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.

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 5.

Consider the full alum formula in the chemical reaction,

Al2(SO4)3 • xH2O + 2 PO4- ^ 2 AlPO4(s)i + 3 SO4- + xH2O (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

Al2(SO4)3 • xH2O + 3 Ca(HCO3)2 ^ 2 Al(OH)31 + 6CO2 + 3 CaSO4 + xH2O

From the previous reactions, the equivalent masses are alum = Al2(SO4)3 • xH2O/ 6 = 57.05 + 3x and calcium bicarbonate = 3Ca(HCO3)2/6 = 81.05. The number of 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,

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

Al 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:

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 [ACa]geqCaCO 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 [ACa]geqCaCO . The quantity of mass remains the same in any system. Thus, referring to Equation (14.20), there will also be the same 162.1 [ACa]geqCaCO 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 [ACa]geqCaCO grams of calcium bicarbonate alkalinity is 162.1 [Ac, ]geqCaCO3/81.05 = 2 [Ac, ]geqCaCO3. And,

Again, [ ACa ]geq is the calcium bicarbonate alkalinity referred to the reaction of alum with calcium bicarbonate and [ACa]geqCaCO is the same bicarbonate alkalinity referred to calcium carbonate with equivalent mas3s 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:

M = /p[Phos]mg , 0 = 9.7(10-5)/p[Phos]mg(57.05 + 3x) ¥

Letting MAl be the total kilograms of alum needed for the removal of phosphorus,

Mai = MAlAlk + M ^ = (57°.-+^([Aca ]geq + 9.7( 10-5 )fP [PhosLg )¥

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 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 7.

The applicable chemical reactions are:

Ca(OH)2 + 3 PO3- ^ 1 Ca5(PO4)3OHW I + ^ OH- (14.28)

To make the previous equations compatible, use 2 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

Let MCaOAlk be the kilograms of lime at fractional purity of PCaO required to react with the calcium carbonate alkalinity. Then,

Now, letting MCaOPhos be the kilograms of lime required to react with the phosphorus,

M — 37.05/p[Phos]mg T, — 4I0(10-3)MPhoslmgV (1430) MCaOPhos — 1CXX)(9.3)PQo V — PCO (14.30)

Let MCaO be the total kilograms of lime needed for the removal of phosphorus. Thus,

MCaO — MCaOAlk + MCa°Phos — (37.05[ACa]geq + 4.0( 10-3) fp[PhOS]mg) )

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 3.

The applicable respective chemical reactions for the satisfaction of the bicarbonate alkalinity are

2FeCl3 + 3Ca(HCO3)2 ^ 2Fe(OH)31 + 6CO2 + 3CaCl2 (14.32)

Fe2(SO4)3 + 3Ca(HCO3)2 ^ 2Fe(OH)31 + 6CO2 + 3CaSO4 (14.33)

Upon satisfaction of the previous reactions, the normal precipitation of the phosphate occurs. The respective reactions are

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, [ACa]geq = 2 [ACa]geqCaO .

Let MpeIIIC]Alk and MFemS^be the kilograms of ferric chloride and ferric sulfate, respectively, required to react with the calcium bicarbonate alkalinity. Also, let the respective purity be Ppeiiicl and PFeIIISO4 . Then,

P FeIIISO4

Now, letting MFeIIIClPhos and MFeIIISO phoj be the kilograms of ferric chloride and ferric sulfate, respectively, required to react with the phosphorus,

M = 66.5MPhos]m „ = 4.3(10-')fv[Phos-\mg,V (1439)

FeIIISO4

Let MFeIIICl and MFeIIISO be the total kilograms of ferric chloride and ferric sulfate, respectively, needed for the removal of phosphorus. Thus,

MFeIIICl = MFeIIICUk + MFeIIIClPhos = ( 54.1 [ Ac,\geq + 3.5( 10^ ) fp [ PhoS\mg ))—

PFeIIICl

MFeIIISO = MFeIIISO Alk + MFeIIISO . Phos

= ( 66.5 [ Aca ]geq + 4.3( 10-3 ) fp [ Phos ]mg )-)— (14.41)

FeIIICl

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Responses

  • SUMMER
    How much quantuty of lime required to neuralize 10 gram alum?
    2 years ago

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