formed are also all at a unit of activity ο The enthalpy change is practically constant with temperature; thus ∆H may be o replaced by ∆H 298 Doing this and integrating the Van’t Hoff equation from KT1 to KT for the equilibrium constant K and from T1 to T2 for the temperature, o ∆H 298 K T = K T exp - ( T – T ) RT T (11.12) This equation expresses the equilibrium constant as a function of temperature © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 518 Friday, June 14, 2002 2:27 PM 518 Physical–Chemical Treatment of Water and Wastewater o 11.1.3 ∆ H 298 ’S FOR PERTINENT CHEMICAL REACTIONS CARBONATE EQUILIBRIA OF THE o Let us now derive the values of the ∆H 298 of the various pertinent chemical reactions in the carbonate equilibria as shown in Eqs (11.1) through (11.4) According to Hess’s law, if the chemical reaction can be written in steps, the enthalpy changes can be obtained as the sum of the steps Thus, consider Equation (11.1) The corresponding reaction is + H + OH HOH − (11.13) Writing in steps to conform to Hess’s law: HOH ( l ) → H + O 2 + H → H ( aq ) 1 − H + O → OH ( aq ) 2 HOH ( l ) o ∆H 298 = +68.317 kcal/gmmol of HOH ( l ) o ∆H 298 = − + − o ∆H 298 = −54.96 kcal/gmmol of OH ( aq ) H ( aq ) + OH ( aq ) o ∆H 298 = +13.36 kcal/gmmol of HOH ( l ) o The values of the ∆H 298’s are obtained from Table 11.2 Note that the values in o the table indicate ∆H 298 of formation having negative values Thus, if the reaction is not a formation but a breakup such as HOH ( l ) → H + O , the sign is positive This reaction indicates that to break the water molecule into its constituent atoms +68.317 kcal/gmmol of energy is required The + sign indicates that the reaction is endothermic requiring energy for the reaction to occur For the ionization of the + − water molecule as represented by HOH ( l ) H (aq) + OH (aq) and using Hess’s law as shown previously, +13.36 kcal/gmol of HOH(l) is required The Hess’s law steps for the rest of Eqs (11.1) through (11.4) are detailed as follows: ∗ o ∗ H CO → H + C + 6O ∆H 298 = +167.0 kcal/gmmol of H CO ( aq ) + H → H ( aq ) − H + C + 6O → HCO ( aq ) ∆H 298 = ∗ H CO ( aq ) + − H ( aq ) + HCO ( aq ) © 2003 by A P Sincero and G A Sincero o o ∆H 298 = – 165.18 kcal/gmmol of HCO ( aq ) o ∆H 298 = +1.82 kcal/gmmol of H CO ( aq ) TX249_frame_C11.fm Page 519 Friday, June 14, 2002 2:27 PM Water Stabilization 519 − HCO ( aq ) → H + C + 6O 2 + o H → H ( aq ) ∆H 298 = 2− o C + 6O → CO ( aq ) 2− ∆H 298 = – 161.63 kcal/gmmol of CO ( aq ) + 2− H ( aq ) + CO ( aq ) HCO ( aq ) o ∆H 298 = +165.18 kcal/gmmol of HCO ( aq ) o ∆H 298 = +3.55 kcal/gmmol of HCO ( aq ) o CaCO ( s ) → Ca + C + 6O ∆H 298 = +288.45 kcal/gmmol of CaCO ( s ) 2+ o Ca → Ca ( aq ) 2+ ∆H 298 = – 129.77 kcal/gmmol of Ca ( aq ) 2− ∆H 298 = – 161.63 kcal/gmmol of CO ( aq ) o 2+ ∆H 298 = – 2.95 kcal/gmmol of CaCO ( s ) C + 6O → CO ( aq ) CaCO ( s ) → Ca ( aq ) + CO ( aq ) 2− o Example 11.5 A softened municipal water supply enters a residence at 15°C and is heated to 60°C in the water heater Compare the values of the equilibrium constants for CaCO3 at these two temperatures If the water was at equilibrium at 25°C, determine if CaCO3 will deposit or not at these two temperatures Solution: o ∆H 298 K T = K T exp - ( T – T ) RT T 2− 2+ –9 K sp,CaCO = { Ca } { CO } = 4.8 ( 10 ) –9 = K T = 4.8 ( 10 ) in gmol units at 25°C o ∆H 298 = – 2.95 kcal/gmmol of CaCO ( s ) = – 2,950 cal /gmmol of CaCO ( s ) cal R = 1.987 gmmol.K° T = 15 + 273 = 288°K T = 25 + 273 = 298°K T = 60 + 273 = 333°K Therefore, 2,950 –9 K T at 15°K = 4.8 ( 10 )exp - ( 288 – 298 ) 1.987 ( 298 ) ( 288 ) –9 = 4.038 ( 10 ) in gmol units © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 520 Friday, June 14, 2002 2:27 PM 520 Physical–Chemical Treatment of Water and Wastewater Therefore, 2,950 –9 K T at 60°K = 4.8 ( 10 )exp - ( 333 – 298 ) 1.987 ( 298 ) ( 333 ) –9 = 8.10 ( 10 ) in gmol units Thus, the value of equilibrium constant is greater at 60°C than at 15°C Ans −9 The value of the equilibrium constant for calcium carbonate at 25°C is 4.8(10 ) 2+ 2− At this condition, the ions Ca and CO ions are given to be in equilibrium; thus, will neither deposit nor dissolve CaCO3 At the temperature of 15°C, the value of −9 −9 the equilibrium constant is 4.038(10 ) This value is less than 4.8(10 ) and will require less of the ionized ion; therefore at 15°C, the water is oversaturated and will deposit CaCO3 Ans −9 At 60°C, the equilibrium constant is 8.10(10 ), which is greater than that at 25°C Thus, at this temperature, the water is undersaturated and will not deposit CaCO3 Ans 11.2 CRITERIA FOR WATER STABILITY AT NORMAL CONDITIONS In the preceding discussions, a criterion for stability was established using the equilibrium constant called Ksp At normal conditions, as especially used in the water works industry, specialized forms of water stability criteria have been developed These are saturation pH, Langelier index, and the precipitation potential of a given water 11.2.1 SATURATION pH AND THE LANGELIER INDEX Because pH is easily determined by simply dipping a probe into a sample, determination of the saturation pH is a convenient method of determining the stability of water The concentrations of any species at equilibrium conditions are in equilibrium with respect to each other Also, for solids, if the condition is at equilibrium no precipitate or scale will form One of the concentration parameters of equilibrium is the hydrogen ion concentration, which can be ascertained by the value of the pH Thus, if the pH of a sample is determined, this can be compared with the equilibrium pH to see if the water is stable or not Therefore, we now proceed to derive the equilibrium pH Equilibrium pH is also called saturation pH In natural systems, the value of the pH is strongly influenced by the carbonate 2− equilibria reactions The CO species of these reactions will pair with a cation, thus “guiding” the equilibrium reactions into a dead end by forming a precipitate For example, the complete carbonate equilibria reactions are as follows: HOH ∗ H CO © 2003 by A P Sincero and G A Sincero + H + OH + − Kw − H + HCO K1 (11.14) (11.15) TX249_frame_C11.fm Page 521 Friday, June 14, 2002 2:27 PM Water Stabilization 521 − HCO ( aq ) Cation ( CO ) c ( s ) ↓ 2− H ( aq ) + CO K2 (11.16) 2− c+ 2Cation + cCO 2− K sp (11.17) c is the charge of the cation that pairs with CO forming the precipitate Cation2(CO3)c(s) We call the formation of this precipitate as the dead end of the carbonate equilibria, since the carbonate species in solution are diminished by the precipitation Let us digress for a moment from our discussion of the saturation pH in order to find the dead end cation for the carbonate system equilibria Several of these cations can possibly pair with the carbonate The pairing will be governed by the value of the solubility product constant, Ksp A small value of the Ksp means that only small values of the concentration of the constituent species are needed to form a product equal to the Ksp This, in turn, means that solids with smaller Ksp’s will easily form the solids Thus, of all the possible cations that can pair with the carbonate, the one with the smallest Ksp value is the one that can form a dead end −5 for the carbonate equilibria reactions Mg forms MgCO3 with a Ksp of 10 Ca forms −9 CaCO3 with a Ksp of 4.8(10 ) Table 11.3 shows other carbonate solids with the respective solubility product constants From the previous table, the smallest of the Ksp’s is that for Hg2CO3 Thus, considering all of the possible candidates that we have written, Hg2CO3 is the one that will form a dead end for the carbonate equilibria; however, of all the possible 2+ cations, Ca is the one that is found in great abundance in nature compared to the rest Thus, although all the other cations have much more smaller Ksp’s than calcium, TABLE 11.3 Solubility Product Constants of Solid Carbonates at 25°C Carbonate Solid BaCO3 CdCO3 CaCO3 CoCO3 CuCO3 FeCO3 PbCO3 MgCO3 MnCO3 Hg2CO3 NiCO3 Ag2CO3 SrCO3 ZnCO3 © 2003 by A P Sincero and G A Sincero Ksp −9 8.1(10 ) −14 2.5(10 ) −9 4.8(10 ) −12 1.0(10 ) −10 1.37(10 ) −11 2.11(10 ) −13 1.5(10 ) −5 1.0(10 ) −11 8.8(10 ) −17 9(10 ) −7 1.36(10 ) −12 8.2(10 ) −10 9.42(10 ) −11 6(10 ) TX249_frame_C11.fm Page 522 Friday, June 14, 2002 2:27 PM 522 Physical–Chemical Treatment of Water and Wastewater they are of no use as dead ends if they not exist The other cation that exists in abundance in natural waters is magnesium In fact, this is the other constituent hardness ion in water Comparing the Ksp’s of the carbonate of these cations, however, CaCO3 is the smaller Thus, as far as the carbonate equilibria reactions are concerned, the calcium ion is the one to be considered to form a dead end in the carbonate system equilibria Cation2(CO3)c(s) is therefore CaCO3(s) For this reason, Equation (11.4) was written in terms of CaCO3 (See Table 11.4) As will be shown later, the saturation pH may conveniently be expressed in terms of total alkalinity and other parameters The alkalinity of water is defined as + its capacity to neutralize any acid added to it When an acid represented by H is − added to a hydroxide represented by OH , the acid will be neutralized according to + − the reaction H + OH HOH Thus, the hydroxide is an alkaline substance When the acid is added to a carbonate, the acid is also neutralized according to the reaction 2− + 2H + CO H CO Carbonate is therefore also an alkaline substance By writing a similar reaction, the bicarbonate ion will also be shown to be an alkaline substance As we know, these species are the components of the carbonate equilibria They also represent as components of the total alkalinity of the carbonate system equilibria They may be added together to produce the value of the total alkalinity To be additive, each of these component alkalinities should be expressed in terms of a common unit A convenient common unit is the gram equivalent − − [OH ] is equal to {OH }/γΟΗ, where γΟΗ is the activity coefficient of the hydroxyl − + − ion {OH } could be eliminated in terms of the ion product of water, Kw = {H }{OH } To establish the equivalence of the component alkalinities, they must all be referred + to a common end point when the acid H is added to the solution From general − chemistry, we learned that this is the methyl orange end point As far as the OH + − ion is concerned, the end point for the reaction H + OH HOH has already been reached well before the methyl orange end point Thus, for the purpose of deter+ − mining equivalents, the reaction for the hydroxide alkalinity is simply H + OH HOH and the equivalent mass of the hydroxide is OH/1 One gram equivalent of the hydroxide is then equal to one gram mole Therefore, − Kw { OH } − − [ OH ] geq = [ OH ] = = -+ γ OH γ OH { H } − (11.18) − [ HCO ] = { HCO }/ γ HCO 3, where γ HCO is the activity coefficient of the bicar− 2− − + bonate ion From Equation (11.3), { HCO } = { H } { CO }/K 2; thus, [ HCO ] = + + 2− + { H } { CO }/ γ HCO K Reaction of the acid H with the bicarbonate given by H + − 2− HCO H CO ends exactly at the methyl orange end point From this reaction, the equivalent weight of the bicarbonate ion is HCO3/1; thus, one gram equivalent is equal to one gram mole Therefore, + 2− { H } { CO } − − [ HCO ] geq = [ HCO ] = -γ HCO K © 2003 by A P Sincero and G A Sincero (11.19) TX249_frame_C11.fm Page 529 Friday, June 14, 2002 2:27 PM Water Stabilization 529 Therefore, + + Kw { H } { CaOH } - = K CaOHc { Ca } (11.36) + γ Ca K w [ Ca ] { CaOH } + [ CaOH ] = - = -+ γ CaOHc γ CaOHc K CaOHc γ H [ H ] (11.37) + + γCaOHc, γCa, and γH are the activity coefficients of CaOH , Ca, and H , respectively o [CaSO ] is calculated as follows: o CaSO 2+ 2+ 2− Ca + SO K CaSO c 2− { Ca } { SO } = K CaSO c o { CaSO } (11.38) 2+ 2− { CaSO } γ Ca [ Ca ] [ SO ] o [ CaSO ] = = -γ CaSO c K CaSO c o The activity coefficient γ CaSO c = 1, because CaSO is not dissociated γ SO4 is the activity coefficient of the sulfate ion + o + o The previous expressions for [ CaCO ], [ CaHCO ], [CaOH ], and [ CaSO ] may 2+ now be substituted into Equation (11.31) and the result solved for [Ca ] to produce o + o [ Ca T ] – [ CaCO ] – [ CaHCO ] 2+ [ Ca ] = 2− γ Ca γ SO [ SO ] γ Ca K w + - + + K (11.39) γ CaOHc K CaOHc γ H [ H ] CaSO c Example 11.8 Analysis of a water sample yields the following results: [TDS] = 140 mg/L, [CaT] = 0.7 mgmol/L, [MgT] = 0.6 mgmol/L, [A]mgeq = 0.4 mg/L, sulfate ion = 0.3 mgmol/L, temperature = 25°C, and pH = 6.7 Calculate the concentration of the calcium ion corrected for the formation of the complex ions Solution: + o [ Ca T ] – [ CaCO ] – [ CaHCO ] 2+ [ Ca ] = γ Ca γ SO [ SO 2− ] γ Ca K w + - + + K CaSO c γ CaOHc K CaOHc γ H [ H ] 4 0.7 –4 [ Ca T ] = - = 7.0 ( 10 ) gmol/L 1000 © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 530 Friday, June 14, 2002 2:27 PM 530 Physical–Chemical Treatment of Water and Wastewater –9 K sp,CaCO 4.8 ( 10 ) o –6 [ CaCO ] = - = = 7.97 ( 10 ) gmol/L – 3.22 K CaCO c 10 γ H K sp,CaCO + + [ CaHCO ] = - [ H ]: γ CaHCO c K K CaHCO c γ H = 0.94 = γ CaHCO c = γ CaOHc K CaHCO c = 10 – 1.26 K = 10 + { H } = 10 – 10.33 – 6.7 Therefore, –9 0.94 [ 4.8 ( 10 ) ] + – 6.7 – 14 [ CaHCO ] = - ( 10 ) = 1.74 ( 10 ) gmol/L – 1.26 0.94 ( 10 ) γ Ca = 0.77 K CaOHc = 10 – 1.49 γ SO4 = 0.77 0.3 2− –6 [ SO ] = = 3.12 ( 10 ) gmol/L ( 1000 ) [ 32.1 + ( 16 ) ] K CaSO c = 10 – 2.31 Therefore, –4 –6 – 14 7.0 ( 10 ) – 7.97 ( 10 ) – 1.74 ( 10 ) 2+ [ Ca ] = – 14 –6 ( 0.77 ) ( 0.77 ) [ 3.12 ( 10 ) ] 0.77 ( 10 ) + + – 1.49 – 6.7 – 2.31 0.94 ( 10 ) ( 0.94 ) ( 10 ) 10 –4 6.92 ( 10 ) –4 = = 6.92 ( 10 ) gmol/L –6 –4 + 1.35 ( 10 ) + 3.78 ( 10 ) = 0.69 mgmol/L = 27.67 mg/L Ans 11.2.3 TOTAL ALKALINITY AS CALCIUM CARBONATE The unit of concentration that we have used for alkalinity is equivalents per unit volume This unit follows directly from the chemical reaction, so this method of expression is fairly easy to understand In practice, however, alkalinity is also expressed in terms of CaCO3 Expressing alkalinities in these terms is a sort of equivalence, although this practice can become very confusing Depending upon the chemical reaction it is involved with, calcium carbonate can have more than one value for its equivalent mass, and this is the source of the confusion Thus, to understand the underpinnings of this method of expression, one must look at the reference chemical reaction As shown previously, the equivalence of all forms of alkalinity was unified by using a common end point—the methyl orange end point that corresponds to a pH © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 531 Friday, June 14, 2002 2:27 PM Water Stabilization 531 − of about 4.5 For example, the OH alkalinity was assumed to go to completion at the methyl orange end point, although the reaction is complete long beforehand at around a pH of 10.8 The same argument held for the carbonate and the bicarbonate alkalinities, which, of course, would be accurate, since their end points are legitimately at the methyl orange end point This was done in order to have a common equivalence point Thus, to express alkalinities in terms of CaCO3, the reaction of calcium carbonate must also be assumed to complete at the same end point Even without assuming, this, of course, happens to be true The reaction is CaCO + 2H + H CO + Ca 2+ (11.40) and CaCO3 has an equivalent mass of CaCO3/2, because the number of reference species is −3 To illustrate the use of this concept, assume 10 gmmol/L of the hydroxide ion and express this concentration in terms of CaCO3 The pertinent reaction is − − + HOH From this reaction, OH has an equivalent mass of OH/1 and OH + H −3 −3 the number of equivalents per liter of the 10 gmol/L of the hydroxide is 10 (OH)/ −3 OH = 10 Thus, expressing the concentration in terms of CaCO3 − [ OH ] = 10 –3 gmmol/L = 10 –3 –3 geq/L = 10 ( CaCO /2 ) g/L as CaCO = 0.05 g/L as CaCO = 50 mg/L as CaCO 11.2.4 PRECIPITATION POTENTIAL Figure 11.1 shows a pipe that is almost completely blocked due to precipitation of CaCO3 Precipitation potential is another criterion for water stability, and application of this concept can help prevent situations like the one shown in this figure Understanding this concept requires prerequisite knowledge of the charge balance All solutions are electrically neutral and negative charges must balance the positive charges Thus, the balance of charges, where concentration must be expressed in terms of equivalents, is − 2− + 2+ [ CO ] eq + [ HCO ] eq + [ OH ] eq = [ H ] eq + [ Ca ] eq (11.41) Expressing in terms of moles, 2− − + 2+ [ CO ] + [ HCO ] + [ OH ] = [ H ] + [ Ca ] (11.42) Now, the amount of calcium carbonate that precipitates is simply the equivalent of the calcium ion that precipitates, Cappt Because the number of moles of Cappt is equal to the number of moles of the carbonate solid CaCO3ppt that precipitates, [ CaCO 3ppt ] = [ Ca ppt ] © 2003 by A P Sincero and G A Sincero (11.43) TX249_frame_C11.fm Page 532 Friday, June 14, 2002 2:27 PM 532 Physical–Chemical Treatment of Water and Wastewater FIGURE 11.1 A water distribution pipe almost completely blocked with precipitated calcium carbonate [CaCO3ppt] is the precipitation potential of calcium carbonate Cappt, in turn, can be 2+ 2+ obtained from the original calcium, Ca before, minus the calcium at equilibrium, Ca after 2+ 2+ [ Ca ppt ] = [ Ca before ] – [ Ca after ] (11.44) 2+ 2+ To use Equation (11.44), [ Ca before ] must first be known To determine [ Ca after ], 2+ 2+ the charge balance equation derived previously will be used [ Ca after ] is the [Ca ] in the charge balance equation, thus − 2− + 2+ [ CO ] + [ HCO ] + [ OH ] = [ H ] + [ Ca after ] (11.45) Making the necessary substitutions of the carbonate system equilibria equations 2+ to Equation (11.45) and solving for [ Ca after ], produces − 2− + 2+ [ CO ] + [ HCO ] + [ OH ] = [ H ] + [ Ca after ] + K sp,CaCO [ H ]K sp,CaCO Kw 2+ + + - + - = [ H ] + [ Ca after ] 2+ 2+ + γ CO3 γ Ca [ Ca after ] γ HCO γ H K γ Ca [ Ca after ] γ OH γ H [ H ] © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 533 Friday, June 14, 2002 2:27 PM Water Stabilization 533 + + [ H ] [ Ca  K sp,CaCO + [ H ] K sp,CaCO 3 Kw 2+ + ] +  [ H ] – -  [ Ca after ] –  - [ H ] + - =  γ HCO γ H K γ Ca  γ OH γ H  γ CO γ Ca 2+ after  + K   + K 2 +  K sp,CaCO3 + + [ H ] K sp,CaCO3 w w –  [ H ] –  +  [ H ] –  + [ H ]  [ H ] +  γ HCO γ H K γ Ca  γ OH γ H γ OH γ H    γ CO γ Ca 2+ [ Ca after ] = + 4[H ] (11.46) 3 + The [H ] in the previous equations is the saturation pH of Equation (11.25) Now, finally, the precipitation potential [CaCO3ppt] is 2+ 2+ [ CaCO 3ppt ] = [ Ca ppt ] = [ Ca before ] – [ Ca after ] (11.47) 11.2.5 DETERMINATION OF PERCENT BLOCKING POTENTIAL OF PIPES Let Volpipe be the volume of the pipe segment upon which the percent blocking potential is to be determined The amount of volume precipitation potential in this volume 2+ 2+ after a time t is ( 100 ( [ Ca before ] – [ Ca after ] )Vol pipe t )/ ρ CaCO t d , where ρ CaCO is the mass density of the carbonate precipitate and td is the detention time of the pipe segment Letting Qpipe be the rate of flow through the pipe, td = Volpipe/Qpipe Substituting this expression for td, the percent blocking potential Pblock after time t is 2+ 2+ 100 ( [ Ca before ] – [ Ca after ] )Vol pipe t Vol pipe ρ CaCO Q pipe P block = - ( 100 ) Vol pipe 2+ (11.48) 2+ 100 ( [ Ca before ] – [ Ca after ] )Q pipe t = - ( 100 ) ρ CaCO Vol pipe (11.49) ρ CaCO = 2.6 g/cc = 2600 g/L Note that since the concentrations are expressed in gram moles per liter, volumes and rates should be expressed in liters and liters per unit time, respectively ρ CaCO should be in g/L Example 11.9 Water of the following composition is obtained after a softening– 2+ −3 − 2− –4 recarbonation process: [Ca ] = 10 gmol/L, [ HCO ] = 10 gmol/L, [ CO ] = −3 3.2(10 ) gmol/L, [ H CO ∗ ] = 10 –9 gmol/L, pH = 8.7, pHs = 10, temperature = 25°C, −3 µ = 5(10 ) Calculate the equilibrium calcium ion concentration precipitation potential Express precipitation potential in gmols/L and mg/L Solution: The saturation pH is given as pHs = 10 The actual pH is 8.7, so the system is not saturated with calcium carbonate and no carbonate will precipitate; the precipitation potential is therefore zero At equilibrium at the Langelier saturation pH, 2+ −3 the calcium ion concentration will remain the same at [Ca ] = 10 gmol/L Ans Example 11.10 Assume that the pH of the treated water in Example 11.9 was raised to cause the precipitation of the carbonate solid The water is distributed through © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 534 Friday, June 14, 2002 2:27 PM 534 Physical–Chemical Treatment of Water and Wastewater a distribution main at a rate of 0.22 m /s Determine the length of time it takes to clog a section of the distribution main km in length, if the diameter is 0.42 m Solution: 2+ 2+ 100 ( [ Ca before ] – [ Ca after ] )Q pipe t P block = - ( 100 ) ρ CaCO Vol pipe −3 2+ [ Ca before ] = 10 gmol/L 2+ [ Ca after ] = Q pipe = 0.22 m /s = 220 L/s ρ CaCO = 2600 g/L π ( 0.42 ) Vol pipe = ( 1000 ) ( 1000 ) = 138,544.24 L Therefore, –3 100 ( 10 – ) ( 220 )t 100 = (100)t = 16,373,410.18 s = 189.51 days Ans ( 2600 ) ( 138,544.24 ) This example shows the importance of controlling the precipitation of CaCO3 Of course, in this particular situation, once the pipe is constricted due to blockage, the consumers would complain and the water utilities would then solve the problem; however, the problem situation may become too severe and the distribution pipe would have to be abandoned 11.3 RECARBONATION OF SOFTENED WATER After the softening process, the pH is so high that reduction is necessary to prevent deposition of scales in distribution pipes This can be accomplished inexpensively using carbon dioxide We will therefore develop the method for determining the carbonic acid necessary to set the water to the equilibrium pH In recarbonation, the available calcium ion in solution is prevented from precipitation Therefore, it remains to determine at what pH will the equilibrium condition be, given this calcium concentration This determination is, in fact, the basis of the Langelier saturation pH Adding carbonic acid will increase the acidity of the solution after it has neutralized any existing alkalinity Let the current pH be pHcur and the pH to which it is to be adjusted (the destination pH) be pHto If pHcur is greater than pHto an acid is needed No matter how insignificant, a natural water will always have an alkalinity in it Alkalinities of surface water can vary from 10 to 800 mg/L (Sincero, 1968) Until it is all consumed, this alkalinity will resist the change in pH Let the current total alkalinity be [Acur]geq in gram equivalents per liter Let the total acidity to be added be [Acadd]geq in gram equivalents per liter © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 535 Friday, June 14, 2002 2:27 PM Water Stabilization 535 TABLE 11.5 Fractional Dissociations, φ aH2 CO3 , of + H from Carbonic Acid Molar Concentration φ aH2 CO3 0.1 0.5 1.0 0.00422 0.00094 0.000668 – pH The hydrogen ion concentration corresponding to pHcur is 10 cur gram moles – pH per liter and that corresponding to pHto is 10 to gram moles per liter Note: The gram moles per liter of hydrogen ion is equal to its number of equivalents per liter – pH – pH Assuming no alkalinity present, the total acid to be added is 10 to − 10 cur gram moles per liter Alkalinity is always present, however, so more acid must be added to counteract the natural alkalinity, [Acur]geq Thus, the total acidity to be added, [Acadd]geq, is −pH −pH 10 to – 10 cur [ A cadd ] geq = [ A cur ] geq + -φa (11.50) where φa is the fractional dissociation of the hydrogen ion from the acid supplied For strong acids, φa is unity; for weak acids, it may be calculated from equilibrium constants (Table 11.5) To determine the equivalent mass of carbon dioxide needed, write the following chemical reactions, noting that carbon dioxide must react with existing alkalinity: CO + H O H CO − (11.51) 2− H CO + 2OH → 2H O + CO From these reactions, the equivalent mass of CO2 is CO2/2 Let us calculate some fractional dissociations of H2CO3 Its dissociation reaction is as follows: − + H + HCO H CO K = 10 −6.35 (11.52) First, assume a 0.1 M concentration of carbonic acid and let x be the concentration + − of H and HCO at equilibrium At equilibrium, the concentration of carbonic acid will be 0.1 − x Thus, substituting into the equilibrium expression produces x – 6.35 - = 10 0.1 – x – 6.35 + ( 10 – 6.35 ) + ( 0.1 ) ( 10 – 6.35 ) – 10 x = - = 0.000211 © 2003 by A P Sincero and G A Sincero (11.53) TX249_frame_C11.fm Page 536 Friday, June 14, 2002 2:27 PM 536 Physical–Chemical Treatment of Water and Wastewater Therefore, 0.000211 φ aH2 CO3 = - = 0.00211 0.1 Note: In the previous calculation, the activity coefficients have been ignored Example 11.11 A water sample from the reactor of a softening plant has a total −3 2+ 2+ alkalinity of [A]geq = 2.74(10 ) geq/L, pH of 10, [Ca ] = 0.7 mgmol/L, [Mg ] = −3 0.6 mgmol/L, and µ = 3.7(10 ) Using the Langelier saturation pH equation, pHs = 8.7 Calculate the amount of carbon dioxide necessary to lower the pH to the saturation value Use φ H2 CO3 = 0.00422 Solution: −pH −pH 10 to – 10 cur [ A cadd ] geq = [ A cur ] geq + -φ –3 [ A cur ] geq = 2.74 ( 10 ) geq/L – 8.7 – 10 10 – 10 –3 –3 [ A cadd ] geq = 2.74 ( 10 ) + = 2.74 ( 10 ) geq/L 0.00422 Ans GLOSSARY Activity—Measure of the effectiveness of a given species in its participation in a reaction Alkalinity—The capacity of a solution to neutralize any acid added to it Activity coefficient—A proportionality constant relating activity and concentration Dead end—For a system in equilibrium, the reaction that causes solids to precipitate Enthalpy—Heat released or absorbed in a chemical reaction at constant pressure Equilibrium constant—The product of the activities of the reactants and products (raised to appropriate powers) of a chemical reaction in equilibrium Ionic strength—A term devised by Lewis and Randall to describe the electric field intensity of a solution Langlier index—The difference between the actual pH and the saturation pH of a solution Precipitation potential—The amount of calcium carbonate that will precipitate when the solution is left by itself from its supersaturating condition Precipitation-potential pH—The pH attained at the precipitation potential condition Proton condition—A condition of balance between species that contain the proton and counteracting species that not contain the proton at a particular end point such as the dead end © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 537 Friday, June 14, 2002 2:27 PM Water Stabilization 537 Recarbonation—The process of adding carbon dioxide to water for the purpose of lowering the pH to the desired value Saturation index—The same as Langelier index Saturation pH—The pH attained at equilibrium with original calcium ion prevented from precipitation Solubility product constant—The term given to the equilibrium constant when products are in equilibrium with solid reactants SYMBOLS aq [CaT] 2+ [ Ca (aq) ] 2+ [ Ca after ] 2+ [ Ca before ] Subscript symbol for “aqueous” Total molar concentration of species containing the calcium atom Molar concentration of the calcium ion Calcium ion concentration after occurrence of precipitation potential Calcium ion concentration before occurrence of precipitation potential [Cappt] Precipitation potential of the calcium ion [CaCO3ppt] Precipitation potential of CaCO3 o CaCO Carbonate complex of calcium ion + CaHCO Bicarbonate complex of calcium ion + CaOH Hydroxide complex of calcium ion [∆carb] Moles of carbon dioxide per unit volume of flow required to bring the pH of water to the saturation pH o CaSO Sulfate complex of calcium ion [∆Decarb] Moles of base per unit volume of flow required to bring the pH of water to the saturation pH ∗ [ ∆H CO ] Change in concentration of carbonic acid needed to lower actual pH to the saturation pH ∗ H CO Mixture of CO2 in water, CO2(aq), and H2CO3; carbonic acid o ∆H Standard enthalpy change o ∆H 298 Standard enthalpy change at 25°C and unit of activity Hs Saturation hydrogen ion concentration corresponding to the saturation pH K Equilibrium constant ∗ K1 Ionization constant of H CO K2 Ionization constant of the bicarbonate ion o K CaCO c Equilibrium constant of CaCO + K CaHCO c Equilibrium constant of CaHCO + KCaOHc Equilibrium constant of CaOH o K CaSO c Equilibrium constant of CaSO Ksp Solubility product constant K sp, CaCO Ksp of CaCO3 KT1 Equilibrium constant at temperature T1 KT Equilibrium constant at temperature T2 Kw Ion product of water © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 538 Friday, June 14, 2002 2:27 PM 538 Physical–Chemical Treatment of Water and Wastewater l LI Pblock pHs Qpipe R s sp sp conduc t td T Volpipe z {} [] γ γCa γ CO3 γH γ HCO γOH µ ρ CaCO Symbol for liquid state of a substance Langelier of Saturation Index Percent blocking potential of a segment of pipe due to potential deposition of CaCO3 Saturation pH Rate of flow through pipe segment in which percent blocking is to be determined Universal gas constant Symbol for solid state of a substance Any species in equilibrium Specific conductance Time to percent blocking potential Detention time in pipe segment for which percent blocking potential is to be determined Absolute temperature Volume of pipe segment in which blocking potential is to be determined Charge of ion Read as “activity of ” Read as “concentration of ” Activity coefficient Activity coefficient of the calcium ion Activity coefficient of the carbonate ion Activity coefficient of the hydrogen ion Activity coefficient of the bicarbonate ion Activity coefficient of the hydroxyl ion Ionic strength Mass density of CaCO3 PROBLEMS 11.1 11.2 11.3 11.4 11.5 11.6 11.7 The pH of a solution is 14 Calculate the hydrogen ion concentration The pH of a solution is Calculate the hydrogen ion concentration The pH of a solution is Calculate the hydroxyl ion concentration The pH of a solution is 14 Calculate the hydroxyl ion concentration The concentration of carbonic acid was analyzed to be 0.2 mgmol/L If −4 the concentration of the bicarbonate ion is 8.93(10 ) gmol/L, what is the pH of the solution if the temperature is 25°C? The concentrations of carbonic acid and bicarbonate were analyzed to be −4 0.2 mgmol/L and 8.93(10 ) gmol/L, respectively If the pH is equal to 7, calculate the value of K1 −4 The concentration of the bicarbonate ion was analyzed to be 8.93(10 ) gmol/L If the pH is equal to and the temperature is 25°C calculate the concentration of carbonic acid © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 539 Friday, June 14, 2002 2:27 PM Water Stabilization 11.8 11.9 11.10 11.11 11.12 11.13 11.14 11.15 11.16 11.17 11.18 11.19 11.20 539 The concentrations of carbonic acid and bicarbonate were analyzed to be −4 0.2 mgmol/L and 8.93(10 ) gmol/L, respectively If the pH is equal to 7, determine the temperature of the solution A sample of water has the following composition: CO2 = 22.0 mg/L, 2+ 2+ + 3− Ca = 80 mg/L, Mg = 12.0 mg/L, Na = 46.0 mg/L, HCO = 152.5 2− mg/L, and SO = 216 mg/L What is the value of z for carbon dioxide? In Problem 11.9, calculate the activity coefficient and the activity in mg/L of the carbon dioxide In Problem 11.9, calculate the activity coefficient and the activity in mg/L of the sulfate ion In Problem 11.9, calculate the activity coefficient and the activity in mg/L of the magnesium ion In Problem 11.9, calculate the activity coefficient and the activity in mg/L of the sodium ion A softened municipal water supply enters a residence at 15°C and is heated to 60°C in the water heater If the water is at equilibrium at 15°C, determine if CaCO3 will deposit or not A softened municipal water supply enters a residence at 15°C and is heated to 60°C in the water heater If the water is at equilibrium at 60°C, determine if CaCO3 will deposit or not Analysis of a water sample yields the following results: [TDS] = 140 2+ 2+ mg/L, [Ca ] = 0.7 mgmol/L, [Mg ] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, temperature = 20°C, and pH = 6.7 The saturation pH was calculated to be equal to 8.67 Calculate the values of K2, K spCaCO , γOH, γ CO3, Kw , and γH Analysis of a water sample yields the following results: [TDS] = 140 2+ 2+ mg/L, [Ca ] = 0.7 mgmol/L, [Mg ] = 0.6 mgmol/L, [A]mgeq = 2.74 geq/L, temperature = 20°C, and pH = 6.7 The saturation pH was calculated to be equal to 8.67 Calculate the values of γ HCO 3, K spCaCO , γOH, γ CO3, Kw , and γH Analysis of a water sample yields the following results: [TDS] = 140 2+ 2+ mg/L, [Ca ] = 0.7 mgmol/L, [Mg ] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, temperature = 20°C, and pH = 6.7 The saturation pH was calculated to be equal to 8.67 Calculate the values of γ HCO , K2, γOH, γ CO3, Kw, and γH Analysis of a water sample yields the following results: [TDS] = 140 2+ 2+ mg/L, [Ca ] = 0.7 mgmol/L, [Mg ] = 0.6 mgmol/L, [A]mgeq = 2.74 geq/L, temperature = 20°C, and pH = 6.7 The saturation pH was calculated to be equal to 8.67 Calculate the values of γ HCO 3, K2, K spCaCO 3, γ CO3, Kw , and γH Analysis of a water sample yields the following results: [TDS] = 140 2+ 2+ mg/L, [Ca ] = 0.7 mgmol/L, [Mg ] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, temperature = 20°C, and pH = 6.7 The saturation pH was calculated to be equal to 8.67 Calculate the values of γ HCO 3, K2, K spCaCO , γOH, Kw , and γH © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 540 Friday, June 14, 2002 2:27 PM 540 Physical–Chemical Treatment of Water and Wastewater 11.21 Analysis of a water sample yields the following results: [TDS] = 140 2+ 2+ mg/L, [Ca ] = 0.7 mgmol/L, [Mg ] = 0.6 mgmol/L, [A]mgeq = 2.74 geq/L, temperature = 20°C, and pH = 6.7 The saturation pH was calculated to be equal to 8.67 Calculate the values of γ HCO 3, K2, K spCaCO , γOH, γ CO3, and γH 11.22 Analysis of a water sample yields the following results: [TDS] = 140 2+ 2+ mg/L, [Ca ] = 0.7 mgmol/L, [Mg ] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, temperature = 20°C, and pH = 6.7 The saturation pH was calculated to be equal to 8.67 Calculate the values of γ HCO 3, K2, K spCaCO , γOH, γ CO3, and Kw 11.23 Analysis of a water sample yields the following results: [TDS] = 140 2+ 2+ mg/L, [Ca ] = 3.2 mgmol/L, [Mg ] = 0.6 mgmol/L, temperature = 20°C, and pH = 6.7 The saturation pH was calculated to be 8.0 Calculate the alkalinity 11.24 Analysis of a water sample yields the following results: [TDS] = 140 2+ mg/L, [Mg ] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, temperature = 20°C, and pH = 6.7 The saturation pH was calculated to be 8.0 Calculate the calcium ion concentration 11.25 Analysis of a water sample yields the following results: [TDS] = 140 2+ 2+ mg/L, [Ca ] = 3.2 mgmol/L, [Mg ] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, temperature = 20°C, and pH = 6.7 Calculate the saturation pH 11.26 Analysis of a water sample yields the following results: [TDS] = 140 2+ 2+ mg/L, [Ca ] = 3.2 mgmol/L, [Mg ] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, temperature = 25°C, and pH = 6.7 Calculate the saturation pH 11.27 Analysis of a water sample yields the following results: [TDS] = 140 mg/L, [CaT] = 0.7 mgmol/L, [MgT] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, temperature = 25°C, and pH = 6.7 The concentration of the calcium after correction for the formation of the complex ion is 0.3 mgmol/L Calculate the sulfate ion concentration 11.28 Analysis of a water sample yields the following results: [TDS] = 140 mg/L, [MgT] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, sulfate ion = 0.3 mgmol/L, temperature = 25°C, and pH = 6.7 The concentration of the calcium after correction for the formation of the complex ion is 0.3 mgmole/L Calculate [CaT] 11.29 Analysis of a water sample yields the following results: [MgT] = 0.6 mgmol/L, sulfate ion = 0.3 mgmol/L, [A]mgeq = 2.74 geq/L, [TDS] = 140 mg/L, [CaT] = 0.7 mgmole/L, and temperature = 25°C The concentration of the calcium after correction for the formation of the complex ion is 0.3 mgmol/L Calculate the pH 11.30 Analysis of a water sample yields the following results: [TDS] = 140 mg/L, [CaT] = 0.7 mgmol/L, [MgT] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, sulfate ion = 0.3 mgmol/L, temperature = 25°C, and pH = 6.7 The concentration of the calcium after correction for the formation of the complex ion is 0.3 mgmol/L Calculate Kw , γCaOHc, KCaOHc, γH, γ SO4, and K CaSO c © 2003 by A P Sincero and G A Sincero TX249_frame_C11.fm Page 541 Friday, June 14, 2002 2:27 PM Water Stabilization 541 11.31 Analysis of a water sample yields the following results: [TDS] = 140 mg/L, [CaT] = 0.7 mgmol/L, [MgT] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, sulfate ion = 0.3 mgmol/L, temperature = 25°C, and pH = 6.7 The concentration of the calcium after correction for the formation of the complex ion is 0.3 mgmol/L Calculate γCa, γCaOHc, KCaOHc, γH, γ SO4, and K CaSO c 11.32 Analysis of a water sample yields the following results: [TDS] = 140 mg/L, [CaT] = 0.7 mgmol/L, [MgT] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, sulfate ion = 0.3 mgmol/L, temperature = 25°C, and pH = 6.7 The concentration of the calcium after correction for the formation of the complex ion is 0.3 mgmol/L Calculate γCa, Kw , KCaOHc, γH, γ SO4, and K CaSO c 11.33 Analysis of a water sample yields the following results: [TDS] = 140 mg/L, [CaT] = 0.7 mgmol/L, [MgT] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, sulfate ion = 0.3 mgmol/L, temperature = 25°C, and pH = 6.7 The concentration of the calcium after correction for the formation of the complex ion is 0.3 mgmol/L Calculate γCa, Kw , γCaOHc, γH, γ SO4, and K CaSO c 11.34 Analysis of a water sample yields the following results: [TDS] = 140 mg/L, [CaT] = 0.7 mgmol/L, [MgT] = 0.6 mgmol/L, [A]mgeq = 2.74 geq/L, sulfate ion = 0.3 mgmol/L, temperature = 25°C, and pH = 6.7 The concentration of the calcium after correction for the formation of the complex ion is 0.3 mgmol/L Calculate γCa, Kw , γCaOHc, KCaOHc, γ SO4, and K CaSO c 11.35 Analysis of a water sample yields the following results: [TDS] = 140 mg/L, [CaT] = 0.7 mgmol/L, [MgT] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, sulfate ion = 0.3 mgmol/L, temperature = 25°C, and pH = 6.7 The concentration of the calcium after correction for the formation of the complex ion is 0.3 mgmol/L Calculate γCa, Kw , γCaOHc, KCaOHc, γH, and K CaSO c 11.36 Analysis of a water sample yields the following results: [TDS] = 140 mg/L, [CaT] = 0.7 mgmol/L, [MgT] = 0.6 mgmol/L, [A]mgeq = 0.4 mgeq/L, sulfate ion = 0.3 mgmol/L, temperature = 25°C, and pH = 6.7 The concentration of the calcium after correction for the formation of the complex ion is 0.3 mgmol/L Calculate γCa, Kw , γCaOHc, KCaOHc, γH, and γ SO4 11.37 Assume that the pH of the treated water in Problem 11.10 was adjusted to cause the precipitation of the carbonate solid, if it was not precipitating already The water is distributed through a distribution main at a rate of 0.22 m /s Determine the length of time it takes to clog a section of the distribution main km in length, if the diameter is 0.42 m Assume −4 2+ [ Ca after ] = 1.47(10 ) gmol/L BIBLIOGRAPHY Al-Rqobah, H E and A Al-Munayyis (1989) Recarbonation process for treatment of distilled water produced by 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