Standard Methods for Examination of Water & Wastewater_12 potx

TX249_frame_C13.fm Page 593 Friday, June 14, 2002 2:32 PM 13 Removal of Iron and Manganese by Chemical Precipitation Iron concentrations as low as 0.3 mg/L and manganese concentrations as low as 0.05 mg/L can cause dirty water complaints At these concentrations, the water may appear clear but imparts brownish colors to laundered goods Iron also affects the taste of beverages such as tea and coffee Manganese flavors tea and coffee with medicinal tastes Some types of bacteria derive their energy by utilizing soluble forms of iron and manganese These organisms are usually found in waters that have high levels of iron and manganese in solution The reaction changes the species from soluble forms into less soluble forms, thus causing precipitation and accumulation of black or reddish brown gelatinous slimes Masses of mucous iron and manganese can clog plumbing and water treatment equipment They also slough away in globs that become iron or manganese stains on laundry Standards for iron and manganese are based on levels that cause taste and staining problems and are set under the Environmental Protection Agency Secondary Drinking Water Standards (EPA SDWA) They are, respectively, 0.3 mg/L for iron and 0.05 mg/L for manganese Iron and manganese are normally found in concentrations not exceeding 10 mg/L and mg/L, respectively, in natural waters Iron and manganese can be found at higher concentrations; however, these conditions are rare Iron concentrations can go as high as 50 mg/L Iron and manganese may be removed by reverse osmosis and ion exchange The unit operation of reverse osmosis was discussed in a previous chapter; the unit process of ion exchange is discussed in a later chapter This chapter discusses the removal of iron and manganese by the unit process of chemical precipitation One manufacturer claimed that these elements could also be removed by biological processes It claimed that a process called the bioferro process encourages the growth of naturally occurring iron assimilating bacteria, such as Gallionella ferruginea, thus reducing iron concentration An experimental result shows a reduction of iron from 6.0 mg/L to less than 0.1 mg/L They also claimed that a companion process called bioman could remove manganese down to 0.08 mg/L This uses naturally occurring manganese bacteria to consume manganese Figure 13.1 is a photograph showing growths of “bioferro” and “bioman” bacteria © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 594 Friday, June 14, 2002 2:32 PM 594 FIGURE 13.1 A photograph of “bioferro” and “bioman” bacteria 13.1 NATURAL OCCURRENCES OF IRON AND MANGANESE Iron and manganese have the electronic configurations of [Ar]3d 4s and [Ar]3d 4s , and are located in Groups VIIIB and VIIB of the Periodic Table, respectively They are both located in the fourth period [Ar] means that these elements have the electronic configuration of the noble gas argon The letters d and s refer to the d and s orbitals; the superscripts indicate the number of electrons that the orbitals contain Thus, the d orbital of iron contains electrons and that of manganese contains electrons Both elements contain electrons in their s orbitals This means that in their most reduced positive state, they acquire oxidation states of 2+ Also, because of the d orbitals, they can form a number of oxidation states The multiplicity of oxidation states give iron and manganese the property of imparting colors such as the imparting of brownish colors to laundered goods Surface waters always contain dissolved oxygen in it Thus, iron and manganese would not exist in their most reduced positive state of 2+ in these waters The reason is that they will simply be oxidized to higher states of oxidation by the dissolved oxygen forming hydroxides and precipitate out Groundwater is a source where these elements could come from Groundwaters occurring deep down in the earth can become devoid of oxygen, thus, any iron or manganese present would have to be reduced Therefore, the waters where removal of iron and manganese could be undertaken are groundwaters and the form of the elements are in the 2+ oxidation states, Fe(II) and Mn(II), respectively © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 595 Friday, June 14, 2002 2:32 PM 595 TABLE 13.1 Solubility Product Constant of Iron and Manganese Precipitation Products Precipitation Product Fe(OH)2 FeCO2 FeS Fe(OH)3 Mn(OH)2 MnCO3 MnS Solubility Product, Ksp −15 3.16(10 ) −11 2.11(10 ) −26 8(10 ) −38 1.1(10 ) −14 4.5(10 ) −11 8.8(10 ) −22 4.3(10 ) 13.2 MODES OF REMOVAL OF IRON AND MANGANESE The best place to investigate for determining the mode by which the elements can be removed is the table of solubility products constants as shown in Table 13.1 In general, a precipitation product that has the lowest Ksp means that the substance is the most insoluble As shown in the table, for iron, the lowest Ksp is that of Fe(OH)3, −38 an Fe(III) iron, and has the value of 1.1(10 ) For manganese, the lowest Ksp is that −22 of MnS, an Mn(II) manganese, and has the value of 1.1(10 ) These Ksp’s indicate that the elements must be removed in the form of ferric hydroxide and manganese sulfide, respectively; however, from the table, manganese −14 can also be removed as Mn(OH)2 at a Ksp = 4.5(10 ) Of course, lime has many uses, while sulfide has only few Sodium sulfide is used in photographic film development; however, lime is used in water and wastewater treatment, as an industrial chemical, as well as being used in agriculture Thus, because of its varied use, lime is much cheaper In addition, using a sulfide to remove iron and manganese would be a new method Its health effect when found in drinking water is not documented On the other hand, lime has been used for years We will therefore use lime as the precipitant for the removal of iron and manganese The probable use of sulfide in removing iron and manganese could be a topic for investigation in applied research 13.3 CHEMICAL REACTIONS OF THE FERROUS AND THE FERRIC IONS The chemical reactions of the ferrous and the ferric ions were already discussed in a previous chapter From the topic in the preceding section, iron is more efficiently removed as ferric hydroxide The natural iron is in the form of Fe(II), so this ferrous must therefore oxidize to the ferric form in order to precipitate as the ferric hydroxide, if, in fact, the iron is to be removed in the ferric form In Chapter 12, this was done using the dissolved oxygen that is relatively abundant in natural waters It must be © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 596 Friday, June 14, 2002 2:32 PM 596 noted, however, that based on the Ksp values, iron can also be removed in the ferrous form as Fe(OH)2 For convenience, the reactions are reproduced next Ferrous: 2+ Fe + 2OH Fe(OH) ( s ) K sp,Fe ( OH )2 = 10 + Fe(OH) ( s ) + OH −14.5 − − K FeOHc = 10 − FeOH + OH Fe(OH) ( s ) − K Fe ( OH )3 c = 10 Fe ( OH ) (13.1) −9.4 (13.2) −5.1 (13.3) Fe(OH) + O + H O → Fe(OH) ↓ (conversion from ferrous to ferric) (13.4) Ferric: 3+ Fe + 3OH Fe(OH) ( s ) − K sp,Fe ( OH )3 = 10 2+ Fe(OH) ( s ) FeOH + 2OH Fe(OH) ( s ) Fe(OH) + OH + Fe(OH) ( s ) + OH − − − − Fe ( OH )4 4+ (13.6) −16.74 K Fe ( OH )4 c = 10 − (13.5) −26.16 K Fe ( OH )2 c = 10 Fe ( OH ) + 4OH 2Fe(OH) ( s ) K FeOHc = 10 −38 (13.7) −5 K Fe2 ( OH )2 c = 10 (13.8) −50.8 (13.9) Remember that the values of the solubility product constants, K sp,Fe ( OH )2 and K sp,Fe ( OH )3 , and all the other equilibrium constants for the complex ions apply only at 25°C The presence of the complex ions increase the solubility of the iron species and therefore increase the concentration of these species in solution For the ferrous and the ferric species, they are spFeII and spFeIII and were derived in Chapter 12, respectively, as − + 2+ [ sp FeII ] = [ Fe ] + [ FeOH ] + [ Fe ( OH ) ] + K sp, Fe ( OH )2 γ H [ H ] K FeOHc γ H [ H + ] K Fe ( OH )3 c K w = - + - + -2 + γ FeOHc K w γ FeII K w γ Fe ( OH )3 c γ H [ H ] 3+ + 2+ − (13.10) 4+ [ sp FeIII ] = [ Fe ] + [ FeOH ] + [ Fe ( OH ) ] + [ Fe ( OH ) ] + [ Fe ( OH ) ] + + K sp, Fe ( OH )3 γ H [ H ] K FeOHc γ [ H + ] K Fe ( OH )2 c γ H [ H ] K Fe ( OH )4 c K w H = - + - + - + -3 + γ Fe ( OH )2 c K w γ FeIII K w γ FeOHc K w γ Fe ( OH )4 c [ H ] + 2K Fe2 ( OH )2 c γ H [ H ] + γ Fe2 ( OH )2 c K w © 2003 by A P Sincero and G A Sincero (13.11) TX249_frame_C13.fm Page 597 Friday, June 14, 2002 2:32 PM Example 13.1 From the respective optimum pH’s of 11.95 and 8.2 for spFeII and spFeIII, calculate the concentrations [spFeII] and [spFeIII], respectively Assume the water contains 140 mg/L of dissolved solids Solution: + K sp,Fe ( OH )2 γ H [ H ] K FeOHc γ H [ H + ] K Fe ( OH )3 c K w [ sp FeII ] = + - + -2 + γ FeOHc K w γ FeII K w γ Fe ( OH )3 c γ H [ H ] K sp, Fe ( OH )2 = 10 −5 µ = 2.5 ( 10 )TDS – 14.5 −5 γ = 10 0.5z i2 ( µ ) – 1+1.14 ( µ ) µ = 2.5 ( 10 ) ( 140 ) = 3.5 ( 10 ) –3 γ H = γ FeOHc = γ Fe ( OH )3 c = 10 –3 0.5 ( ) [ 3.5 ( 10 ) ] – –3 1+1.14 [ 3.5 ( 10 ) ] = 0.94 –3 γ FeII = 10 0.5 ( ) [ 3.5 ( 10 ) ] – –3 1+1.14 [ 3.5 ( 10 ) ] K FeOHc = 10 – 9.4 = 0.77K w = ( 10 K Fe(OH )3 c = 10 – 14 ) – 5.1 Therefore, – 14.5 – 9.4 – 5.1 – 14 – 11.95 – 11.95 ( 0.94 ) [ 10 ] ( 10 ) ( 0.94 ) [ 10 ] ( 10 ) ( 10 ) 10 −−−−−−−−−−−– − −−−− −−−−−−−−−−−−–−−− −− [ sp FeII ] = −−−−−−−−−−−−14−2−−− + −−−−−−−−−−−−14−−−− + −−−−−−−−−−−−−11.95− −−−−−−−−−−−−−− −−− − − −−− – ( 0.94 ) ( 0.94 ) [ 10 ] ( 0.94 ) ( 10 ) ( 0.77 ) ( 10 ) – 11 –8 –8 –7 = 4.57 ( 10 ) + 4.47 ( 10 ) + 8.01 ( 10 ) = 1.0 ( 10 ) gmol/L = 0.0056 mg/L + + Ans K sp,Fe ( OH )3 γ H [ H ] K FeOHc γ [ H + ] K Fe ( OH )2 c γ H [ H ] K Fe ( OH )4 c K w H [ sp FeIII ] = + - + + -3 + γ Fe ( OH )2 c K w γ FeIII K w γ FeOHc K w γ Fe ( OH )4 c [ H ] + 2K Fe2 ( OH )2 c γ H [ H ] + -4 γ Fe2 ( OH )2 c K w K sp,Fe ( OH )3 = 10 – 38 γ FeIII = 10 –3 0.5 ( ) [ 3.5 ( 10 ) ] – –3 1+1.14 [ 3.5 ( 10 ) ] = 0.56 K FeOHc = 10 –3 γ FeOHc = 10 0.5 ( ) [ 3.5 ( 10 ) ] – –3 1+1.14 [ 3.5 ( 10 ) ] = 0.77 –3 γ Fe ( OH )2 c = 10 0.5 ( ) [ 3.5 ( 10 ) ] – –3 1+1.14 [ 3.5 ( 10 ) ] © 2003 by A P Sincero and G A Sincero = 0.94 K Fe ( OH )2 c = 10 – 16.74 – 26.16 TX249_frame_C13.fm Page 598 Friday, June 14, 2002 2:32 PM K Fe(OH) c = 10 –5 γ Fe(OH) c = γ Fe(OH) c = 0.94 K Fe2 (OH) c = 10 – 50.8 –3 γ Fe2 (OH) c = 10 0.5 ( ) [ 3.5 ( 10 ) ] – –3 1+1.14 [ 3.5 ( 10 ) ] = 0.36 Therefore, ( 10 –38 ) ( 0.94 ) [ 10 ] ( 10 –26.16 ) ( 0.94 ) [ 10 –8.2 ] ( 10 –16.74 ) ( 0.94 ) [ 10 –8.2 ] [ sp FeIII ] = −−−−−−−−−−−–−−−−−− + −−−−−−−−−−−−−−−−−− + −−−−−−−−−−−–−−−−− −−−−−−−−−−− −−−−− − −−−−−−−−−−−−−−2 −−− −−−−−−−−−−− 14 −−−− − −− 14 – 14 ( 0.94 ) ( 10 ) ( 0.56 ) ( 10 ) ( 0.77 ) ( 10 ) – 8.2 – 8.2 ( 10 ) ( 10 ) ( 10 ) ( 0.94 ) [ 10 ] + + – 8.2 – 14 –5 – 14 ( 0.94 ) [ 10 – 50.8 ] ( 0.36 ) ( 10 ) – 19 – 84 2.09 ( 10 ) 2.43 ( 10 ) 1.08 ( 10 ) 1.0 ( 10 ) 3.92 ( 10 ) = - + - + - + - + –9 – 57 – 43 – 29 – 15 5.6 ( 10 ) 7.7 ( 10 ) 9.4 ( 10 ) 5.93 ( 10 ) 3.6 ( 10 ) – 43 – 63 – 25 = 3.73 ( 10 – 21 ) + 3.16 ( 10 = 2.84 ( 10 – 11 ) gmol/L = 0.0000016 mg/L – 15 ) + 1.15 ( 10 13.3.1 PRACTICAL OPTIMUM pH RANGE OF FERROUS AND FERRIC – 11 ) + 1.69 ( 10 FOR THE – 11 ) + 1.08 ( 10 – 27 ) Ans REMOVAL As shown in Chapter 12, at 25°C and at a solids concentration of 140 mg/L, the optimum pH’s correspond to 11.95 and 8.2 (or around 12 and 8), respectively, for ferrous and ferric The respective concentrations for spFeII and spFeII at these conditions as obtained in the previous example are [spFeII] = 0.0056 mg/L and [spFeIII] = 0.0000016 mg/L A pH range exists, however, at which units used for the removal of the elements can be operated and effect good results This range is called the practical optimum pH range Tables 13.2 through 13.5 show the respective concentrations of spFeII and spFeIII at other conditions of pH and total solids The values for [spFeII] were obtained using Equation (13.10) and the values for [spFeIII] were obtained using Equation (13.11) Note that these equations require the values of the activity coefficients of the ions The activity coefficients are needed by the equations and, since activity coefficients are functions of the dissolved solids, dissolved solids are used as parameters in the tables, in addition to pH The previous tables indicate that the total solids (or equivalently, the activities of the ions) not have a significant effect on the optimum pH values, which for spFeII remain at about 12.0 and for spFeIII remain at about 8.0 For practical purposes, however, the practical optimum pH for spFeII ranges from 11 to 13 and for spFeIII, it ranges from 5.0 to 13.0 Note that for the range of pH for spFeII, it is assumed the element is to be removed as Fe(OH)2 If it is to be removed as Fe(OH)3, the pH of the solution during its oxidation by dissolved oxygen or any oxidizer need not be adjusted since the practical optimum pH for the precipitation of ferric hydroxide varies over a wide range from 5.0 to 13.0 and already includes the range for the ferrous removal © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 599 Friday, June 14, 2002 2:32 PM TABLE 13.2 Concentration of spFeII as a ° Function of pH at 25°C pH, Dissolved Solids = 140 mg/L [spFeII], mg/L 10 11 12 13 14 15 5.0(10 ) +16 2.0(10 ) +14 2.0(10 ) +12 2.0(10 ) +10 2.0(10 ) +8 2.0(10 ) +6 2.0(10 ) +4 2.0(10 ) +2 2.2(10 ) +0 4.2(10 ) −1 2.4(10 ) −2 2.3(10 ) −3 7.3(10 ) −2 5.1(10 ) −1 5.0(10 ) +0 5.0(10 ) +18 TABLE 13.3 Concentration of spFeII as a ° Function of pH at 25°C pH, Dissolved Solids = 35,000 mg/L [spFeII], mg/L 10 11 12 13 14 15 5.0(10 ) +16 5.0(10 ) +14 5.0(10 ) +12 5.0(10 ) +10 5.0(10 ) +8 5.0(10 ) +6 5.0(10 ) +4 5.0(10 ) +2 5.2(10 ) +0 7.2(10 ) −1 2.7(10 ) −2 2.4(10 ) −2 1.5(10 ) −1 1.3(10 ) +0 1.3(10 ) +1 1.3(10 ) © 2003 by A P Sincero and G A Sincero +18 TX249_frame_C13.fm Page 600 Friday, June 14, 2002 2:32 PM TABLE 13.4 Concentration of spFeIII as a ° Function of pH at 25°C pH, Dissolved Solids = 140 mg/L [spFeII], mg/L 10 11 12 13 14 15 1.2(10 ) +5 8.6(10 ) +3 1.3(10 ) +0 5.3(10 ) −2 5.5(10 ) −3 1.4(10 ) −4 1.1(10 ) −5 1.0(10 ) −6 1.6(10 ) −6 6.0(10 ) −5 6.0(10 ) −4 6.0(10 ) −3 6.0(10 ) −2 6.0(10 ) −1 6.0(10 ) +0 6.0(10 ) +10 TABLE 13.5 Concentration of spFeIII as a ° Function of pH at 25°C pH, Dissolved Solids = 35,000 mg/L [spFeIII], mg/L 10 11 12 13 14 15 1.2(10 ) +7 1.2(10 ) +4 1.3(10 ) +1 2.3(10 ) −1 1.2(10 ) −3 1.1(10 ) −5 1.1(10 ) −7 2.0(10 ) −7 9.4(10 ) −6 9.4(10 ) −5 9.4(10 ) −4 9.4(10 ) −3 9.4(10 ) −2 9.4(10 ) −1 9.4(10 ) +0 9.4(10 ) © 2003 by A P Sincero and G A Sincero +10 TX249_frame_C13.fm Page 601 Friday, June 14, 2002 2:32 PM The previous optimum values of pH apply only at 25°C As indicated in the formulas, equilibrium constants are being used to compute these values The values of the equilibrium constants vary with temperature, so the optimum range at other temperatures would be different Equilibrium constants at other temperatures can be calculated using the Van’t Hoff equation which, however, requires the values of the standard enthalpy changes At present, these values are unavailable making values of optimum pH range at other temperatures impossible to calculate For this reason, the pH range found above must be modified to a conservative range Hence, adopt the following: for ferrous removal as Fe(OH)2, 11.5 ≤ optimum pH ≤ 12.5 and for ferrous removal as Fe(OH)3, 5.5 ≤ optimum pH ≤ 12.5 13.4 CHEMICAL REACTIONS OF THE MANGANOUS ION [Mn(II)] − Manganese can be removed as Mn(OH)2 using a suitable OH source Upon introduction of the hydroxide source, however, it is not only this solid that is produced Manganese forms complex ions with the hydroxide The complex equilibrium reactions are as follows (Snoeyink and Jenkins, 1980): Mn(OH) + 2+ Mn + OH 2+ Mn + 2OH Mn(OH) ( s ) − − K MnOHc = 10 – 3.4 K sp,Mn(OH) = 4.5 ( 10 (13.12) – 14 ) (13.13) Mn(OH) Mn + 2OH − K Mn(OH) c = 10 – 6.8 (13.14) − Mn 2+ + 3OH − K Mn(OH) c = 10 – 7.8 (13.15) 2+ Mn(OH) The values of the equilibrium constants given above are at 25°C The complexes + − − are Mn(OH) , Mn (OH) , and Mn (OH) Also note that the OH ion is a participant in these reactions This means that the concentrations of each of these complex ions are determined by the pH of the solution In the application of the previous equations in an actual treatment of water, conditions must be adjusted to allow maximum precipitation of the solid represented by Mn(OH)2(s) To allow for this maximum precipitation, the concentrations of the complex ions must be held to the minimum The pH corresponding to this condition is the optimum pH From the previous reactions, the 2+ equivalent mass of Mn is Mn/2 = 27.45 13.4.1 DETERMINATION OF THE OPTIMUM pH Let spMn represent the collection of species standing in solution containing the Mn(II) species Thus, 2+ + − [ sp Mn ] = [ Mn ] + [ Mn(OH) ] + [ Mn(OH) ] + [ Mn(OH) ] (13.16) All the concentrations in the right-hand side of the above equation will now be expressed in terms of the hydrogen ion concentration This will result in expressing © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 602 Friday, June 14, 2002 2:32 PM [spMn] in terms of the hydrogen ion Differentiating the resulting equation of [spMn] + with respect to [H ] and equating the result to zero will produce the minimum concentration of spMn and, thus, the optimum pH determined Using the equations and equilibrium constants of Eqs (13.12) through (13.15), along with the ion product of water, we proceed as follows: + + 2+ K sp, Mn ( OH )2 K sp, Mn ( OH )2 { H } K sp, Mn ( OH )2 γ H [ H ] { Mn } [ Mn ] = = - = = -2 − γ Mn γ Mn K w γ Mn K w γ Mn { OH } (13.17) 2+ + + 2+ − K sp, Mn ( OH )2 γ H [ H ] { Mn ( OH ) } { Mn } { OH } + [ Mn ( OH ) ] = = = γ Mn ( OH )c γ Mn ( OH )c K MnOHc γ Mn ( OH )c K MnOHc K w (13.18) 2+ − K sp, Mn ( OH ) { Mn(OH) } { Mn } { OH } 0 [ Mn(OH) ] = = { Mn(OH) } = = γ Mn(OH) c=1 K Mn ( OH )2 c K Mn ( OH )2 c (13.19) − 2+ − K sp, Mn ( OH )2 K w { Mn(OH) } { Mn } { OH } − [ Mn(OH) ] = - = = -+ γ Mn ( OH )3 c γ Mn ( OH )3 c K Mn ( OH )3 c γ Mn ( OH )3 c K Mn ( OH )3 c γ H [ H ] (13.20) γMn, γMnOHc, γMn(OH)2c, and γMn(OH)3c are, respectively, the activity coefficients + − of Mn(II) and the complexes Mn(OH) , Mn (OH) , and Mn ( OH ) K sp,Mn(OH) is the solubility product constant of the solid Mn(OH)2(s) KMnOHc, K Mn(OH) c , and + K Mn(OH) c are, respectively, the equilibrium constants of the complexes Mn(OH) , − Mn (OH) , and Mn ( OH ) Equations (13.17) through (13.20) may now be substituted into Equation (13.16) to produce + + K sp, Mn ( OH )2 γ H [ H ] K sp, Mn ( OH )2 γ H [ H ] K sp, Mn ( OH ) [ sp Mn ] = + - + 2 γ Mn ( OH )c K MnOHc K w K Mn ( OH )2 c γ Mn K w K sp, Mn ( OH )2 K w + -+ γ Mn ( OH )3 c K Mn ( OH )3 c γ H { H } + (13.21) + Differentiating with respect to [H ], equating to zero, rearranging, and changing H + to H opt , the concentration of the hydrogen ion at optimum conditions, the following expression is produced: + + 2 γ H [ H opt ] γ H [ H opt ] Kw - + - = -2 γ Mn ( OH )c K MnOHc K w γ Mn ( OH )3 c K Mn ( OH )3 c γ H γ Mn K w The value of [Hopt] may be solved by trial error © 2003 by A P Sincero and G A Sincero (13.22) TX249_frame_C13.fm Page 611 Friday, June 14, 2002 2:32 PM 2+ 2+ These tables show that, except for Fe and Mn , the equivalent masses are the same in both the ferrous and manganese reactions Now, let MCaOFe = kilograms of lime used for the ferrous reaction MNaOHFe = kilograms of caustic used for the ferrous reaction M Cl2 Fe = kilograms of chlorine used for the ferrous reaction MNaClOFe = kilograms of sodium hypochlorite used in the ferrous reaction M Ca ( ClO )2 Fe = kilograms of calcium hypochlorite used in the ferrous reaction M KMnO Fe = kilograms of potassium permanganate used in the ferrous reaction M O3 Fe = kilograms of ozone used in the ferrous reaction M O2 Fe = kilograms of dissolved oxygen used in the ferrous reaction MCaOMn = kilograms of lime used for the manganous reaction MNaOHMn = kilograms of caustic used for the manganous reaction M Cl2 Mn = kilograms of chlorine used for the manganous reaction MNaClOMn = kilograms of sodium hypochlorite used in the manganous reaction M Ca ( ClO )2 Mn = kilograms of calcium hypochlorite used in the manganous reaction M KMnO4 Mn = kilograms of potassium permanganate used in the manganous reaction M O3 Mn = kilograms of ozone used in the manganous reaction M O2 Mn = kilograms of dissolved oxygen used in the manganous reaction ∆ [ Fe ] mg = mg/L of ferrous to be removed ∆ [ Mn ] mg = mg/L of manganous removed V = cubic meters of water treated Therefore, 28.05∆ [ Fe ] mg V M CaOFe = - = 0.0010∆ [ Fe ] mg V 1000 ( 27.9 ) (13.38) 40∆ [ Fe ] mg V M NaOHFe = - = 0.0014∆ [ Fe ] mg V 1000 ( 27.9 ) (13.39) 35.5∆ [ Fe ] mg V M Cl2 Fe = = 0.00064∆ [ Fe ] mg V 1000 ( 55.8 ) (13.40) 37.25∆ [ Fe ] mg V M NaClOFe = - = 0.00067∆ [ Fe ] mg V 1000 ( 55.8 ) (13.41) 35.78∆ [ Fe ] mg V M Ca ( ClO )2 Fe = - = 0.00064∆ [ Fe ] mg V 1000 ( 55.8 ) (13.42) 52.67∆ [ Fe ] mg V M KMnO Fe = - = 0.00094∆ [ Fe ] mg V 1000 ( 55.8 ) (13.43) 8∆ [ Fe ] mg V M O3 Fe = = 0.00014∆ [ Fe ] mg V 1000 ( 55.8 ) (13.44) © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 612 Friday, June 14, 2002 2:32 PM 8∆ [ Fe ] mg V M O2 Fe = = 0.00014∆ [ Fe ] mg V 1000 ( 55.8 ) (13.45) 28.05∆ [ Mn ] mg V M CaOMn = - = 0.0010∆ [ Mn ] mg V 1000 ( 27.45 ) (13.46) 40.0∆ [ Mn ] mg V M NaOHMn = = 0.0015∆ [ Mn ] mg V 1000 ( 27.45 ) (13.47) 35.5∆ [ Mn ] mg V M Cl2 Mn = = 0.0013∆ [ Mn ] mg V 1000 ( 27.45 ) (13.48) 37.25∆ [ Mn ] mg V M NaClOMn = - = 0.0014∆ [ Mn ] mg V 1000 ( 27.45 ) (13.49) 35.78∆ [ Mn ] mg V M Ca ( ClO )2 Mn = - = 0.0013∆ [ Mn ] mg V 1000 ( 27.45 ) (13.50) 52.67∆ [ Mn ] mg V M KMnO Mn = - = 0.0019∆ [ Mn ] mg V 1000 ( 27.45 ) (13.51) 8.0∆ [ Mn ] mg V M O3 Mn = - = 0.00029∆ [ Mn ] mg V 1000 ( 27.45 ) (13.52) 8.0∆ [ Mn ] mg V M O2 Mn = - = 0.00029∆ [ Mn ] mg V 1000 ( 27.45 ) (13.53) Example 13.4 A raw water contains 2.5 mg/L Mn Calculate the kilograms of lime per cubic meter needed to meet the limit concentration of 0.05 mg/L, if removal is to be done at the high pH range Solution: M CaOMn = 0.0010∆ [ Mn ] mgV = 0.0010 ( 2.5 – 0.05 ) ( ) = 2.45 ( 10 –3 ) kg/m Ans Example 13.5 A raw water contains 40 mg/L iron Calculate the kilograms of lime per cubic meter needed to meet the limit concentration of 0.3 mg/L, if removal is to be done at the high pH range Solution: M CaOFe = 0.0010∆ [ Fe ] mg V = 0.0010 ( 40 – 0.3 ) ( ) = 0.040 kg/m © 2003 by A P Sincero and G A Sincero Ans TX249_frame_C13.fm Page 613 Friday, June 14, 2002 2:32 PM − 13.8 ALKALINITY EXPRESSED IN OH AND ACIDITY + EXPRESSED IN H Knowledge of the equivalent mass is needed in the succeeding discussions on the chemical requirements for pH adjustments, which entails adding acids or bases As acids are added to water and wastewater to lower the pH, they need to react with existing alkalinities first before the pH changes For the purpose of determining the equivalent masses of the acids as a result of the reaction due to this addition, the constituent species of the alkalinity need to be expressed in terms of a single unifying species Once this unifying species has been established, the process of determining the equivalent mass becomes simpler This particular species happens to be the hydroxide ion Similar statements also hold with respect to bases added to raise the pH They need to react first with existing acidities before the pH can increase, and their equivalent masses also need to be determined by referring to a single unifying species This single unifying species is the hydrogen ion We will discuss the hydroxide ion first In the carbonate system, the total alkalinity species has been identified as composed of the constituent species carbonate ion, bicarbonate ion, and hydroxide ion To express all these constituent species in terms of a single parameter, total alkalinity, they need to be expressed in terms of the number of equivalents Thus, if the concentrations are in terms of equivalents per liter, they can all be added to sum up to the total alkalinity, of course, also in terms of equivalents per liter All the concentrations are expressed in terms of equivalents, so any one species may represent the total alkalinity For example, the total alkalinity may be expressed in terms of equivalents of the carbonate ion, or in terms of equivalents of the bicarbonate ion, or in terms of the equivalents of the hydroxide ion The number of equivalents of the hydroxide ion is equal to its number of moles, so it is convenient to expresses total alkalinity in terms of the equivalent hydroxide ion, and the reaction of total alkalinity − may then be simply represented by the reaction of the hydroxide ion OH Without going into a detailed discussion, the total acidity may also be expressed in terms of the equivalent hydrogen ion Thus, the reaction of total acidity may be + represented by the H ion 13.9 CHEMICAL REQUIREMENTS FOR pH ADJUSTMENTS The formulas for the kilograms of carbon dioxide, M CO2 pH , needed to lower the pH and the kilograms of lime, M CaOpH, needed to raise the pH were derived in Chapter 12 For convenience, they are reproduced below: – pH M CO2 pH – pH  10 to – 10 cur  = 22  [ A cadd ] geq = [ A cur ] geq + - V φa   – pH cur – pH to 10 – 10 M CaOpH = 28.05 [ A add ] geq V = 28.05  [ A ccur ] geq + -  V   φb © 2003 by A P Sincero and G A Sincero (13.54) (13.55) TX249_frame_C13.fm Page 614 Friday, June 14, 2002 2:32 PM where, [Acadd] = gram equivalents of acidity to be added [Aadd] = gram equivalents of base to be added [Acur]geq = gram equivalents of current alkalinity [Accur]geq = gram equivalents of current acicity pHto = the destination pH pHcur = current pH = cubic meters of water treated V φa = fractional dissociation of the hydrogen ion from the acid provided φb = fractional dissociation of the hydroxide ion from the base provided Let us derive the formulas for calculating the quantities of sulfuric acid, hydrochloric acid, and nitric acid and the formulas for calculating the quantities of caustic soda and soda ash that may be needed to lower and to raise the pH, respectively To find the equivalent masses of the acids, they must be reacted with the hydroxyl ion Reaction with this ion is necessary, since total alkalinity may be represented by the hydroxyl ion Remember that the acids must first consume all the existing alkalinity represented − in the overall by the OH before they can lower the pH Thus, proceed as follows: − 2− H SO + 2OH → 2Η O + SO − HCl + OH → Η O + Cl (13.56) − − (13.57) − HNO + OH → Η O + NO (13.58) From these equations, the equivalent masses are sulfuric acid = Η2SO4/2 = 49.05, hydrochloric acid = ΗCl/1 = 36.5, and nitric acid = ΗΝΟ3 /1 = 63.01 To find the equivalent masses of the caustic soda and soda ash, they must be reacted with an acid, which may be represented by the hydrogen ion Again, remember that the bases specified previously must first consume all the existing acidity of + the water represented in the overall by H before they can raise the pH Thus, + NaOH + H → H O + Na + + (13.59) Na CO + 2H → H CO + 2Na + (13.60) From the equations, the equivalent masses are caustic soda = NaOH/1 = 40 and soda ash = Na2CO3/2 = × 23 + 12 + 48/2 = 53 Let M H2 SO4 pH , MHClpH, and M HNO pH be the kilograms of sulfuric acid, hydrochloric acid, or nitric acid used to lower the pH from the current pH to pHto Gleaning from Equation (13.54) and the respective equivalent masses of H2SO4, HCl, and HNO3 and the cubic meters, V , of water treated, – pH – pH  10 to – 10 cur  M H2 SO4 pH = 49.05  [ A cadd ] geq = [ A cur ] geq + - V φa   © 2003 by A P Sincero and G A Sincero (13.61) TX249_frame_C13.fm Page 615 Friday, June 14, 2002 2:32 PM – pH M HClpH – pH  10 to – 10 cur  = 36.5  [ A cadd ] geq = [ A cur ] geq + - V φa   – pH (13.62) – pH  10 to – 10 cur  M HNO3 pH = 63.01  [ A cadd ] geq = [ A cur ] geq + - V φa   (13.63) And, also, gleaning from Equation (13.55) and the respective equivalent masses of NaOH and Na2CO3 and the cubic meters of water treated, V , – pH – pH  10 cur – 10 to  = 40.0  [ A ccur ] geq + - V φb   M NaOHpH – pH (13.64) – pH  10 cur – 10 to  M Na2 CO3 pH = 53.0  [ A ccur ] geq + - V φb   (13.65) where ΜNaOHpH and M Na2 CO3 pH are the kilograms of caustic soda or soda ash, respectively, used to raise the pH from the current pH to pHto They are strong acids, therefore, the φa’s for H2SO4, HCl, and HNO3 are unity The hydrogen ion resulting from the second ionization of sulfuric is very small so it can be neglected Also, because NaOH is a strong base, its φb is equal to unity The φb for Na2CO3 is not as straightforward, and we need to calculate it Sodium carbonate ionizes completely, as follows: + 2− Na CO → 2Na + CO (13.66) The carbonate ion then proceeds to react with water to produce the hydroxide ion, as follows (Holtzclaw and Robinson, 1988): 2− CO + HOH − HCO + OH − –4 K b = 1.4 ( 10 ) (13.67) Now, proceed to calculate the fractional dissociation of the hydroxide ion from sodium carbonate Begin by assuming a concentration for the carbonate of 0.1 M This will then produce, also, 0.1 M of the carbonate ion By its subsequent reaction with water, however, its concentration at equilibrium will be smaller Let x be the − 2− − concentrations of HCO and OH at equilibrium Then the concentration of CO is 0.1 − x (at equilibrium) Thus, x –4 K b = 1.4 ( 10 ) = 0.1 – x x = 0.00367 © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 616 Friday, June 14, 2002 2:32 PM TABLE 13.10 φa or φb for Acids and Bases Used in Water and Wastewater Treatment Acid or Base φa or φb 0.1 M 1.0 1.0 1.0 0.00422 1.0 0.0367 0.125 H2SO4 HCl HNO3 H2CO3 NaOH Na2CO3 Ca(OH)2 0.5 M 1.0 1.0 1.0 0.00094 1.0 0.0166 0.025 1.0 M 1.0 1.0 1.0 0.000668 1.0 0.0118 0.0125 Therefore, 0.00367 φ b, Na2 CO3 = = 0.0367 0.1 Note that, in the previous calculation, the activity coefficients have been ignored Lastly, find the φb for lime The chemical reaction for lime after it has been slaked is Ca ( OH ) 2+ Ca + 2OH − –6 K sp,Ca ( OH )2 = 7.9 ( 10 ) (13.68) 2+ Begin by assuming a concentration 0.1 M Let x be the concentrations of Ca at − equilibrium; the corresponding concentration of OH will then be 2x at equilibrium Thus, –6 K sp,Ca ( OH )2 = 7.9 ( 10 ) = x ( 2x ) = 4x x = 0.0125 Therefore, 0.0125 φ b, Ca ( OH )2 = - = 0.125 0.1 Table 13.10 shows some other values of φa and φb –4 –4 –4 – 1.4 ( 10 ) + [ 1.4 ( 10 ) ] + ( 1.0 ) [ 1.4 ( 10 ) ] x = - = 0.0083 © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 617 Friday, June 14, 2002 2:32 PM Example 13.6 A raw water containing mg/L of manganese has a pH 4.0 To remove the manganese, the pH needs to be raised to 6.0 The current acidity is 2+ 30 mg/L as CaCO3 [Ca ] = 0.7 mgmol/L Assume the temperature is 25°C and φb of 0.0367 Calculate the amount of soda ash needed Solution: – pH – pH  10 cur – 10 to  M Na2 CO3 pH = 53.0  [ A ccur ] geq + - V φb   10 – pH to = 10 –6 10 –pHcur = 10 gmols/L; –4 gmols/L; V = 1.0 m Therefore, –4 –6  30 10 – 10  M Na2 CO3 pH = 53.0  + - ( ) = 0.175 kg/m 1000 ( 50 ) 0.0367   Ans 13.10 SLUDGE PRODUCTION For convenience, the chemical reactions responsible for the production of sludge solids are summarized next These equations have been derived at various points 2+ − Fe + 2OH → Fe ( OH ) ( s ) ↓ (13.69) 2+ − + Fe + Cl + 3H O → Fe ( OH ) ( s ) ↓ + Cl + 3H (13.70) 1 + + 2+ Fe + KMnO + H O → Fe ( OH ) ( s ) ↓ + MnO ↓ + K + H 3 3 (13.71) 15 2+ + Fe + O + - H O → Fe ( OH ) ( s ) ↓ + 2H 6 (13.72) 2+ + Fe + H O + O → Fe ( OH ) ( s ) ↓ + 2H (13.73) 2+ − Mn + 2OH → Mn ( OH ) ( s ) ↓ 2+ (13.74) − Mn + Cl + 2H O → Mn O ( s ) ↓ + 2Cl + 4H + 2 + 2+ Mn + KMnO + H O → MnO ( s ) ↓ + K + + H 3 3 © 2003 by A P Sincero and G A Sincero (13.75) (13.76) TX249_frame_C13.fm Page 618 Friday, June 14, 2002 2:32 PM TABLE 13.11 Summary of Equivalent Masses Fe Equation (13.69) Eqs (13.70) through (13.73) Eqs (13.74) through (13.78) 2+ 27.9 55.8 — Mn 2+ — — 27.45 Fe(OH)2(s) Fe(OH)3(s) Mn(OH)2(s) MnO2(s) 44.9 — — — 106.8 — — — 44.45 — — 43.45 2+ + Mn + O + H O → Mn O ( s ) ↓ + 2H (13.77) The solids produced from the previous reactions are Fe(OH)2(s), Fe(OH)3(s), Mn(OH)2(s), and MnO2(s) From Equation (13.69), the equivalent mass of Fe(OH)2(s) 2+ is Fe(OH)2/2 = 44.9 and that of Fe is Fe/2 = 27.9 From Eqs (13.70) through (13.73), the equivalent mass of Fe(OH)3(s) is Fe(OH)3(s)/1 = 106.8; the corresponding 2+ equivalent mass of Fe is Fe/1 = 55.8 Equations (13.74) through (13.77) produce 2+ the following equivalent masses: Mn = Mn/2 = 27.45; Mn(OH)2(s) = Mn(OH)2(s)/2 = 44.45 and MnO2(s) = MnO2(s)/2 = 43.45 Table 13.11 summarizes the equivalent masses Let M Fe ( OH )2 FeRem and M Fe ( OH )3 FeRem be the kilogram mass of ferrous hydroxide and ferric hydroxide produced from ∆[Fe]mg milligrams per liter of ferrous iron removed, respectively M Fe ( OH )2 FeRem is produced at the high pH range and M Fe ( OH )3 FeRem is produced at the low pH range Also, let M Mn ( OH )2 MnRem and M MnO MnRem be the kilogram mass of manganous hydroxide and manganic oxide produced from ∆[Mn]mg milligrams per liter of manganese removed, respectively Also, M Mn ( OH )2 MnRem is produced at the high pH range and M MnO MnRem is produced at the low pH range Let the volume of water treated be V cubic meters Using these information and the equivalent masses derived above produce the following: 44.9∆ [ Fe ] mg V M Fe ( OH )2 FeRem = = 0.0016∆ [ Fe ] mg V 1000 ( 27.9 ) (13.78) 106.8∆ [ Fe ] mg V M Fe ( OH )3 FeRem = - = 0.0019∆ [ Fe ] mg V 1000 ( 55.8 ) (13.79) 44.45∆ [ Mn ] mg V M Mn ( OH )2 MnRem = - = 0.0016∆ [ Mn ] mg V 1000 ( 27.45 ) (13.80) 43.45∆ [ Mn ] mg V M MnO2 MnRem = - = 0.0016∆ [ Mn ] mg V 1000 ( 27.45 ) (13.81) © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 619 Friday, June 14, 2002 2:32 PM Example 13.7 A raw water contains 2.5 mg/L Mn Calculate the solids produced per cubic meter of water treated The effluent is to contain 0.05 mg/L of manganese and the removal is to be done at the high pH range Solution: M Mn ( OH )2 MnRem = 0.0016∆ [ Mn ] mg V –3 = 0.0016 ( 2.5 – 0.05 ) ( ) = 3.92 ( 10 ) kg/m Ans GLOSSARY Complex ions—An ion formed in which a central atom or ion is bonded to a number of surrounding atoms, ions, or molecules Electronic configuration—The arrangement by which electrons occupy orbitals of an atom Equivalent mass—Mass divided by the number of reference species High pH range—The optimum pH range from 11.5 to 12.5 Low pH range—The optimum pH range from 5.5 to 12.5 Orbitals—A volume of space in an atom where electrons of specific energy occupy Oxidation—The removal of electrons from the valence shell of an atom Practical optimum pH range—A range of pH values at which unit process can be operated to produce satisfactory results Reduction—The addition of electrons to the valence shell of an atom Reference species—The number of moles of electrons involved or the number of positive (or negative) moles of charges or oxidation states in a reactant involved in a given chemical reaction Sludge—A mixture of solids and liquids in which the mixture behaves more like a solid than a liquid SYMBOLS Aadd [Aadd]geq Acadd [Acadd]geq Acur [Acur]geq ∆[Fe]mg + FeOH 2+ FeOH + Fe ( OH ) − Fe ( OH ) − Fe ( OH ) Alkalinity to be added to water Gram equivalents per liter of alkalinity to be added to water Acidity to be added to water Gram equivalents per liter of acidity to be added to water Current alkalinity of water Gram equivalents per liter of current alkalinity of water mg/L of ferrous to be removed Monohydroxo Fe(II) complex ion Monohydroxo Fe(III) complex ion Dihydroxo Fe(III) complex ion Trihydroxo Fe(II) complex ion Tetrahydroxo Fe(III) complex ion © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 620 Friday, June 14, 2002 2:32 PM 4+ Fe2 ( OH ) KFeOHc K Fe ( OH )2 c K Fe ( OH )3 c K Fe ( OH )4 c K Fe2 ( OH )2 c KMnOHc K Mn ( OH )2 c K Mn ( OH )3 c K sp,Fe ( OH )2 K sp,Fe ( OH )3 K sp,Mn ( OH )2 M Ca ( ClO )2 Fe M Ca ( ClO )2 Mn MCaOFe MCaOMn M Cl2 Fe M Cl2 Mn M Fe ( OH )2 FeRem M Fe ( OH )3 FeRem M KMnO Fe M KMnO Mn M Mn ( OH )2 MnRem M MnO MnRem MNaClOFe MNaClOMn MNaOHFe MNaOHMn ∆[Mn]mg M O2 Fe M O2 Mn M O3 Fe M O3 Mn + Mn(OH) Mn ( OH ) − Mn ( OH ) ΜCaOpH M CO2 pH MHClpH M H2 SO4 pH M Na2 CO3 pH MNaOHpH spFeII Dihydroxo 2-Fe(III) complex ion + 2+ Equilibrium constant of FeOH and FeOH + Equilibrium constant of Fe ( OH ) − Equilibrium constant of Fe ( OH ) − Equilibrium constant of Fe ( OH ) 4+ Equilibrium constant of Fe2 ( OH ) + Equilibrium constant of Mn ( OH ) Equilibrium constant of Mn ( OH ) − Equilibrium constant of Mn ( OH ) Solubility product constant of Fe ( OH ) ( s ) Solubility product constant of Fe ( OH ) ( s ) Solubility product constant of Mn ( OH ) ( s ) Kilograms of calcium hypochlorite used in the ferrous reaction Kilograms of calcium hypochlorite used in the manganous reaction Kilograms of lime used for the ferrous reaction Kilograms of lime used for the manganous reaction Kilograms of chlorine used for the ferrous reaction Kilograms of chlorine used for the manganous reaction Kilograms of Fe(OH)2 precipitated from removal of ferrous Kilograms of Fe(OH)3 precipitated from removal of ferrous Kilograms of potassium permanganate used in the ferrous reaction Kilograms of potassium permanganate used in the manganous reaction Kilograms of Mn(OH)2 precipitated from removal of manganese Kilograms of MnO2 precipitated from removal of manganese Kilograms of sodium hypochlorite used in the ferrous reaction Kilograms of sodium hypochlorite used in the manganous reaction Kilograms of caustic used for the ferrous reaction Kilograms of caustic used for the manganous reaction mg/L of manganous to be removed Kilograms of dissolved oxygen used in the ferrous reaction Kilograms of dissolved oxygen used in the manganous reaction Kilograms of ozone used in the ferrous reaction Kilograms of ozone used in the manganous reaction Monohydroxo Mn(II) complex ion Dihydroxo Mn(II) complex ion Trihydroxo Mn(II) complex ion Kilograms of lime needed to raise pH Kilograms of carbon dioxide needed to lower pH Kilograms of hydrochloric acid needed to lower pH Kilograms of sulfuric acid needed to lower pH Kilograms of soda ash needed to raise pH Kilograms of caustic soda needed to raise pH Species containing the Fe(II) species © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 621 Friday, June 14, 2002 2:32 PM spFeIII spMn V γH γFeII γFeIII γFeOHc γ Fe ( OH )2 c γ Fe ( OH )3 c γ Fe ( OH )4 c γ Fe2 ( OH )2 c γMn γMn(OH)c γ Mn ( OH )3 c Species containing the Fe(III) species Species containing the Mn(II) species Cubic meters of water treated Activity coefficient of the hydrogen ion Activity coefficient of the ferrous ion Activity coefficient of the ferric ion + 2+ Activity coefficient of FeOH and FeOH + Activity coefficient of Fe ( OH ) − Activity coefficient of Fe ( OH ) − Activity coefficient of Fe ( OH ) 4+ Activity coefficient of Fe2 ( OH ) Activity coefficient of Mn(II) ion + Activity coefficient of Mn(OH) − Activity coefficient of Mn ( OH ) PROBLEMS 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 −6 The concentration spFeIII is determined to be 1.59(10 ) mg/L Assuming the dissolved solids are 140 mg/L, calculate the optimum pH −3 The concentration spFeII is determined to be 6.981(10 ) mg/L Assuming the dissolved solids are 140 mg/L, calculate the optimum pH −3 The concentration spFeII is determined to be 6.981(10 ) mg/L The dissolved solids are 140 mg/L and the temperature is 25°C Calculate the optimum pH −3 The concentration spFeII is determined to be 6.981(10 ) mg/L The dissolved solids are 140 mg/L and the temperature is 25°C Although the solubility product is known at this temperature, calculate it using values of the other parameters The optimum pH is 11.95 −3 The concentration spFeII is determined to be 6.981(10 ) mg/L The dissolved solids are 140 mg/L and the temperature is 25°C Although the activity coefficient of the hydrogen ion can be obtained from the knowledge of the dissolved solids content, calculate it using values of the other parameters The optimum pH is 11.95 −3 The concentration spFeII is determined to be 6.981(10 ) mg/L The dissolved solids are 140 mg/L and the temperature is 25°C Although the + activity coefficient of FeOH ion can be obtained from the knowledge of the dissolved solids content, calculate it using values of the other parameters The optimum pH is 11.95 The optimum pH for precipitating Mn(OH)2 is 11.97 and the dissolved solids content of the water is 140 mg/L The temperature is 25°C Calculate the activity coefficient of the hydrogen ion The data on dissolved solids may be used to calculate other parameters but not for the activity coefficient of the hydrogen ion The optimum pH for precipitating Mn(OH)2 is 11.97 and the dissolved solids content of the water is 140 mg/L The temperature is 25°C Calculate © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 622 Friday, June 14, 2002 2:32 PM 13.9 13.10 13.11 13.12 13.13 13.14 13.15 13.16 13.17 13.18 the ion product of water The given temperature may be used to calculate other parameters but not for the ion product The optimum pH for precipitating Mn(OH)2 is 11.97 and the dissolved solids content of the water is 140 mg/L The temperature is 25°C Calculate the activity coefficient of the Mn(II) ion The data on dissolved solids may be used to calculate other parameters but not for the activity coefficient of the manganese ion The optimum pH for precipitating Mn(OH)2 is 11.97 and the dissolved solids content of the water is 140 mg/L The temperature is 25°C Calculate the − equilibrium constant of Mn ( OH ) The given temperature may used to calculate other parameters but not for the equilibrium constant of − Mn ( OH ) The optimum pH for precipitating Mn(OH)2 is 11.97 and the dissolved solids content of the water is 140 mg/L The temperature is 25°C Calcu− late the activity coefficient of the Mn ( OH ) complex ion The data on dissolved solids may be used to calculate other parameters but not for the activity coefficient of Mn(OH)3 The optimum pH for precipitating Mn(OH)2 is 11.97 and the dissolved solids content of the water is 140 mg/L The temperature is 25°C Calculate + the equilibrium constant of Mn(OH) The given temperature may used to + calculate other parameters but not for the equilibrium constant of Mn(OH) The optimum pH for precipitating Mn(OH)2 is 11.97 and the dissolved solids content of the water is 140 mg/L The temperature is 25°C Calcu+ late the activity coefficient of the Mn(OH) complex ion The data on dissolved solids may be used to calculate other parameters but not for the + activity coefficient of Mn(OH) At a pH of 11.97 and a dissolved solids of 140 mg/L, the concentration of spMn is 0.0179 mg/L Calculate the solubility product constant of Mn(OH)2(s) The temperature is 25°C This given temperature may be used to calculate other parameters but not for the solubility product constant At a pH of 11.97 and dissolved solids of 140 mg/L, the concentration of spMn is 0.0179 mg/L Calculate the activity coefficient of the hydrogen ion The temperature is 25°C The data on dissolved solids may be used to calculate other parameters but not to calculate the activity coefficient of the hydrogen ion At a pH of 11.97 and dissolved solids of 140 mg/L, the concentration of + spMn is 0.0179 mg/L Calculate the equilibrium constant of Mn(OH) The temperature is 25°C This given temperature may be used to calculate + other parameters but not the equilibrium constant of Mn(OH) At a pH of 11.97 and dissolved solids of 140 mg/L, the concentration of + spMn is 0.0179 mg/L Calculate the activity coefficient of Mn(OH) The temperature is 25°C The data on dissolved solids may be used to calculate + other parameters but not to calculate the activity coefficient of Mn(OH) At a pH of 11.97 and dissolved solids of 140 mg/L, the concentration of − spMn is 0.0179 mg/L Calculate the equilibrium constant of Mn ( OH ) © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 623 Friday, June 14, 2002 2:32 PM 13.19 13.20 13.21 13.22 13.23 13.24 13.25 13.26 13.27 13.28 13.29 13.30 13.31 13.32 13.33 13.34 The temperature is 25°C This given temperature may be used to calculate − other parameters but not the equilibrium constant of Mn ( OH ) At a pH of 11.97 and dissolved solids of 140 mg/L, the concentration of − spMn is 0.0179 mg/L Calculate the activity coefficient of Mn ( OH ) The temperature is 25°C The data on dissolved solids may be used to calculate − other parameters but not to calculate the activity coefficient of Mn ( OH ) At a pH of 11.97 and dissolved solids of 140 mg/L, the concentration of spMn is 0.0179 mg/L Calculate the ion product of water The temperature is 25°C This given temperature may be used to calculate other parameters but not the ion product of water The kilograms of lime per cubic meter needed to meet the limit concentration of 0.05 mg/L is 0.00071 This removal is done at the high pH range Calculate the concentration of manganese in the raw water The kilograms of lime per cubic meter needed to meet the limit concentration of 0.05 mg/L is 0.0014 This removal is done at the high pH range and the concentration of manganese in the raw water is 2.5 mg/L Calculate the volume of water treated A raw water contains 2.5 mg/L Mn Calculate the kilograms of caustic used per cubic meter needed to meet the limit concentration of 0.05 mg/L A raw water contains 2.5 mg/L Mn Calculate the kilograms of chlorine per cubic meter needed to meet the limit concentration of 0.05 mg/L A raw water contains 2.5 mg/L Mn Calculate the kilograms of sodium hypochlorite per cubic meter needed to meet the limit concentration of 0.05 mg/L A raw water contains 2.5 mg/L Mn Calculate the kilograms of calcium hypochlorite per cubic meter needed to meet the limit concentration of 0.05 mg/L A raw water contains 2.5 mg/L Mn Calculate the kilograms of ozone per cubic meter needed to meet the limit concentration of 0.05 mg/L A raw water contains 2.5 mg/L Mn Calculate the kilograms of dissolved oxygen per cubic meter needed to meet the limit concentration of 0.05 mg/L A raw water contains 40 mg/L ferrous Calculate the kilograms of caustic used per cubic meter needed to meet the limit concentration of 0.3 mg/L A raw water contains 40 mg/L ferrous Calculate the kilograms of chlorine per cubic meter needed to meet the limit concentration of 0.3 mg/L A raw water contains 40 mg/L ferrous Calculate the kilograms of sodium hypochlorite per cubic meter needed to meet the limit concentration of 0.3 mg/L A raw water contains 40 mg/L ferrous Calculate the kilograms of calcium hypochlorite per cubic meter needed to meet the limit concentration of 0.3 mg/L A raw water contains 40 mg/L ferrous Calculate the kilograms of ozone per cubic meter needed to meet the limit concentration of 0.3 mg/L A raw water contains 40 mg/L ferrous Calculate the kilograms of dissolved oxygen per cubic meter needed to meet the limit concentration of 0.3 mg/L © 2003 by A P Sincero and G A Sincero TX249_frame_C13.fm Page 624 Friday, June 14, 2002 2:32 PM 13.35 A raw water containing mg/L of manganese has a pH 4.0 To remove the manganese, the pH needs to be raised to 6.0 The current acidity is 30 mg/L as CaCO3 Calculate the amount of lime needed 13.36 A raw water containing mg/L of manganese has a pH 4.0 To remove the manganese, the pH needs to be raised to 6.0 The current acidity is 30 mg/L as CaCO3 Calculate the amount of caustic needed 13.37 A raw water contains 2.5 mg/L Mn Calculate the solids produced per cubic meter of water treated The effluent is to contain 0.05 mg/L of manganese and the removal is to be done at the low pH range using chlorine to oxidize the manganous ion 13.38 A raw water contains 2.5 mg/L Mn Calculate the solids produced per cubic meter of water treated The effluent is to contain 0.05 mg/L of manganese and the removal is to be done at the low pH range using dissolved oxygen to oxidize the manganous ion 13.39 A raw water contains 2.5 mg/L Mn Calculate the solids produced per cubic meter of water treated The effluent is to contain 0.05 mg/L of manganese and the removal is to be done at the low pH range using ozone to oxidize the manganous ion BIBLIOGRAPHY Admakin, L A., et al (1998) Prospects for utilization of activated anthracites in technology for cleaning the waste waters Koks i Khimiya/Coke Chem., 8, 30–32 American Water Works Association (1990) Alternative Oxidants for the Removal of Soluble Iron and Manganese Am Water Works Assoc., Denver American Water Works Association (1984) Softening, Iron Removal, and Manganese Removal Am Water Works Assoc., Denver AWI MARKETING Anglian House, Ambury Road, Huntingdon, Cambs PE18 6NZ, England As obtained from the Internet, July 1998 Burns, F L (1997) Case study: Automatic reservoir aeration to control manganese 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for the removal of iron as Fe(OH)2 and for the removal of + manganese as Mn(OH)2 is the formation of the complex ions These ions are FeOH + − − and Fe (OH) for the... added to water Gram equivalents per liter of alkalinity to be added to water Acidity to be added to water Gram equivalents per liter of acidity to be added to water Current alkalinity of water Gram... Wahlberg (1997) Upgrading of waterworks with a new biooxidation process for removal of manganese and iron Water Science Technol., Proc 1997 Int Conf on Upgrading of Water and Wastewater Syst., May 25–28,