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 pro- duction are all discussed. The chemical precipitation method employed uses alum, lime, and the ferric salts, FeCl 3 and Fe 2 (SO 4 ) 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]3 s 2 3 p 3 . [Ne] means that the neon configuration is filled. The valence configuration represented by the 3, the M shell, shows five electrons in the orbitals: 2 electrons in the s orbitals and 3 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 phosphates containing the ion, hydrogen phosphate containing the ion, and the phosphates containing the 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 eutroph- ication in receiving streams. Thus, concentrations of orthophosphates should be controlled through removal before discharging the wastewater into receiving bodies 14 H 2 PO 4 − HPO 4 2− PO 4 3− TX249_frame_C14.fm Page 627 Friday, June 14, 2002 2:34 PM © 2003 by A. P. Sincero and G. A. Sincero 628 of water. The orthophosphates of concern in wastewater are sodium phosphate (Na 3 PO 4 ), sodium hydrogen phosphate (Na 2 HPO 4 ), sodium dihydrogen phosphate (NaH 2 PO 4 ), and ammonium hydrogen phosphate [(NH 4 ) 2 HPO 4 ]. 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 (H 4 P 2 O 7 , oxidation state = 5 + ), tripolyphosphoric acid (H 5 P 3 O 10 , oxidation state = 5 + ), and metaphosphoric acid [(HPO 3 ) 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 (NaPO 3 ) 6 , sodium dipolyphosphate (Na 4 P 2 O 7 ), and sodium tripolyphosphate (Na 5 P 3 O 10 ). 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 FIGURE 14.1 Structural formulas of various forms of phosphates. Orthophosphoric acid Trimetaphosphoric acid Tripolyphosphoric acid Organic backbone Organic phosphate Dipolyphosphoric acid or pyrophosphoric acid TX249_frame_C14.fm Page 628 Friday, June 14, 2002 2:34 PM © 2003 by A. P. Sincero and G. A. Sincero 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 trimet- aphosphoric 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 3 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 K sp 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 Ca 5 (PO 4 ) 3 (OH) ( s ) or Ca 3 (PO 4 ) 2 . Of these two precipitates, Ca 5 (PO 4 ) 3 (OH) ( s ) has the smaller K sp of 10 − 55.9 ; thus, it will be used as the criterion for the precipitation of phosphates. Ca 5 (PO 3 ) 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 AlPO 4( s ) and FePO 4 . 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 Solubility Product, K sp at 25 ° C Ca 5 (PO 3 ) 3 (OH) ( s ) (10 − 55.9 ) Ca 3 (PO 4 ) 2 (10 − 25 ) AlPO 4( s ) (10 − 21 ) FePO 4 (10 − 21.9 ) TX249_frame_C14.fm Page 629 Friday, June 14, 2002 2:34 PM © 2003 by A. P. Sincero and G. A. Sincero 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: (14.1) 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): (14.2) (14.3) (14.4) Let represent the species in solution containing the PO 4 species of the orthophosphates, using alum as the precipitant. Therefore, (14.5) Express Equation (14.5) in terms of using Eqs. (14.2) through (14.4). This will enable to be expressed in terms of [Al 3 + ] using Equation (14.1) and . Proceed as follows: (14.6) (14.7) Al 3+ PO 4 3− AlPO 4 s() ↓+ AlPO 4 s() ↓ Al 3 PO 4 3− K sp,AlPO 4 + 10 −21 = PO 4 3− H + HPO 4 2− + HPO 4 2− PO 4 3− H + Κ HPO 4 +⇒ 10 −12.3 = HPO 4 2− H + H 2 PO 4 − + H 2 PO 4 − HPO 4 2− H + Κ H 2 PO 4 +⇒ 10 −7.2 = H 2 PO 4 − H + H 3 PO 4 + H 3 PO 4 H 2 PO 4 − H + Κ H 3 PO 4 +⇒ 10 −2.1 = sp PO 4 Al sp PO 4 Al []PO 4 3− []HPO 4 2− []H 2 PO 4 − []H 3 PO 4 []+++= PO 4 3− [] sp PO 4 Al [] K sp,AlPO 4 HPO 4 2− [] HPO 4 2− {} γ HPO 4 PO 4 3− {}H + {} γ HPO 4 K HPO 4 γ PO 4 γ H PO 4 3− []H + [] γ HPO 4 K HPO 4 == = H 2 PO 4 − [] H 2 PO 4 − {} γ H 2 PO 4 HPO 4 2− {}H + {} γ H 2 PO 4 K H 2 PO 4 γ PO 4 γ H 2 PO 4 3− []H + [] 2 γ H 2 PO 4 K H 2 PO 4 K HPO 4 == = TX249_frame_C14.fm Page 630 Friday, June 14, 2002 2:34 PM © 2003 by A. P. Sincero and G. A. Sincero (14.8) Equations (14.6) through (14.8) may now be substituted into Equation (14.5) to produce (14.9) But, from Equation (14.1), = / = / = / . Substituting, (14.10) , γ Al , γ H , , and are, respectively, the activity coefficients of the phos- phate, aluminum, hydrogen, hydrogen phosphate, and dihydrogen phosphate ions. , , and are, respectively, the equilibrium constants of the hydro- gen phosphate and dihydrogen phosphate ions and phosphoric acid. [Al 3+ ] 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 AlPO 4(s) , but it will also react with the OH − to precipitate Al(OH) 3(s) . Also, Al 3+ will form complexes Al(OH) 2+ , , , , and in addition to the Al(OH) 3(s) . All these interactions complicate our objective of eliminating [Al 3+ ]. Consider, however, the equilibrium of Al(OH) 3(s) , which is as follows: (14.11) = 10 −33 may be compared with = 10 −21 . From = 10 −33 , the concentration of Al 3+ needed to precipitate Al(OH) 3 may be calculated to be 2.0(10 −9 ) gmol/L. Doing similar calculation from = 10 −21 , the concentration of Al 3+ needed to precipitate AlPO 4 is 3.0(10 −11 ) gmol/L. These two concentrations H 3 PO 4 []H 3 PO 4 {} H 2 PO 4 − {}H + {} K H 3 PO 4 γ PO 4 γ H 3 PO 4 3− []H + [] 3 K H 3 PO 4 K H 2 PO 4 K HPO 4 == = sp PO 4 Al []PO 4 3− [] γ PO 4 γ H PO 4 3− []H + [] γ HPO 4 K HPO 4 γ PO 4 γ H 2 PO 4 3− []H + [] 2 γ H 2 PO 4 K H 2 PO 4 K HPO 4 ++= γ PO 4 γ H 3 PO 4 3− []H + [] 3 K H 3 PO 4 K H 2 PO 4 K HPO 4 + PO 4 3− []PO 4 3− {} γ PO 4 K sp,AlPO 4 γ PO 4 Al 3+ {}K sp,AlPO 4 γ PO 4 γ Al Al 3+ [] sp PO 4 Al [] K sp,AlPO 4 γ PO 4 γ Al Al 3+ [] γ H K sp,AlPO 4 H + [] γ HPO 4 K HPO 4 γ Al Al 3+ [] γ H 2 K sp,AlPO 4 H + [] 2 γ H 2 PO 4 K H 2 PO 4 K HPO 4 γ Al Al 3+ [] ++= γ H 3 K sp,AlPO 4 H + [] 3 K H 3 PO 4 K H 2 PO 4 K HPO 4 γ Al Al 3+ [] + γ PO 4 γ HPO 4 γ H 2 PO 4 K HPO 4 K H 2 PO 4 K H 3 PO 4 Al 7 (OH) 17 4+ Al 13 (OH) 34 5+ Al(OH) 4 − Al 2 (OH) 2 4+ Al(OH) 3 s() Al 3+ 3OH − K sp,Al(OH) 3 + 10 −33 = K sp,Al(OH) 3 K sp,AlPO 4 K sp,Al(OH) 3 K sp,AlPO 4 TX249_frame_C14.fm Page 631 Friday, June 14, 2002 2:34 PM © 2003 by A. P. Sincero and G. A. Sincero are so close to each other that it may be concluded that Al(OH) 3 and AlPO 4 are coprecipitating. This finding allows us to eliminate [Al 3+ ]. Thus, Substituting into Equation (14.10), (14.12) Equation (14.12) portrays the equilibrium relationship between and [H + ]. Example 14.1 Calculate when the pH is 10. Assume the water con- tains 140 mg/L of dissolved solids. Solution: Al 3+ [] Al 3+ {} γ Al K sp,Al OH() 3 γ Al OH − {} 3 K sp,Al OH() 3 γ H 3 H + [] 3 γ Al K w 3 .== = sp PO 4 Al [] K sp,AlPO 4 K w 3 γ PO 4 K sp,Al(OH) 3 γ H 3 H + [] 3 K sp,AlPO 4 K w 3 γ HPO 4 K HPO 4 K sp,Al(OH) 3 γ H 2 H + [] 2 += K sp,AlPO 4 K w 3 γ H 2 PO 4 K H 2 PO 4 K HPO 4 K sp,Al(OH) 3 γ H H + [] K sp,AlPO 4 K w 3 K H 3 PO 4 K H 2 PO 4 K HPO 4 K sp,Al(OH) 3 ++ sp PO 4 [] sp PO 4 Al [] sp PO 4 Al [] K sp,AlPO 4 K w 3 γ PO 4 K sp,Al(OH) 3 γ H 3 H + [] 3 K sp,AlPO 4 K w 3 γ HPO 4 K HPO 4 K sp,Al(OH) 3 γ H 2 H + [] 2 += K sp,AlPO 4 K w 3 γ H 2 PO 4 K H 2 PO 4 K HPO 4 K sp,Al(OH) 3 γ H H + [] K sp,AlPO 4 K w 3 K H 3 PO 4 K H 2 PO 4 K HPO 4 K sp,Al(OH) 3 ++ K sp,AlPO 4 10 −21 = µ 2.5 10 5– ()TDS γ 10 0.5z i 2 µ () 1+1.14 µ () – == µ 2.5 10 5– ()140()3.5 10 3– () γ PO 4 10 0.5 3() 2 – 3.5 10 3– ()[] 1+1.14 3.5 10 3– ()[] – 0.56== = = K sp,Al(OH) 3 10 33– = γ H γ H 2 PO 4 10 − 0.5 1() 2 3.5 10 3– ()[] 1+1.14 3.5 10 3– ()[] 0.94== = γ HPO 4 = 10 − 0.5 2() 2 3.5 10 3– ()[] 1+1.14 3.5 10 3– ()[] 0.77 K HPO 4 = 10 12.3– K H 2 PO 4 = 10 7.2– K H 3 PO 4 = 10 2.1– = TX249_frame_C14.fm Page 632 Friday, June 14, 2002 2:34 PM © 2003 by A. P. Sincero and G. A. Sincero Therefore, The answer of 9.17(10 +6 ) 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 p H RANGE Equation (14.12) shows that is a function of the hydrogen ion concentration. This means that the concentration of the species containing the PO 4 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 calcu- lated. 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 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: (14.13) sp PO 4 Al [] 10 21– ()10 14– () 3 0.56()10 33– ()0.94() 3 10 10– [] 3 10 21– ()10 14– () 3 0.77()10 12.3– ()10 33– ()0.94() 2 10 10– [] 2 += 10 21– ()10 14– () 3 0.94()10 7.2– ()10 12.3– ()10 33– ()0.94()10 10– [] + 10 21– ()10 14– () 3 10 2.1– ()10 7.2– ()10 12.3– ()10 −33 () + 1.0 10 63– () 4.65 10 64– () 1.0 10 63– () 3.41 10 66– () 1.0 10 63– () 2.79 10 63– () 1.0 10 63– () 2.51 10 55– () +++= 295.76 gmols/L 9.17 10 6 () mg/L as P== sp PO 4 Al [] sp PO 4 Al sp PO 4 Al 5Ca 2+ 3PO 4 3− OH − Ca 5 PO 4 () 3 OH s() ↓++ Ca 5 PO 4 () 3 OH s() ↓ 5Ca 2+ 3PO 4 3− OH − ++K sp,apatite 10 55.9– = TX249_frame_C14.fm Page 633 Friday, June 14, 2002 2:34 PM © 2003 by A. P. Sincero and G. A. Sincero TABLE 14.2 Concentration of as a Function of pH at 25°C (mg/L as P) pH, Dissolved Solids = 140 mg/L (mg/L) 0 6.6(10 −5 ) 1 7.1(10 −5 ) 2 1.2(10 −4 ) 3 6.5(10 −4 ) 4 5.9(10 −3 ) 5 5.9(10 −2 ) 6 6.3(10 −1 ) 7 1.1(10 +1 ) 8 5.4(10 +2 ) 9 4.8(10 +4 ) 10 4.8(10 +6 ) 11 5.8(10 +8 ) 12 8.3(10 +10 ) 13 4.0(10 +13 ) 14 2.3(10 +16 ) 15 3.5(10 +19 ) TABLE 14.3 Concentration of as a Function of pH at 25°C (mg/L as P) pH, Dissolved Solids == == 35,000 mg/L (mg/L) 0 6.6(10 −5 ) 1 8.0(10 −5 ) 2 2.1(10 −4 ) 3 1.5(10 −3 ) 4 1.5(10 −2 ) 5 1.5(10 −1 ) 6 2.0(10 −0 ) 7 8.9(10 +1 ) 8 7.6(10 +3 ) 9 7.5(10 +5 ) 10 8.2(10 +7 ) 11 1.6(10 +10 ) 12 9.2(10 +12 ) 13 8.5(10 +15 ) 14 8.5(10 +18 ) 15 8.5(10 +21 ) sp PO 4 Al [sp PO 4 Al ] sp PO 4 Al [sp PO 4 Al ] TX249_frame_C14.fm Page 634 Friday, June 14, 2002 2:34 PM © 2003 by A. P. Sincero and G. A. Sincero 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 be the species in solution containing the PO 4 species, using the calcium in lime as the precipitant. 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), Substituting in Equation (14.9) and simplifying, (14.14) γ Ca and K w are, respectively, the activity coefficient of the calcium ion and ion product of water. The [Ca 2+ ] in the denominator is analogous to [Al 3+ ] in the case of alum; however, the present case is different in that [Ca 2+ ] cannot be eliminated in a straightforward manner. [Al 3+ ] was easily eliminated because Al(OH) 3(s) was precipitating along with AlPO 4 . Ca(OH) 2(s) , which could be precipitating, has a K sp value of 7.9(10 −6 ). This is much, much greater than K sp,apatite = 10 −55.9 and calcium hydroxide will not be precipitating along with Ca 5 (PO 4 ) 3 OH (s) . The other possible precipitate is CaCO 3 that is produced when lime reacts with the natural alkalinity of the water; however, calcium carbonate has a K sp value of 4.8(10 −9 ). Again, this value is much, much greater than K sp,apatite and calcium carbonate will not be precipitating along with Ca 5 (PO 4 ) 3 OH (s) , either. We will, therefore, let the equation stand and express as a function of [Ca 2+ ], along with [H + ] and the constants. Example 14.2 Calculate expressed as mg/L of P when the pH is 8. Assume the water contains 140 mg/L of dissolved solids and that [Ca 2+ ] = 130 mg/L as CaCO 3 . sp PO 4 Ca sp PO 4 Ca [] PO 4 3− [] PO 4 3− {} γ PO 4 K sp,apatite Ca 2+ {} 5 OH − {} 1/3 γ PO 4 K sp,apatite 1/3 γ H 1/3 H + [] 1/3 γ Ca 1/3 Ca 2+ [] 5/3 K w 1/3 γ PO 4 == = sp PO 4 Ca [] K sp,apatite 1/3 γ H 1/3 H + [] 1/3 γ Ca 1/3 Ca 2+ [] 5/3 K w 1/3 γ PO 4 K sp,apatite 1/3 γ H 4/3 H + [] 4/3 γ HPO 4 K HPO 4 γ Ca 1/3 Ca 2+ [] 5/3 K w 1/3 += K sp,apatite 1/3 γ H 7/3 H + [] 7/3 γ H 2 PO 4 K H 2 PO 4 K HPO 4 γ Ca 1/3 Ca 2+ [] 5/3 K w 1/3 + K sp,apatite 1/3 γ H 10/3 H + [] 10/3 K H 3 PO 4 K H 2 PO 4 K HPO 4 γ Ca 1/3 Ca 2+ [] 5/3 K w 1/3 + sp PO 4 Ca [] sp PO 4 Ca [] TX249_frame_C14.fm Page 635 Friday, June 14, 2002 2:34 PM © 2003 by A. P. Sincero and G. A. Sincero Solution: Therefore, sp PO 4 Ca [] K sp,apatite 1/3 γ H 1/3 H + [] 1/3 γ Ca 1/3 Ca 2+ [] 5/3 K w 1/3 γ PO 4 K sp,apatite 1/3 γ H 4/3 H + [] 4/3 γ HPO 4 K HPO 4 γ Ca 1/3 Ca 2+ [] 5/3 K w 1/3 += K sp,apatite 1/3 γ H 7/3 H + [] 7/3 γ H 2 PO 4 K H 2 PO 4 K HPO 4 γ Ca 1/3 Ca 2+ [] 5/3 K w 1/3 + K sp,apatite 1/3 γ H 10/3 H + [] 10/3 K H 3 PO 4 K H 2 PO 4 K HPO 4 γ Ca 1/3 Ca 2+ [] 5/3 K w 1/3 + K sp,apatite 10 55.9– γ H γ H 2 PO 4 10 0.5 1() 2 3.5 10 3– ()[] 1+1.14 3.5 10 3– ()[] – 0.94=== = γ Ca γ HPO 4 10 0.5 2() 2 3.5 10 3– ()[] 1+1.14 3.5 10 3– ()[] – 0.77== = CA 2+ [] 130 1000 40.1() 3.24 10 3– ()gmol/L== γ PO 4 10 0.5 3() 2 3.5 10 3– ()[] 1+1.14 3.5 10 3– ()[] – 0.56== K HPO 4 10 12.3– K H 2 PO 4 10 7.2– K H 3 PO 4 10 2.1– === sp PO 4 Ca [] 10 55.9– () 1/3 0.94() 1/3 10 8– [] 1/3 0.77() 1/3 3.24 10 3– ()[] 5/3 10 14– () 1/3 0.56() = 10 55.9– () 1/3 0.94() 4/3 10 8– [] 4/3 0.77()10 12.3– ()0.77() 1/3 3.24 10 3– ()[] 5/3 10 14– () 1/3 + 10 55.9– () 1/3 0.94() 7/3 10 8– [] 7/3 0.94()10 7.2– ()10 12.3– ()0.77() 1/3 3.24 10 3– ()[] 5/3 10 14– () 1/3 + 10 55.9– () 1/3 0.94() 10/3 10 8– [] 10/3 10 2.1– ()10 7.2– ()10 12.3– ()0.77() 1/3 3.24 10 3– ()[] 5/3 10 14– () 1/3 + 4.91 10 22– () 7.85 10 10– () 4.62 10 30– () 5.41 10 22– () 4.34 10 38– () 4.16 10 29– () 4.08 10 46– () 3.52 10 31– () +++= 9.58 10 9– ()gmol/L= 6.25 10 13– ()8.54 10 9– ()1.04 10 9– ()1.16 10 15– ()+++= 2.97 10 4– ()mg/L Ans= TX249_frame_C14.fm Page 636 Friday, June 14, 2002 2:34 PM © 2003 by A. P. Sincero and G. A. Sincero [...]... TX249_frame_C14.fm Page 639 Friday, June 14, 2002 2:34 PM from pH 7.0 and above The effect of dissolved solids has reduced this range to 8 and above; however, a concentration of 35,000 mg/L of total solids is already very high and would not be encountered in the normal treatment of water and wastewater This concentration is representative of the dissolved solids concentration of sea water 14. 5 CHEMICAL. .. 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 4 assuming that... 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... 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... 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... optimum and the optimum pH range will be determined by preparing a table as was done in the case of alum Tables 14. 4 through 14. 6 show the results This table was prepared using Equation (14. 14) Tables 14. 4 and 14. 5 reveal very important information The efficiency of removal of phosphorus increases as the concentration of calcium increases from 0 to 130 mg/L Remember that we disallowed the use of the Ksp of. .. 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... mg/L of dissolved solids and 100 mg/L of alkalinity expressed as CaCO3 The phosphorus is to be 3 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... 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. .. phosphorus is to be 3 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 . encountered in the normal treatment of water and wastewater. This concentration is representative of the dissolved solids concentra- tion of sea water. 14. 5 CHEMICAL REACTION OF THE PHOSPHATE ION. Equation (14. 5) which, along with Eqs. (14. 2) through (14. 4), can be manipulated to produce Equation (14. 9). From Equation (14. 13), Substituting in Equation (14. 9) and simplifying, (14. 14) γ Ca and. case of alum. Tables 14. 4 through 14. 6 show the results. This table was prepared using Equation (14. 14). Tables 14. 4 and 14. 5 reveal very important information. The efficiency of removal of phosphorus