TX249_frame_C14.fm Page 627 Friday, June 14, 2002 2:34 PM 14 Removal of Phosphorus by Chemical Precipitation Phosphorus is a very important element that has attracted much attention because of its ability to cause eutrophication in bodies of water For example, tributaries from as far away as the farmlands of New York feed the Chesapeake Bay in Maryland and Virginia Because of the use of phosphorus in fertilizers for these farms, the bay receives an extraordinarily large amount of phosphorus input that has triggered excessive growths of algae in the water body Presently, large portions of the bay are eutrophied Without any doubt, all coves and little estuaries that are tributaries to this bay are also eutrophied Thus, it is important that discharges of phosphorus be controlled in order to avert an environmental catastrophe In fact, the eutrophication of the Chesapeake Bay and the clogging of the Potomac River by blue greens are two of the reasons for the passage of the Federal Water Pollution Control Act Amendments of 1972 This chapter discusses the removal of phosphorus by chemical precipitation It first discusses the natural occurrence of phosphorus, followed by a discussion on the modes of removal of the element The chemical reactions of removal, unit operations of removal, chemical requirements, optimum pH range of operation, and sludge production are all discussed The chemical precipitation method employed uses alum, lime, and the ferric salts, FeCl3 and Fe2(SO4)3 14.1 NATURAL OCCURRENCE OF PHOSPHORUS The element phosphorus is a nonmetal It belongs to Group VA in the Periodic Table in the third period Its electronic configuration is [Ne]3s 23p [Ne] means that the neon configuration is filled The valence configuration represented by the 3, the M shell, shows five electrons in the orbitals: electrons in the s orbitals and electrons in the p orbitals This means that phosphorus can have a maximum oxidation state of +5 The commonly observed oxidation states are 3−, 3+, and 5+ Phosphorus is too active a nonmetal to be found free in nature Our interest in its occurrence is the form that makes it as fertilizer to plants As a fertilizer, it must be in the form of orthophosphate Phosphorus occurs in three phosphate forms: orthophosphate, condensed phosphates (or polyphosphates), and organic phosphates Phosphoric acid, being triprotic, forms three series of salts: dihydrogen − 2− phosphates containing the H PO ion, hydrogen phosphate containing the HPO 3− ion, and the phosphates containing the PO ion These three ions collectively are called orthophosphates As orthophosphates, the phosphorus atom exists in its highest possible oxidation state of 5+ As mentioned, phosphorus can cause eutrophication in receiving streams Thus, concentrations of orthophosphates should be controlled through removal before discharging the wastewater into receiving bodies © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 628 Friday, June 14, 2002 2:34 PM 628 of water The orthophosphates of concern in wastewater are sodium phosphate (Na3PO4), sodium hydrogen phosphate (Na2HPO4), sodium dihydrogen phosphate (NaH2PO4), and ammonium hydrogen phosphate [(NH4)2HPO4] They cause the problems associated with algal blooms When phosphoric acid is heated, it decomposes, losing molecules of water forming the P–O–P bonds.The process of losing water is called condensation, thus the term condensed phosphates and, since they have more than one phosphate group in the molecule, they are also called polyphosphates Among the acids formed from the condensation of phosphoric acid are dipolyphosphoric acid or pyrophosphoric acid (H4P2O7, oxidation state = 5+), tripolyphosphoric acid (H5P3O10, oxidation state = 5+), and metaphosphoric acid [(HPO3)n, oxidation state = 5+] Condensed phosphates undergo hydrolysis in aqueous solutions and transform into the orthophosphates Thus, they must also be controlled Condensed phosphates of concern in wastewater are sodium hexametaphosphate (NaPO3)6, sodium dipolyphosphate (Na4P2O7), and sodium tripolyphosphate (Na5P3O10) When organic compounds containing phosphorus are attacked by microorganisms, they also undergo hydrolysis into the orthophosphate forms Thus, as with all the other phosphorus species, they have to be controlled before wastewaters are discharged Figure 14.1 shows the structural formulas of the various forms of phosphates Note that the oxygen not bonded to hydrogen in orthophosphoric acid, trimetaphosphoric Orthophosphoric acid Trimetaphosphoric acid Dipolyphosphoric acid or pyrophosphoric acid Tripolyphosphoric acid Organic backbone Organic phosphate FIGURE 14.1 Structural formulas of various forms of phosphates © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 629 Friday, June 14, 2002 2:34 PM 629 acid, dipolyphosphoric acid, and tripolyphosphoric acid has a single bond with the central phosphorus atom Oxygen has six electrons in its valence shell, therefore, this indicates that phosphorus has shared two of its own valence electrons to oxygen without oxygen sharing any of its electrons to phosphorus The acceptance by oxygen of these phosphorus electrons, completes its valence orbitals to the required eight electrons for stability All of the hydrogen atoms are ionizable This means that the largest negative charge for the complete ionization of orthophosphoric acid is 3−; that of trimetaphosphoric acid is also 3− Dipolyphosphoric acid will have 4− as the largest negative charge and tripolyphosphoric acid will have 5− as the largest negative charge The charges in organic phosphates depend upon the organic backbone the phosphates are attached to and how many of the phosphate radicals are being attached The phosphorus concentration in domestic wastewaters varies from to 15 mg/L and that in lake surface waters, from 0.01 to 0.04 mg/L all measured as P These values include all the forms of phosphorus 14.2 MODES OF PHOSPHORUS REMOVAL Again, as in previous chapters, the best place to investigate for determining the modes of removal is the table of solubility products constants as shown in Table 14.1 A precipitation product that has the lowest Ksp means that the substance is the most insoluble As shown in the table, the phosphate ion can be precipitated using a calcium precipitant producing either Ca5(PO4)3(OH)(s) or Ca3(PO4)2 Of these two −55.9 precipitates, Ca5(PO4)3(OH)(s) has the smaller Ksp of 10 ; thus, it will be used as the criterion for the precipitation of phosphates Ca5(PO3)3(OH)(s) is also called calcium hydroxy apatite As shown in the table, the other mode of precipitation possible is through precipitating the phosphate ion as AlPO4(s) and FePO4 The precipitant normally used in this instances are alum and the ferric salts (ferric chloride and ferric sulfate), respectively TABLE 14.1 Solubility Product Constants for Phosphate Precipitation Precipitation Product Ca5(PO3)3(OH)(s) Ca3(PO4)2 AlPO4(s) FePO4 © 2003 by A P Sincero and G A Sincero Solubility Product, Ksp at 25°C −55.9 (10 ) −25 (10 ) −21 (10 ) −21.9 (10 ) TX249_frame_C14.fm Page 630 Friday, June 14, 2002 2:34 PM 630 14.3 CHEMICAL REACTION OF THE PHOSPHATE ION WITH ALUM To precipitate the phosphate ion as aluminum phosphate, alum is normally used The chemical reaction is shown next: 3− 3+ Al + PO AlPO ( s ) ↓ 3− AlPO ( s ) ↓ Al + PO K sp,AlPO4 = 10 (14.1) −21 As shown in these reactions, the phosphorus must be in the phosphate form The reaction occurs in water, so the phosphate ion originates a series of equilibrium orthophosphate reactions with the hydrogen ion This series is shown as follows (Snoeyink and Jenkins): 3− PO + H + 2− HPO 2− ⇒ HPO 2− HPO + H Κ HPO4 = 10 −12.3 (14.2) H PO − − + − + ⇒ H PO H PO + H 3− PO + H + 2− + Κ H2 PO4 = 10 −7.2 − + Κ H3 PO4 = 10 −2.1 HPO + H (14.3) H PO ⇒ H PO H PO + H (14.4) Let sp PO4 Al represent the species in solution containing the PO4 species of the orthophosphates, using alum as the precipitant Therefore, 3− 2− − [ sp PO4 Al ] = [ PO ] + [ HPO ] + [ H PO ] + [ H PO ] (14.5) 3− Express Equation (14.5) in terms of [ PO ] using Eqs (14.2) through (14.4) This 3+ will enable [ sp PO4 Al ] to be expressed in terms of [Al ] using Equation (14.1) and K sp,AlPO4 Proceed as follows: + 2− 3− + γ PO4 γ H [ PO ] [ H ] { HPO } { PO } { H } [ HPO ] = = - = -γ HPO γ HPO K HPO γ HPO4 K HPO4 3− 2− + + − 2− γ PO4 γ H [ PO ] [ H ] { H PO } { HPO } { H } [ H PO ] = = - = -γ H2 PO4 γ H2 PO4 K H2 PO4 γ H2 PO4 K H2 PO4 K HPO − © 2003 by A P Sincero and G A Sincero 3− (14.6) (14.7) TX249_frame_C14.fm Page 631 Friday, June 14, 2002 2:34 PM + + − γ PO4 γ H [ PO ] [ H ] { H PO } { H } [ H PO ] = { H PO } = = -K H3 PO4 K H3 PO4 K H2 PO4 K HPO 3− (14.8) Equations (14.6) through (14.8) may now be substituted into Equation (14.5) to produce + + γ PO4 γ H [ PO ] [ H ] γ PO4 γ H [ PO ] [ H ] [ sp PO4 Al ] = [ PO ] + + -γ HPO K HPO γ H2 PO4 K H2 PO4 K HPO4 3− 3− 3− + γ PO4 γ H [ PO ] [ H ] + -K H3 PO4 K H2 PO4 K HPO 3− (14.9) But, from Equation (14.1), [ PO ] = { PO }/ γ PO4 = K sp,AlPO4 /γ PO4 { Al } = K sp,AlPO4 / 3+ γ PO4 γ Al [ Al ] Substituting, 3− 3− 3+ + + γ H K sp,AlPO4 [ H ] γ H K sp,AlPO4 [ H ] K sp,AlPO4 [ sp PO4 Al ] = - + + -3+ 3+ 3+ γ PO4 γ Al [ Al ] γ HPO K HPO γ Al [ Al ] γ H2 PO4 K H2 PO4 K HPO4 γ Al [ Al ] + γ H K sp,AlPO4 [ H ] + 3+ K H3 PO4 K H2 PO4 K HPO γ Al [ Al ] (14.10) γ PO4, γAl, γH, γ HPO , and γ H2 PO4 are, respectively, the activity coefficients of the phosphate, aluminum, hydrogen, hydrogen phosphate, and dihydrogen phosphate ions K HPO , K H2 PO4, and K H3 PO4 are, respectively, the equilibrium constants of the hydrogen phosphate and dihydrogen phosphate ions and phosphoric acid 3+ [Al ] needs to be eliminated for the equation to be expressed solely in terms of + [H ] When alum is added to water, it will unavoidably react with the existing natural alkalinity For this reason, the aluminum ion will not only react with the phosphate ion − to precipitate AlPO4(s), but it will also react with the OH to precipitate Al(OH)3(s) Also, 3+ 2+ − 4+ 5+ Al will form complexes Al(OH) , Al (OH) 17 , Al 13 (OH) 34 , Al(OH) , and 4+ Al (OH) in addition to the Al(OH)3(s) All these interactions complicate our objective 3+ of eliminating [Al ] Consider, however, the equilibrium of Al(OH)3(s), which is as follows: Al(OH) ( s ) −33 3+ Al + 3OH − K sp,Al(OH) = 10 −21 −33 (14.11) −33 K sp,Al(OH) = 10 may be compared with K sp,AlPO4 = 10 From K sp,Al(OH) = 10 , 3+ the concentration of Al needed to precipitate Al(OH)3 may be calculated to be −9 −21 2.0(10 ) gmol/L Doing similar calculation from K sp,AlPO4 = 10 , the concentration 3+ −11 of Al needed to precipitate AlPO4 is 3.0(10 ) gmol/L These two concentrations © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 632 Friday, June 14, 2002 2:34 PM are so close to each other that it may be concluded that Al(OH)3 and AlPO4 are 3+ coprecipitating This finding allows us to eliminate [Al ] Thus, + 3+ K sp,Al ( OH )3 K sp,Al ( OH )3 γ H [ H ] { Al } - [ Al ] = = = -3 − γ Al γ Al K w γ Al { OH } 3+ Substituting into Equation (14.10), 3 K sp,AlPO4 K w K sp,AlPO4 K w [ sp PO4 Al ] = -3 + -3 + + γ PO4 K sp,Al(OH) γ H [ H ] γ HPO4 K HPO4 K sp,Al(OH) γ H [ H ] 3 K sp,AlPO4 K w K sp,AlPO4 K w + + + γ H2 PO4 K H2 PO4 K HPO4 K sp,Al(OH) γ H [ H ] K H3 PO4 K H2 PO4 K HPO K sp,Al(OH) (14.12) + Equation (14.12) portrays the equilibrium relationship between [ sp PO4 ] and [H ] Example 14.1 Calculate [ sp PO4 Al ] when the pH is 10 Assume the water contains 140 mg/L of dissolved solids Solution: 3 K sp,AlPO4 K w K sp,AlPO4 K w [ sp PO4 Al ] = -3 + -3 + + γ PO4 K sp,Al(OH) γ H [ H ] γ HPO4 K HPO4 K sp,Al(OH) γ H [ H ] 3 K sp,AlPO4 K w K sp,AlPO4 K w + + + γ H PO K H PO K HPO K sp,Al(OH) γ H [ H ] K H3 PO4 K H2 PO4 K HPO4 K sp,Al(OH) 4 K sp,AlPO4 = 10 −21 µ = 2.5 ( 10 )TDS γ = 10 –5 0.5z i ( µ ) – 1+1.14 ( µ ) µ = 2.5 ( 10 ) ( 140 ) = 3.5 ( 10 ) –5 –3 γ PO4 = 10 K sp,Al(OH) = 10 – 33 γ HPO4 = 10 γ H = γ H2 PO4 = 10 –3 – 0.5 ( ) [ 3.5 ( 10 ) ] – -–3 1+1.14 [ 3.5 ( 10 ) ] = 0.56 –3 0.5 ( ) [ 3.5 ( 10 ) ] − –3 1+1.14 [ 3.5 ( 10 ) ] = 0.94 –3 0.5 ( ) [ 3.5 ( 10 ) ] − –3 1+1.14 [ 3.5 ( 10 ) ] = 0.77 K HPO = 10 © 2003 by A P Sincero and G A Sincero – 12.3 K H2 PO4 = 10 – 7.2 K H3 PO4 = 10 – 2.1 TX249_frame_C14.fm Page 633 Friday, June 14, 2002 2:34 PM Therefore, – 14 – 21 – 14 – 21 ( 10 ) ( 10 ) ( 10 ) ( 10 ) [ sp PO4 Al ] = + – 33 – 10 – 12.3 – 33 – 10 ( 0.56 ) ( 10 ) ( 0.94 ) [ 10 ] ( 0.77 ) ( 10 ) ( 10 ) ( 0.94 ) [ 10 ] – 14 – 21 ( 10 ) ( 10 ) + -– 7.2 – 12.3 – 33 – 10 ( 0.94 ) ( 10 ) ( 10 ) ( 10 ) ( 0.94 ) [ 10 ] – 14 – 21 ( 10 ) ( 10 ) + -– 2.1 – 7.2 – 12.3 −33 ( 10 ) ( 10 ) ( 10 ) ( 10 ) – 63 – 63 – 63 – 63 1.0 ( 10 ) 1.0 ( 10 ) 1.0 ( 10 ) 1.0 ( 10 ) = + + + -– 64 – 66 – 63 – 55 4.65 ( 10 ) 3.41 ( 10 ) 2.79 ( 10 ) 2.51 ( 10 ) = 295.76 gmols/L = 9.17 ( 10 ) mg/L as P +6 The answer of 9.17(10 ) mg/L emphasizes a very important fact: phosphorus cannot be removed at alkaline conditions It will be shown in subsequent discussions that the solution conditions must be acidic for effective removal of phosphorus using alum 14.3.1 DETERMINATION OF THE OPTIMUM pH RANGE Equation (14.12) shows that [ sp PO4 Al ] is a function of the hydrogen ion concentration This means that the concentration of the species containing the PO4 species of the orthophosphates is a function of pH If the equation is differentiated and the result equated to zero, however, an optimum value cannot be guaranteed to be found A range of pH values can, however, be assigned and the corresponding values of sp PO4 Al calculated By inspection of the result, the optimum range can be determined Tables 14.2 and 14.3 show the results of assigning this range of pH and values of sp PO4 Al calculated using Equation (14.12) These tables show that optimum removal of phosphorus using alum results when the unit is operated at pH values less than 5.0 Note: The dissolved solids content has only a negligible effect on the resulting concentrations 14.4 CHEMICAL REACTION OF THE PHOSPHATE ION WITH LIME Calcium hydroxy apatite contains the phosphate and hydroxyl groups Using calcium hydroxide as the precipitant, the chemical reaction is shown below: 2+ 3− 5Ca + 3PO + OH Ca ( PO ) OH ( s ) ↓ © 2003 by A P Sincero and G A Sincero − Ca ( PO ) OH ( s ) ↓ 2+ 3− 5Ca + PO + OH − K sp,apatite = 10 – 55.9 (14.13) TX249_frame_C14.fm Page 634 Friday, June 14, 2002 2:34 PM TABLE 14.2 Concentration of sp PO4 Al as a Function of pH at 25°C (mg/L as P) pH, Dissolved Solids = 140 mg/L [sp PO4 Al ](mg/L) 10 11 12 13 14 15 6.6(10 ) −5 7.1(10 ) −4 1.2(10 ) −4 6.5(10 ) −3 5.9(10 ) −2 5.9(10 ) −1 6.3(10 ) +1 1.1(10 ) +2 5.4(10 ) +4 4.8(10 ) +6 4.8(10 ) +8 5.8(10 ) +10 8.3(10 ) +13 4.0(10 ) +16 2.3(10 ) +19 3.5(10 ) −5 TABLE 14.3 Concentration of sp PO4 Al as a Function of pH at 25°C (mg/L as P) pH, Dissolved Solids = 35,000 mg/L [sp PO4 Al ] (mg/L) 10 11 12 13 14 15 6.6(10 ) −5 8.0(10 ) −4 2.1(10 ) −3 1.5(10 ) −2 1.5(10 ) −1 1.5(10 ) −0 2.0(10 ) +1 8.9(10 ) +3 7.6(10 ) +5 7.5(10 ) +7 8.2(10 ) +10 1.6(10 ) +12 9.2(10 ) +15 8.5(10 ) +18 8.5(10 ) +21 8.5(10 ) © 2003 by A P Sincero and G A Sincero −5 TX249_frame_C14.fm Page 635 Friday, June 14, 2002 2:34 PM These equations also show that the phosphorus must be in the phosphate form As in the case of alum, the phosphate ion produces the set of reactions given by Eqs (14.2) through (14.4) Let sp PO4 Ca be the species in solution containing the PO4 species, using the calcium in lime as the precipitant [ sp PO4 Ca ] will be the same as given by Equation (14.5) which, along with Eqs (14.2) through (14.4), can be manipulated to produce Equation (14.9) From Equation (14.13), 1/3 K sp,apatite 1/3 1/3 + 1/3 { Ca 2+ } { OH − } K sp,apatite γ H [ H ] { PO } 3− [ PO ] = - = - = 1/3 2+ 5/3 1/3 γ PO4 γ PO4 γ Ca [ Ca ] K w γ PO4 3− Substituting in Equation (14.9) and simplifying, + 1/3 + 4/3 K sp,apatite γ H [ H ] K sp,apatite γ H [ H ] [ sp PO4 Ca ] = - + -1/3 1/3 2+ 5/3 1/3 2+ 5/3 1/3 γ Ca [ Ca ] K w γ PO4 γ HPO K HPO γ Ca [ Ca ] K w 1/3 1/3 1/3 4/3 + 7/3 K sp,apatite γ H [ H ] + -1/3 2+ 5/3 1/3 γ H2 PO4 K H2 PO4 K HPO4 γ Ca [ Ca ] K w 1/3 7/3 + 10/3 K sp,apatite γ H [ H ] + 1/3 2+ 5/3 1/3 K H3 PO4 K H2 PO4 K HPO γ Ca [ Ca ] K w 1/3 10/3 (14.14) γCa and Kw are, respectively, the activity coefficient of the calcium ion and ion product of water 2+ 3+ The [Ca ] in the denominator is analogous to [Al ] in the case of alum; however, 2+ the present case is different in that [Ca ] cannot be eliminated in a straightforward 3+ manner [Al ] was easily eliminated because Al(OH)3(s) was precipitating along with −6 AlPO4 Ca(OH)2(s), which could be precipitating, has a Ksp value of 7.9(10 ) This −55.9 is much, much greater than Ksp,apatite = 10 and calcium hydroxide will not be precipitating along with Ca5(PO4)3OH(s) The other possible precipitate is CaCO3 that is produced when lime reacts with the natural alkalinity of the water; however, −9 calcium carbonate has a Ksp value of 4.8(10 ) Again, this value is much, much greater than Ksp,apatite and calcium carbonate will not be precipitating along with Ca5(PO4)3OH(s), either We will, therefore, let the equation stand and express [ sp PO4 Ca ] 2+ + as a function of [Ca ], along with [H ] and the constants Example 14.2 Calculate [ sp PO4 Ca ] expressed as mg/L of P when the pH is 2+ Assume the water contains 140 mg/L of dissolved solids and that [Ca ] = 130 mg/L as CaCO3 © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 636 Friday, June 14, 2002 2:34 PM Solution: + 1/3 + 4/3 K sp,apatite γ H [ H ] K sp,apatite γ H [ H ] [ sp PO4 Ca ] = - + -2+ 5/3 1/3 2+ 5/3 1/3 1/3 1/3 γ Ca [ Ca ] K w γ PO4 γ HPO K HPO γ Ca [ Ca ] K w 1/3 1/3 1/3 4/3 + 7/3 K sp,apatite γ H [ H ] + -1/3 2+ 5/3 1/3 γ H2 PO4 K H2 PO4 K HPO γ Ca [ Ca ] K w 1/3 7/3 + 10/3 K sp,apatite γ H [ H ] + 1/3 2+ 5/3 1/3 K H3 PO4 K H2 PO4 K HPO γ Ca [ Ca ] K w 1/3 10/3 K sp,apatite = 10 γ H = γ H2 PO4 = 10 – 55.9 γ Ca = γ HPO = 10 –3 0.5 ( ) [ 3.5 ( 10 ) ] – –3 1+1.14 [ 3.5 ( 10 ) ] = 0.94 –3 0.5 ( ) [ 3.5 ( 10 ) ] – –3 1+1.14 [ 3.5 ( 10 ) ] = 0.77 130 2+ –3 [ CA ] = = 3.24 ( 10 ) gmol/L 1000 ( 40.1 ) γ PO4 = 10 K HPO = 10 –3 0.5 ( ) [ 3.5 ( 10 ) ] – –3 1+1.14 [ 3.5 ( 10 ) ] – 12.3 = 0.56 K H2 PO4 = 10 – 7.2 K H3 PO4 = 10 – 2.1 Therefore, – 55.9 1/3 1/3 – 1/3 ( 10 ) ( 0.94 ) [ 10 ] [ sp PO4 Ca ] = 1/3 – 5/3 – 14 1/3 ( 0.77 ) [ 3.24 ( 10 ) ] ( 10 ) ( 0.56 ) – 55.9 1/3 4/3 – 4/3 ( 10 ) ( 0.94 ) [ 10 ] + – 12.3 1/3 – 5/3 – 14 1/3 ( 0.77 ) ( 10 ) ( 0.77 ) [ 3.24 ( 10 ) ] ( 10 ) – 55.9 1/3 7/3 – 7/3 – 55.9 1/3 10/3 ( 10 ) ( 0.94 ) [ 10 ] + – 7.2 – 12.3 1/3 – 5/3 – 14 1/3 ( 0.94 ) ( 10 ) ( 10 ) ( 0.77 ) [ 3.24 ( 10 ) ] ( 10 ) – 10/3 ( 10 ) ( 0.94 ) [ 10 ] + – 2.1 – 7.2 – 12.3 1/3 – 5/3 – 14 1/3 ( 10 ) ( 10 ) ( 10 ) ( 0.77 ) [ 3.24 ( 10 ) ] ( 10 ) – 22 – 30 – 38 – 46 4.91 ( 10 ) 4.62 ( 10 ) 4.34 ( 10 ) 4.08 ( 10 ) = + + + -– 10 – 22 – 29 – 31 7.85 ( 10 ) 5.41 ( 10 ) 4.16 ( 10 ) 3.52 ( 10 ) –9 = 9.58 ( 10 ) gmol/L = 6.25 ( 10 – 13 –4 = 2.97 ( 10 ) mg/L © 2003 by A P Sincero and G A Sincero –9 –9 ) + 8.54 ( 10 ) + 1.04 ( 10 ) + 1.16 ( 10 Ans – 15 ) TX249_frame_C14.fm Page 643 Friday, June 14, 2002 2:34 PM 14.7 EFFECT OF THE Ksp’s ON THE PRECIPITATION OF PHOSPHORUS −36 The worst of the precipitants are the ferric salts Compare K sp,Fe ( OH )3 = 1.1(10 ) and −21.9 K sp,FePO4 = 10 These two Ksp’s produce comparable concentrations of the ferric ion to precipitate either Fe(OH)3 or FePO4 Thus, at high pH conditions, the phos3+ phate ion would have a big competitor in the form of the hydroxide ion An Fe − available in solution is grabbed by the OH ion to form the ferric hydroxide, leaving 3+ less amount of Fe to precipitate ferric phosphate The ferric salts are, therefore, a poor performer for removing phosphorus Also, to be effective requires adjusting the pH to the pH of almost mineral acidity of less than −33 −21 Now, compare K sp,Al ( OH )3 = 10 and K sp,AlPO4 = 10 for the case of using alum This situation is similar to the case of the ferric salts The concentrations of the aluminum ion to precipitate either the aluminum hydroxide or the aluminum phosphate are comparable This means that the hydroxyl ion is also a competitor in the phosphate precipitation The process operates poorly at high pH, although it has a better pH range of equal to or less than compared to that of the ferric salts In the case of using lime to precipitate phosphorus, we know that calcium hydroxide has a much, much larger Ksp than that of the apatite, the required precip−6 itate The former has a Ksp of K sp,Ca ( OH )2 = 7.9(10 ) Compare this to that of apatite −55.9 which is Ksp,apatite = 10 Thus, adding lime to the water means that a large amount of the calcium ion is available to precipitate the apatite, Ca5(PO4)3OH(s) Also, no other precipitate competes with this precipitation Therefore, using lime is the best method for removing phosphorus Adding a quantity of lime to the point of saturation, Ca(OH)2 has the potential of removing phosphorus “100%.” This process must, however, be investigated In the overall, the process using lime entails adjusting the pH to greater than 7, which is much, much better than the range for the use of either alum or ferric salts 14.8 UNIT OPERATIONS FOR PHOSPHORUS REMOVAL The removal of phosphorus involves provision for the addition of chemicals both for adjusting the pH and for precipitating the phosphate, mixing them with the raw water, settling the resulting precipitates, and filtering those precipitates that escape removal by settling The precipitates formed are still in suspension; thus, it is also necessary for them to flocculate by adding a flocculation tank 14.9 CHEMICAL REQUIREMENTS The chemicals required are alum, lime, and the ferric salts FeCl3 and Fe2(SO4)3 These are in addition to the acid or bases needed to adjust the pH The formulas used to calculate the amounts of these chemicals for pH adjustment were already derived in Chapters 11, 12, and 13 The chemical requirements to be discussed here will only be for alum, lime, and the ferric salts © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 644 Friday, June 14, 2002 2:34 PM The amount of alum needed to precipitate the phosphate is composed of the alum required to satisfy the natural alkalinity of the water and the amount needed to precipitate the phosphate Satisfaction of the natural alkalinity will bring the equilibrium of aluminum hydroxide Remember, however, that even if these quantities of alum were provided, the concentration of phosphorus that will be discharged from the effluent of the unit still has to conform to the equilibrium reaction that depends upon the pH level at which the process was conducted The optimum pH, we have found, is equal to or less Consider the full alum formula in the chemical reaction, 3− 2− Al ( SO ) ⋅ xH O + PO → AlPO ( s ) ↓ + SO + xH O (14.19) Note: Only one arrow is used and not the forward and backward arrows The reaction is not in equilibrium As mentioned before, the previous reaction for phosphate removal takes place in conjunction with the Al(OH)3 equilibrium Bicarbonate alkalinity is always present in natural waters, so this equilibrium is facilitated by alum reacting with calcium bicarbonate so that, in addition to the amount of alum required for the precipitation of phosphorus, more is needed to neutralize the bicarbonate The reaction is represented by Al ( SO ) ⋅ xH O + Ca ( HCO ) → Al ( OH ) ↓ + CO + CaSO + xH O (14.20) From the previous reactions, the equivalent masses are alum = Al2(SO4)3 ⋅ xH2O/ = 57.05 + 3x and calcium bicarbonate = 3Ca(HCO3)2/6 = 81.05 The number of 3− moles of phosphorus in PO is one; thus, from the balanced chemical reaction, the equivalent mass of phosphorus is 2P/6 = 10.33 Although the alkalinity found in natural waters is normally calcium bicarbonate, it is conceivable that natural alkalinities could also be associated with other cations Thus, to be specific, as shown by Equation (14.20), the alkalinity discussed here is calcium bicarbonate, with the understanding that “other forms” of natural alkalinities may be equated to calcium bicarbonate alkalinity when expressed in equivalent concentrations Let ACa be this calcium bicarbonate alkalinity The quantity of alum needed to react with the calcium bicarbonate alkalinity is exactly the quantity of alum needed to establish the aluminum hydroxide equilibrium Let [ACa]geq be the gram equivalents per liter of this bicarbonate alkalinity in the raw water of volume V cubic meters and let MAlAlk be the kilograms of alum at a fractional purity of PAl required to react with it Then, [ A Ca ] geq ( 57.05 + 3x )V M AlAlk = P Al (14.21) Remember that alkalinity is also expressed analytically in terms of calcium carbonate in which the equivalent mass is considered 50 Let us digress from our main discussion and explore this matter further in order to determine how the various © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 645 Friday, June 14, 2002 2:34 PM methods of expressing alkalinities are related How will the analytical result be converted to an equivalent concentration to be used in Equation (14.21)? Consider the following reactions: + 2+ CaCO3 + 2H → H2CO3 + CA + (14.22) 2+ 2Ca(HCO3)2 + 2H → H2CO3 + 2Ca (14.23) From these reactions, the equivalent mass of calcium bicarbonate in relation to the equivalent mass of calcium carbonate of 50 is 2Ca(HCO3)2/2 = 162.1 (much different in comparison with 81.05) Now, assume [ A Ca ] geqCaCO3 gram equivalents per liter of calcium bicarbonate alkalinity being computed on the basis of Equation (14.23) This computation, in turn, is obtained from an analytical result based on Equation (14.22) The number of grams of the bicarbonate alkalinity is then equal to 162.1 [ A Ca ] geqCaCO3 The quantity of mass remains the same in any system Thus, referring to Equation (14.20), there will also be the same 162.1 [ A Ca ] geqCaCO3 grams of the bicarbonate there To find the number of equivalents, however, it is always equal to the mass divided by the equivalent mass Therefore, based on Equation (14.20), the corresponding number of equivalents of the 162.1 [ A Ca ] geqCaCO3 grams of calcium bicarbonate alkalinity is 162.1 [ A Ca ] geqCaCO3/81.05 = [ A Ca ] geqCaCO3 And, [ A Ca ] geq = [ A Ca ] geqCaCO3 (14.24) Again, [ A Ca ] geq is the calcium bicarbonate alkalinity referred to the reaction of alum with calcium bicarbonate and [ A Ca ] geqCaCO3 is the same bicarbonate alkalinity referred to calcium carbonate with equivalent mass of 50 This equation converts analytical results expressed as calcium carbonate to a form that Equation (14.21) can use Now, let [Phos]mg be the milligram per liter of phosphorus as P in phosphate and let MAlPhos be the kilograms of alum at fractional purity of PAl required to precipitate this phosphate phosphorus Letting fP represent the fractional removal of phosphorus, the following equation is then produced: –5 f P [ Phos ] mg 9.7 ( 10 ) f P [ Phos ] mg ( 57.05 + 3x ) V M AlPhos = - ( 57.05 + 3x ) V = 1000 ( 10.33 )P Al P Al (14.25) Letting MAl be the total kilograms of alum needed for the removal of phosphorus, ( 57.05 + 3x ) –5 M Al = M AlAlk + M AlPhos = ( [ A Ca ] geq + 9.7 ( 10 ) f P [ Phos ] mg ) V P Al (14.26) Now, let us tackle the problem of calculating the chemical requirements for removing calcium apatite Although calcium carbonate has a very high Ksp that it would not be precipitating with the apatite, the natural alkalinity would have to be © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 646 Friday, June 14, 2002 2:34 PM neutralized nonetheless, producing the carbonate The formation of the calcium carbonate solid will only occur, however, after the apatite formation has completed Thus, the amount of lime needed to precipitate the phosphate is composed of the lime needed to precipitate the phosphate in apatite and the lime needed to neutralize the natural alkalinity of the water Note: Even if these quantities were provided, the concentration of phosphorus that would be discharged from the effluent of the unit still has to conform to the equilibrium reaction that depends upon the pH level at which the process was conducted The optimum pH, we have found, is equal to or greater than The applicable chemical reactions are: Ca(OH)2 + Ca(HCO3)2 → H2O + CaCO3(s) ↓ 3− Ca ( OH ) + PO − Ca ( PO ) OH ( s ) ↓ + OH 5 (14.27) (14.28) To make the previous equations compatible, use as the number of reference species, based on Ca(OH)2 Therefore the equivalent masses are lime = CaO/2 = 37.05; calcium bicarbonate = Ca(HCO3)2/2 = 81.05; and phosphorus = 3/5P/2 = 9.3 In this particular case, because the equivalent mass of calcium bicarbonate is also equal to 81.05, [ A Ca ] geq = [ A Ca ] geqCaCO Let MCaOAlk be the kilograms of lime at fractional purity of PCaO required to react with the calcium carbonate alkalinity Then, 37.05 [ A Ca ] geq V M CaOAlk = P CaO (14.29) Now, letting MCaOPhos be the kilograms of lime required to react with the phosphorus, –3 37.05 f P [ Phos ] mg 4.0 ( 10 ) f P [ Phos ] mg V M CaOPhos = - V = -1000 ( 9.3 )P CaO P CaO (14.30) Let MCaO be the total kilograms of lime needed for the removal of phosphorus Thus, V –3 M CaO = M CaOAlk + M CaOPhos = ( 37.05 [ A Ca ] geq + 4.0 ( 10 ) f P [ Phos ] mg ) P CaO (14.31) Last, let us tackle the calculation of the amount of ferric salts needed to precipitate the phosphate This amount is composed of the amount needed to neutralize the natural alkalinity of the water and the amount needed to precipitate the phosphate Again, remember, that even if these quantities were provided, the concentration of phosphorus that will be discharged from the effluent of the unit still has to conform to the equilibrium reaction that depends upon the pH level at which the process was conducted The optimum pH, we have found, is equal to or less than © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 647 Friday, June 14, 2002 2:34 PM The applicable respective chemical reactions for the satisfaction of the bicarbonate alkalinity are 2FeCl3 + 3Ca(HCO3)2 → 2Fe(OH)3 ↓ + 6CO2 + 3CaCl2 (14.32) Fe2(SO4)3 + 3Ca(HCO3)2 → 2Fe(OH)3 ↓ + 6CO2 + 3CaSO4 (14.33) Upon satisfaction of the previous reactions, the normal precipitation of the phosphate occurs The respective reactions are 3− 2FeCl + 2PO → 2FePO ↓ + Cl 3− − (14.34) 2− Fe ( SO ) + 2PO → 2FePO ↓ + 3SO (14.35) Note that the coefficients of Equation (14.34) have been adjusted to make the equation compatible with Equation (14.32) and all the rest of the above equations From these equations, the equivalent masses are ferric chloride = 2FeCl3/6 = 54.1; calcium bicarbonate = 3Ca(HCO3)2/6 = 81.05; ferric sulfate = Fe2(SO4)3/6 = 66.65; and phosphorus = 2P/6 = 15.5 Since the equivalent mass is still 81.05, [ A Ca ] geq = [ A Ca ] geqCaO Let MFeIIIClAlk and M FeIIISO4 Alk be the kilograms of ferric chloride and ferric sulfate, respectively, required to react with the calcium bicarbonate alkalinity Also, let the respective purity be PFeIIICl and P FeIIISO4 Then, 54.1 [ A Ca ] geq V M FeIIIClAlk = -P FeIIICl (14.36) 66.65 [ A Ca ] geq V M FeIIISO4 Alk = P FeIIISO4 (14.37) Now, letting M FeIIIClPhos and M FeIIISO Phos be the kilograms of ferric chloride and ferric sulfate, respectively, required to react with the phosphorus, –3 3.5 ( 10 ) f P [ Phos ] mg V 54.1 f P [ Phos ] mg M FeIIIClPhos = V = -1000 ( 15.5 )P FeIIICl P FeIIICl (14.38) –3 66.5 f P [ Phos ] mg 4.3 ( 10 ) f P [ Phos ] mg V M FeIIISO Phos = - V = -1000 ( 15.5 )P FeIIISO4 P FeIIISO4 (14.39) Let M FeIIICl and M FeIIISO be the total kilograms of ferric chloride and ferric sulfate, respectively, needed for the removal of phosphorus Thus, V –3 M FeIIICl = M FeIIIClAlk + M FeIIIClPhos = ( 54.1 [ A Ca ] geq + 3.5 ( 10 ) f P [ Phos ] mg ) P FeIIICl (14.40) M FeIIISO = M FeIIISO Alk + M FeIIISO Phos V –3 = ( 66.5 [ A Ca ] geq + 4.3 ( 10 ) f P [ Phos ] mg ) -P FeIIISO © 2003 by A P Sincero and G A Sincero (14.41) TX249_frame_C14.fm Page 648 Friday, June 14, 2002 2:34 PM 14.10 SLUDGE PRODUCTION Phosphorus is normally removed from wastewaters and, as such, the solids produced would include the original suspended solids or turbidity Therefore, two sources contribute to the production of solids in phosphorus removal: the original suspended solids and the solids produced from chemical reaction If V is the cubic meters of water treated and spss are the solids removed at a fractional efficiency of fss, the kilograms of solids produced from the suspended solids is (fss[spss]mg/1000) V , where [spss]mg is in mg/L For convenience, the chemical reactions responsible for production of chemical sludge are reproduced below They were derived previously 3− 2− Al2(SO4)3 ⋅ xH2O + PO → 2AlPO4(s) ↓ + SO + xH2O (14.42) Al2(SO4)3 ⋅ xH2O + 3Ca(HCO3)2 → 2Al(OH)3 ↓ + 6CO2 + 3CaSO4 + xH2O (14.43) Ca(OH)2 + Ca(HCO3)2 → H2O + CaCO3(s) ↓ 3− Ca(OH)2 + PO − Ca5(PO4)3OH(s) ↓ + OH 5 (14.44) (14.45) 2FeCl3 + 3Ca(HCO3)2 → 2Fe(OH)3 ↓ + 6CO2 + 3CaCl2 (14.46) Fe2(SO4)3 + 3Ca(HCO3)2 → 2Fe(OH)3 ↓ + 6CO2 + 3CaSO4 (14.47) 3− − 2FeCl3 + PO → 2FePO4 ↓ + 6Cl 3− (14.48) 2− Fe2(SO4)3 + PO → 2FePO4 ↓ + SO (14.49) Referring to the previous reactions, the solids produced from the removal of phosphorus are AlPO4(s), Al(OH)3, CaCO3, Ca5(PO4)3OH(s), Fe(OH)3, and FePO4 The equivalent masses are aluminum phosphate = 2AlPO4 /6 = 40.67; aluminum hydroxide is 2Al(OH)3/6 = 26.0; calcium carbonate = CaCO3/2 = 50; calcium hydroxyapatite = 1/5Ca5(PO4)3OH/2 = 50.25; ferric hydroxide = 2Fe(OH)3/6 = 35.6; and ferric phosphate = 2FePO4/6 = 50.27 The equivalent mass of phosphorus varies For Eqs (14.42), (14.48), and (14.49), the equivalent mass of phosphorus = 2P/6 = 10.33; from Equation (14.45), equivalent mass = 3/5P/2 = 9.3 For calcium carbonate, the equivalent mass in all the reactions is 3Ca(HCO3)2/6 or Ca(HCO3)2/2 = 81.05 Thus, [ A Ca ] geq = [ A Ca ] geqCaCO 3 (14.50) From treating V m of water with [Phos]mg mg/L of phosphate phosphorus being removed, let M AlPO = the kilograms of aluminum phosphate solids produced; © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 649 Friday, June 14, 2002 2:34 PM M Ca5 ( PO4 )3 OH = kilograms of apatite solids produced; and M FePO4 = kilograms of ferric phosphate solids produced Also, from reacting with [Acur]geq gram moles per liter of natural alkalinity in V cubic meters of water, let M Al ( OH )3 = kilograms of alum solids produced; M CaCO3 = kilograms calcium carbonate solids produced; and M Fe ( OH )3 = kilograms of ferric hydroxide solids produced Finally, let Mss = kilograms of solids produced from the original suspended solids Considering the respective equivalent masses for the above equations, the following equations are produced: 40.67 f P [ Phos ] mg V –3 M AlPO = - = 3.94 ( 10 ) f P [ Phos ] mg V 1000 ( 10.33 ) (14.51) 50.25 f P [ Phos ] mg V –3 M Ca5 ( PO4 )3 OH = - = 5.40 ( 10 ) f P [ Phos ] mg V 1000 ( 9.3 ) (14.52) 50.27 f P [ Phos ] mg V –3 M FePO4 = - = 4.87 ( 10 ) f P [ Phos ] mg V 1000 ( 10.33 ) (14.53) M Al ( OH )3 = 26.0 [ A Ca ] geq V (14.54) M CaCO = 50 [ A Ca ] geq V (14.55) M Fe ( OH )3 = 35.6 [ A Ca ] geq V (14.56) f ss [ sp ss ] mg M ss = V 1000 (14.57) Now, let M AlPO4 Solids = total solids produced in removing phosphorus using alum; M CaOPO Solids = total solids produced in removing phosphorus using lime; and M FeIIIPO Solids = total solids produced in removing phosphorus using the ferric salts M AlPO Solids = M AlPO + M Al ( OH )3 + Mss; M CaOPO Solids = M Ca5 ( PO4 )3 OH + M CaCO3 + Mss; and M FeIIIPO4 Solids = M FePO4 + M Fe ( OH )3 + Mss Therefore, f ss [ sp ss ] mg –3 M AlPO Solids = 3.94 ( 10 ) f P [ Phos ] mg + 26.0 [ A Ca ] geq + V 1000 (14.58) f ss [ sp ss ] mg –3 M CaOPO Solids = 5.40 ( 10 ) f P [ Phos ] mg + 50 [ A Ca ] geq + V 1000 (14.59) f ss [ sp ss ] mg –3 M FeIIIPO4 Solids = 4.87 ( 10 ) f P [ Phos ] mg + 35.6 [ A Ca ] geq + V 1000 (14.60) Example 14.4 A wastewater contains 10 mg/L of phosphorus expressed as P The pH is 8.0 Assume the wastewater contains 100 mg/L of alkalinity expressed © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 650 Friday, June 14, 2002 2:34 PM as CaCO3 The phosphorus is to be removed using lime If the wastewater flow is 0.75 m /sec, calculate the kilograms of solids produced per day Assume the suspended solids content is 200 mg/L Solution: f ss [ sp ss ] mg –3 M CaOPO Solids = 5.40 ( 10 ) f P [ Phos ] mg + 50 [ A Ca ] geq + V 1000 Assume all of the phosphorus and suspended solids removed Therefore, fP = 1; fss = ( 100 ) [ A Ca ] geq = [ A Ca ] geqCaCO = - = 0.004 geq/L ( 1000 ) ( 50 ) Therefore, ( ) ( 200 ) –3 M CaOPO Solids = 5.40 ( 10 ) ( ) ( 10 ) + 50 ( 0.004 ) + - ( 0.75 ) ( 60 ) ( 60 ) ( 24 ) 1000 = 29,419.2 kg/d Ans GLOSSARY Calcium hydroxy apatite—Ca5(PO3)3(OH)(s) Condensation—The formation of water molecules when substances react Condensed phosphates—Substances containing phosphates characterized by the presence of the P–O–P bonds formed when phosphoric acid combines in a process called condensation Metaphosphoric acid—A condensation product of phosphoric acid forming a chain of (HPO3)n Optimum pH—The pH at which maximum precipitation occurs or at which the concentration of the species to be removed is at the lowest Optimum pH range—A range of pH at which the quantity of precipitates produced or the concentration of the species to be removed remaining in solution is acceptable Orbitals—A volume of atomic space in which a maximum of two electrons occupy − 2− Orthophosphates—The group of species containing the of H PO , HPO , and 3− PO groups Oxidation states—A condition of being an atom in relation to the extent of the atom losing or gaining electrons in its valence shell Polyphosphates—A compound containing two or more phosphate groups in the molecule Shell—A volume of atomic space holding electronic orbitals Solubility product constant—The product of the concentrations of ions, raised to appropriate powers, in equilibrium with their solid © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 651 Friday, June 14, 2002 2:34 PM Valence configuration—The arrangement of electrons in an atomic shell that accounts for the chemical reactivity of the atom Valence shell—The shell of an atom responsible for chemical reactions SYMBOLS Aca Calcium bicarbonate alkalinity [ACa]geq Gram equivalents per liter of calcium bicarbonate alkalinity [ A Ca ] geqCaCO Gram equivalents per liter of alkalinity being computed on the basis of CaCO3 having an equivalent mass of 50 2+ Al(OH) Hydroxo Al(III) ion 4+ Al ( OH ) 17 17-hydroxo, 7-Al(III) complex ion 5+ Al 13 ( OH ) 34 34-hydroxo, 13-Al(III) complex ion − Al ( OH ) Tetrahydroxo Al(III) complex ion 4+ Al ( OH ) Dihydroxo 2-Al(III) complex ion f ss Fractional removal of original suspended of raw water K HPO Equilibrium constant of the hydrogen phosphate ion K H2 PO4 Equilibrium constant of the dihydrogen phosphate ion K H3 PO4 Equilibrium constant of phosphoric acid K sp,Al ( OH )3 Solubility product constant of aluminum hydroxide K sp,AlPO4 Solubility product constant of aluminum phosphate K sp,apatite Solubility product constant of calcium hydroxy apatite phosphate K sp,Ca ( OH )2 Solubility product constant of calcium hydroxide K sp,Fe ( OH )3 Solubility product constant of ferric hydroxide K sp,FePO4 Solubility product constant of ferric phosphate Ion product of water Kw MAl Total kilograms of alum needed for the removal of phosphorus MAlAlk Kilograms of alum that react with the calcium bicarbonate alkalinity Kilograms of alum solids produced in the removal of phosphorus M Al ( OH )3 M AlPhos Kilograms of alum that react with phosphate phosphorus in raw water M AlPO Kilograms of aluminum phosphate solids produced in the removal of phosphorus M AlPO4 Solids Total solids produced in removing phosphorus using alum as the precipitant M CaOPO Solids Total solids produced in removing phosphorus using lime as the precipitant (including suspended solids in raw water) M Ca5 ( PO4 )3 OH Kilograms of apatite solids produced in the removal of phosphorus M CaCO3 Kilograms calcium carbonate solids produced in the removal of phosphorus MCaO Kilograms of lime needed for the removal of phosphorus MCaOAlk Kilograms of lime that react with the calcium bicarbonate alkalinity of the raw water MCaOPhos Kilograms of lime that react with phosphate phosphorus in raw water MFeIIICl Total kilograms of ferric chloride used for phosphate removal MFeIIIClAlk Kilograms of ferric chloride that react with the calcium bicarbonate of the raw water © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 652 Friday, June 14, 2002 2:34 PM MFeIIIClPhos M Fe ( OH )3 M FePO4 M FeIIIPO4 Solids Mss M FeIIISO M FeIIISO Alk M FeIIISO Phos p PAl PCaO PFeIIICl P FeIIISO [Phos]mg s sp PO4 Al sp PO4 Ca sp PO4 FeIII [spss]mg V x γ Ca γH γ HPO4 γ H2 PO4 γ PO4 Kilograms of ferric chloride that reacts with the phosphate phosphorus of the raw water Kilograms of ferric hydroxide solids produced in the removal of phosphorus Kilograms of ferric phosphate solids produced in the removal of phosphorus Total solids produced in removing phosphorus using the ferric salts as precipitants (including suspended solids in raw water) Kilograms of solids produced from the original suspended solids in the removal of phosphorus Total kilograms of ferric sulfate used for phosphate removal Kilograms of ferric sulfate that react with the calcium bicarbonate alkalinity of the raw water Kilograms of ferric sulfate that react with the phosphate phosphorus of the raw water The p orbital of the valence shell Fractional purity of alum Fractional purity of lime Fractional purity of ferric chloride Fractional purity of ferric sulfate Milligram per liter of phosphate phosphorus in raw water The s orbital of the valence shell Species in solution containing the PO4 species of the orthophosphates, using alum as the precipitant Species in solution containing the PO4 species of the orthophosphates, using lime as the precipitant Species in solution containing the PO4 species of the orthophosphates, using the ferric salts FeCl3 and Fe2(SO4)3 as precipitants Milligrams per liter of suspended solids in raw water Cubic meters of water treated Water of hydration of alum Activity coefficient of the calcium ion Activity coefficient of the hydrogen ion Activity coefficient of the hydrogen phosphate ion Activity coefficient of the dihydrogen phosphate ion Activity coefficient of the phosphate ion PROBLEMS 14.1 A plant is conducting phosphorus removal at a pH of 5.0 [ sp PO4 Al ] is calculated to be 0.112 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L Calculate K sp,AlPO4 assuming it cannot be determined from the given temperature © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 653 Friday, June 14, 2002 2:34 PM A plant is conducting phosphorus removal at a pH of 5.0 [ sp PO4 Al ] is calculated to be 0.112 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L Calculate Kw assuming it cannot be determined from the given temperature 14.3 A plant is conducting phosphorus removal at a pH of 5.0 [ sp PO4 Al ] is calculated to be 0.112 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L Calculate K sp,Al ( OH )3 assuming it cannot be determined from the given temperature 14.4 A plant is conducting phosphorus removal at a pH of 5.0 [ sp PO4 Al ] is calculated to be 0.112 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L Calculate γ H assuming it cannot be determined from the given dissolved solids content of the raw water 14.5 A plant is conducting phosphorus removal at an average temperature of approximately 25°C [ sp PO4 Al ] is calculated to be 0.112 mg/L as P The dissolved solids content of the raw water is 140 mg/L Calculate the pH at which the unit is being operated 14.6 A plant is conducting phosphorus removal at an average temperature of approximately 25°C and at a pH of 5.0 [ sp PO4 Al ] is calculated to be 0.112 mg/L as P The dissolved solids content of the raw water is 140 mg/L Calculate the γ PO4 assuming it cannot be determined from the given dissolved solids content of the raw water 14.7 A plant is conducting phosphorus removal at an average temperature of approximately 25°C and at a pH of 5.0 [ sp PO4 Al ] is calculated to be 0.112 mg/L as P The dissolved solids content of the raw water is 140 mg/L Calculate the γ HPO assuming it cannot be determined from the given dissolved solids content of the raw water 14.8 A plant is conducting phosphorus removal at an average temperature of approximately 25°C and at a pH of 5.0 [ sp PO4 Al ] is calculated to be 0.112 mg/L as P The dissolved solids content of the raw water is 140 mg/L Calculate the K HPO4 assuming it cannot be determined from the given temperature 14.9 A plant is conducting phosphorus removal at an average temperature of approximately 25°C and at a pH of 5.0 [ sp PO4 Al ] is calculated to be 0.112 mg/L as P The dissolved solids content of the raw water is 140 mg/L Calculate the γ H2 PO4 assuming it cannot be determined from the given dissolved solids content of the raw water 14.10 A plant is conducting phosphorus removal at an average temperature of approximately 25°C and at a pH of 5.0 [ sp PO4 Al ] is calculated to be 0.112 mg/L as P The dissolved solids content of the raw water is 140 mg/L Calculate the K H2 PO4 assuming it cannot be determined from the given temperature 14.11 A plant is conducting phosphorus removal at an average temperature of approximately 25°C and at a pH of 5.0 [ sp PO4 Al ] is calculated to be 0.112 mg/L 14.2 © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 654 Friday, June 14, 2002 2:34 PM 14.12 14.13 14.14 14.15 14.16 14.17 14.18 14.19 14.20 as P The dissolved solids content of the raw water is 140 mg/L Calculate the K H3 PO4 assuming it cannot be determined from the given temperature Calculate [ sp PO4 Ca ] expressed as P when the pH is 7.0 Assume the water 2+ contains 140 mg/L of dissolved solids and that [Ca ] = 130 mg/L as CaCO3 What is the precipitant used? A plant is conducting phosphorus removal at a pH of 7.0 [ sp PO4 Ca ] is calculated to be 0.0126 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L, and the concentration of the calcium ion is 130 mg/L Calculate Ksp,apatite assuming that it cannot be determined from the given temperature A plant is conducting phosphorus removal at a pH of 7.0 [ sp PO4 Ca ] is calculated to be 0.0126 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L, and the concentration of the calcium ion is 130 mg/L Calculate Kw assuming that it cannot be determined from the given temperature A plant is conducting phosphorus removal at a pH of 7.0 [ sp PO4 Ca ] is calculated to be 0.0126 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L, and the concentration of the calcium ion is 130 mg/L What is the precipitant used? Calculate γ Ca assuming that it cannot be determined from the given dissolved solids content of the raw water A plant is conducting phosphorus removal at a pH of 7.0 [ sp PO4 Ca ] is calculated to be 0.0126 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L What is the precipitant used? Calculate [Ca] A plant is conducting phosphorus removal at a pH of 7.0 [ sp PO4 Ca ] is calculated to be 0.0126 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L, and the concentration of the calcium ion is 130 mg/L Calculate γH assuming that it cannot be determined from the given dissolved solids content of the raw water A plant is conducting phosphorus removal [ sp PO4 Ca ] is calculated to be 0.0126 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L, and the concentration of the calcium ion is 130 mg/L Calculate the pH A plant is conducting phosphorus removal at a pH of 7.0 [ sp PO4 Ca ] is calculated to be 0.0126 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L, and the concentration of the calcium ion is 130 mg/L Calculate γ PO4 assuming that it cannot be determined from the given dissolved solids content of the raw water A plant is conducting phosphorus removal at a pH of 7.0 [ sp PO4 Ca ] is calculated to be 0.0126 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L, and the concentration of the calcium ion is 130 mg/L © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 655 Friday, June 14, 2002 2:34 PM 14.21 14.22 14.23 14.24 14.25 14.26 14.27 14.28 14.29 Calculate γ HPO assuming that it cannot be determined from the given dissolved solids content of the raw water A plant is conducting phosphorus removal at a pH of 7.0 [ sp PO4 Ca ] is calculated to be 0.0126 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L, and the concentration of the calcium ion is 130 mg/L Calculate K HPO assuming that it cannot be determined from the given temperature A plant is conducting phosphorus removal at a pH of 7.0 [ sp PO4 Ca ] is calculated to be 0.0126 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L, and the concentration of the calcium ion is 130 mg/L Calculate γ H2 PO4 assuming that it cannot be determined from the given dissolved solids content of the raw water A plant is conducting phosphorus removal at a pH of 7.0 [ sp PO4 Ca ] is calculated to be 0.0126 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L, and the concentration of the calcium ion is 130 mg/L Calculate K H2 PO4 assuming that it cannot be determined from the given temperature A plant is conducting phosphorus removal at a pH of 7.0 [ sp PO4 Ca ] is calculated to be 0.0126 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L, and the concentration of the calcium ion is 130 mg/L Calculate K H3 PO4 assuming that it cannot be determined from the given temperature Calculate [ sp PO4 FeIII ] expressed as P when the pH is 3.0 Assume the water contains 140 mg/L of dissolved solids A plant is conducting phosphorus removal at a pH of 3.0 [ sp PO4 FeIII ] is calculated to be 0.141 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L What precipitant is used? Calculate K sp,FePO4 assuming that it cannot be determined from the given temperature A plant is conducting phosphorus removal at a pH of 3.0 [ sp PO4 FeIII ] is calculated to be 0.141 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L Calculate Kw assuming that it cannot be determined from the given temperature A plant is conducting phosphorus removal at a pH of 3.0 [ sp PO4 FeIII ] is calculated to be 0.141 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L Calculate K sp, ( FeOH )3 assuming that it cannot be determined from the given temperature A plant is conducting phosphorus removal at a pH of 3.0 [ sp PO4 FeIII ] is calculated to be 0.141 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 656 Friday, June 14, 2002 2:34 PM 14.30 14.31 14.32 14.33 14.34 14.35 14.36 14.37 14.38 water is 140 mg/L Calculate γH assuming that it cannot be determined from the given dissolved solids content of the raw water A plant is conducting phosphorus removal [ sp PO4 FeIII ] is calculated to be 0.141 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 + mg/L Calculate [H ] A plant is conducting phosphorus removal at a pH of 3.0 [ sp PO4 FeIII ] is calculated to be 0.141 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L Calculate γ PO4 assuming that it cannot be determined from the given dissolved solids content of the raw water A plant is conducting phosphorus removal at a pH of 3.0 [ sp PO4 FeIII ] is calculated to be 0.141 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L Calculate γ HPO assuming that it cannot be determined from the given dissolved solids content of the raw water A plant is conducting phosphorus removal at a pH of 3.0 [ sp PO4 FeIII ] is calculated to be 0.141 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L Calculate K HPO assuming that it cannot be determined from the given temperature A plant is conducting phosphorus removal at a pH of 3.0 [ sp PO4 FeIII ] is calculated to be 0.141 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L Calculate γ H2 PO4 assuming that it cannot be determined from the given dissolved solids content of the raw water A plant is conducting phosphorus removal at a pH of 3.0 [ sp PO4 FeIII ] is calculated to be 0.141 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L Calculate K H2 PO4 assuming that it cannot be determined from the given temperature A plant is conducting phosphorus removal at a pH of 3.0 [ sp PO4 FeIII ] is calculated to be 0.141 mg/L as P and the unit is operated at approximately an average temperature of 25°C The dissolved solids content of the raw water is 140 mg/L Calculate K H3 PO4 assuming that it cannot be determined from the given temperature A wastewater contains 10 mg/L of phosphorus expressed as P The pH is 8.0 Assume the wastewater contains 140 mg/L of dissolved solids and 100 mg/L of alkalinity expressed as CaCO3 The phosphorus is to be removed using lime If the wastewater flow is 0.75 m /sec, calculate the kilograms of lime per day needed to react with the alkalinity A wastewater contains 10 mg/L of phosphorus expressed as P The pH is 8.0 Assume the wastewater contains 140 mg/L of dissolved solids and 100 mg/L of alkalinity expressed as CaCO3 The phosphorus is to be removed using lime If the wastewater flow is 0.75 m /sec, calculate the kilograms of lime per day needed to react with the phosphorus © 2003 by A P Sincero and G A Sincero TX249_frame_C14.fm Page 657 Friday, June 14, 2002 2:34 PM 14.39 A wastewater contains 10 mg/L of phosphorus expressed as P The pH is 10.0 Assume the wastewater contains 100 mg/L of alkalinity expressed as CaCO3 The phosphorus is to be removed using ferric ferric chloride If the wastewater flow is 0.75 m /sec, calculate the kilograms of solids produced per day Assume the suspended solids content is 200 mg/L BIBLIOGRAPHY Baker, M J., D W Blowes, and C J Ptacek (1998) Laboratory development of permeable reactive mixtures for the removal of phosphorus from onsite wastewater disposal systems Environ Science Technol 32, 15, 2308–2316 Banister, S S., A R Pitman, and W A Pretorius (1998) Solubilisation of N and P during primary sludge acid fermentation and precipitation of the resultant P Water S.A 24, 4, 337–342 Barralet, J., S Best, and W Bonfield (1998) Carbonate substitution in precipitated hydroxyapatite: An investigation into the effects of reaction temperature and bicarbonate ion concentration J Biomedical Materials Res 41, 1, 79–86 Battistoni, P., et al (1998) Phosphate removal in real anaerobic supernatants: Modelling and performance of a fluidized bed reactor Water Science Technol Wastewater: Nutrient Removal, Proc 1998 19th Biennial Conf Int Assoc on Water Quality, Part 1, June 21–26, Vancouver, Canada, 38, 1, 275–283 Elsevier Science Ltd Exeter, England Booker, N A and R B Brooks (1994) Scale-up of the rapid sewage treatment SIROFLOCTM process Process Safety Environ Protection: Trans Inst Chem Engineers, Part B, 72, 2, 109–112 Inst of Chem Engineers, Rugby, England Fytianos, K., E Voudrias, and N Raikos (1998) Modelling of phosphorus removal from aqueous and wastewater samples using ferric iron Environ Pollut 101, 1, 123–130 Holtzclaw Jr., H F and W R Robinson (1988) General Chemistry D C Heath and Company, Lexington, MA, A7 Jiang, J Q and N J D Graham (1998) Pre-polymerised inorganic coagulants and phosphorus removal by coagulation—A review Water S.A 24, 3, 237–244 Jonsson, L (1997) Experiences of nitrogen and phosphorus removal in deep-bed filters at Henriksdal Sewage Works in Stockholm Water Science Technol., Proc 1997 Int Conf on Upgrading of Water and Wastewater Syst., May 25–28, Kalmir, Sweden, 37, 9, 193–200, Elsevier Science Ltd., Exeter, England Maurer, M., et al (1999) Kinetics of biologically induced phosphorus precipitation in wastewater treatment Water Res 33, 2, 484–493 Morse, G K., et al (1998) Review: Phosphorus removal and recovery technologies Science Total Environment 212, 1, 69–81 Piotrowski, J and R Onderko (1998) Phosphorus control and reduction at a corrugating medium mill Proc 1998 TAPPI Int Environ Conf Exhibit, Part 1, April 5–8, Vancouver, Canada, 1, 101–115 TAPPI Press, Norcross, GA Snoeyink, V L and D Jenkins (1980) Water Chemistry John Wiley & Sons, New York, 280, 301 Strikland, J (1998) Development and application of phosphorus removal from wastewater using biological and metal precipitation techniques J Chartered Inst Water Environment Manage 12, 1, 30–37 Ugurlu, A and B Salman (1998) Phosphorus removal by fly ash Environment Int 24, 8, 911–918 © 2003 by A P Sincero and G A Sincero ... liter of suspended solids in raw water Cubic meters of water treated Water of hydration of alum Activity coefficient of the calcium ion Activity coefficient of the hydrogen ion Activity coefficient of. .. than the range for the use of either alum or ferric salts 14.8 UNIT OPERATIONS FOR PHOSPHORUS REMOVAL The removal of phosphorus involves provision for the addition of chemicals both for adjusting... removal of phosphorus Total kilograms of ferric sulfate used for phosphate removal Kilograms of ferric sulfate that react with the calcium bicarbonate alkalinity of the raw water Kilograms of ferric