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imately (10 ). This means that the hydroxide is very insoluble. Thus, with the ions of the coagulant and the bicarbonate dispersed in the water, Al3+ "grabs" whatever 0H_ there is to form the precipitate, Al(0H)3, and the reaction portrayed above ensues.

A very important point must be discussed with respect to the previous coagulation reaction, in comparison with those found in the literature. The environmental engineering literature normally uses the equilibrium arrows, ^ , instead of the single forward arrow, ^, as written previously. Equilibrium arrows indicate that a particular reaction is in equilibrium, which would mean for the present case, that alum is produced in the backward reaction. Alum, however, is never produced by mixing aluminum hydroxide, carbon dioxide, calcium sulfate, and water, the species found on the right-hand side of the previous equation. Once Al2(S04)3 • 14H20 is mixed with Ca(HC03)2, the alum is gone forever producing the aluminum hydroxide precipitate—it cannot be recovered. After the formation of the precipitate, any backward reaction would be for the complex reactions and not for the formation of Al2(S04)3 • xH20, as would be inferred if the above reaction were written with the equilibrium arrows.

In addition, coagulation is a process of expending the coagulant. In the process of expenditure, the alum must react to produce its products. This means that what must "exist" is the forward arrow and not any backward arrow. Portraying the backward arrow would mean that the alum is produced, but it is known that it is not produced but expended. During expenditure, no equilibrium must exist. To reiterate, the coagulation reaction should be represented by the forward arrow and not by the equilibrium arrows.

As shown in Equation (12.46), an alkaline substance is needed to react with the alum. The bicarbonate alkalinity is used, since it is the alkalinity that is always found in natural waters. In practice, its concentration must be determined to ascertain if enough is present to satisfy the optimum alum dose. If found deficient, then lime is normally added to satisfy the additional alkalinity requirement. As we have found, the reaction is optimum at a pH of 5.32 at 25°C when the dissolved solids concentration is 140 mg/L.

In any given application, it is uncertain whether the bicarbonate alone or in combination is needed, therefore, the most practical way of expressing the alkalinity requirement is through the use of equivalents. Using this method, the number of equivalents of the alum used is equal to the number of equivalents of the alkalinity required; in fact, it is equal to the same number of equivalents of any species participating in the chemical reaction. All that is needed, therefore, is to find the number of equivalent masses of the alum and the alkalinity needed is equal to this number of equivalent masses.

In the previous reaction, the number of references is 6. Thus, the equivalent mass of alum is Al2(SO4)3 • xH2O/6 = 57.05 + 3x and that of the calcium bicarbonate species is 3Ca(HCO3)2/6 = 81.05. The other alkalinity sources that can be used are lime, caustic soda, and soda ash. Lime is used in the discussion that follows. Also, alkalinity requirements are usually expressed in terms of CaCO3. Therefore, we also express the reactions of alum in terms of calcium carbonate. The respective chemical reactions are:

Al2(SO4)3 • xH2O + 3Ca(OH)2 ^ 2Al(OH)31 + 3CaSO4 + xH2O (12.47)

Al2(SO4)3 • xH2O + 3CaCO3 + 3HOH ^ 2Al(OH)31 + 3CaSO4 + 3CO2 + xH2O

Ca(OH)2 is actually slaked lime: CaO + H2O. Note that in order to find the equivalent masses, the same number of molecules of alum in the balanced chemical reaction as used in Equation (12.46) should be used in Eqs. (12.47) and (12.48); otherwise, the equivalent masses obtained are equivalent to each other. From the reactions, the equivalent mass of lime (CaO) is 3CaO/6 = 28.05 and that of calcium carbonate is 3CaCO3/6 = 50.

It is impossible to determine the optimum dose of alum using chemical reaction. This value must be obtained through the jar test. Let [Alopt]mg and [Alopt]geq be the milligrams per liter and gram equivalents per liter of optimum alum dose, respectively and let V be the cubic meters of water or wastewater treated. Also, let MCaOkgeqAl and MCaOAl be the kilogram equivalents and kilogram mass of lime, respectively, used at a fractional purity of PCaO. In the case of Ca(HCO3)2, the respective symbols are MCa(HCO3)2kgeqAl and MCa(HCO3)2Al at a fractional purity of Pc^hco,),. PCa(HCO3)2 = 1 in natural waters. Note the Al is one of the subscripts. This is to differentiate when ferrous and ferric are used as the coagulants.

The number of kilogram equivalents of alum needed is ([Alopt]geq(1000)/1000) V = [Alopt]geqV = ([Alopt]mg/1000(57.05 + 3x)) V. Let MAlkgeq and MAl be the kilogram equivalents and kilograms of alum used, respectively, at a fractional purity of PAl. Thus,

The most general notion is that MAlkgeq is reacted by all the alkalinities. This, of course, is not necessarily true in practice, but is conceptually correct. Let f AlCa(HC0 )2 be the fraction of MAlkgeq reacted by calcium bicarbonate and fAlCa0, the fraction reacted by lime. If calcium bicarbonate and lime are the only alkalinities reacting with the alum, f AlCa(HC03)2 + f AlCao = 1. We now have f AlCa(HCO3)2[ Alopt ]

M ca(Hco3 )2kgeqAl — " 1000(537Il5 + 3x)g : (12.51) 0.081 f AlCa(HCO3)2 [ Al opt ] mg

Example 12.4 A raw water containing 140 mg/L of dissolved solids is subjected to a coagulation treatment using alum. If the optimum alum dose as determined by a jar test is 40 mg/L, calculate the alkalinity of calcium bicarbonate required. If the natural alkalinity is 100 mg/L as CaCO3, is there enough alkalinity to neutralize the alum dose. Note that, as will be explained later, the equivalent mass of alkalinity expressed as CaCO3 is CaCO3 /2.

Solution:

MCa(HCO3)2kgeqAl — 1000(57.05 + 3x) V f AlCa(HC03)2 — 1 : — 1 m3 let x — 18

Therefore,

MCa(HCO3)2kgeqAl = 1000[5^()+) 3 * = ^t10^ kgeq/m3 A°S

100 3 3 100 mg/L as CaCO3 = 10qq(050) = 2.0(10-3) kgeq/m3

Therefore, enough alkalinity is present. Ans

12.9.2 Key to Understanding Subscripts

At this juncture, we will digress from our discussion and address the question of understanding the subscripts of the mass and fractional variables. Consider first, the mass variables MCa(HC03)2kgeqAl, MCa(HC03)2Al, and Mcsoai.

As discussed in Chapter 10, if T is used as the first subscript, then it refers to the total of the type of species that follows it. In our example variables, this T is not present. For example in MCaCHCOo^kgeqAb the T is not present; but the first subscript is calcium bicarbonate, which is the type of species involved, and there can be no other calcium bicarbonate. Thus, in this instance, the T is not needed. The second subscript in our example is kgeq. From previous conventions, this would be the reason for the existence of the calcium bicarbonate; however, kgeq cannot be a reason for the existence of calcium bicarbonate, since it is a unit of measurement. Therefore, if we see a unit of measurement used as a subscript, that simply indicates the unit of measurement of the type of mass. Thus, kgeq is the unit of the calcium bicarbonate. The last subscript, Al, is the one which is the reason for the existence of the calcium bicarbonate.

M Ca(Hœ3)2kgeqAl then stands for the mass of calcium bicarbonate expressed in kilogram equivalents, with alum used as the coagulant. By the same token, MCa(HCO ) Al is the mass of calcium bicarbonate, however, with no unit specified for the type of mass. Al, again, indicates that alum is the coagulant used. The unit in this symbol cannot be ascertained, but must be established by convention, and the convention we have used must be in kilograms. Finally, then, by using this mass convention, MCaOAl stands for the kilograms of lime used with alum as the coagulant. Other symbols of mass variables can be interpreted similarly.

Now, consider the fractional variables such as f AlCa(HC^^ and f AlCaO. Take f AlCa(Hœ3)2 first. By convention, the first subscript Al is the type of fraction of the species involved. In this instance, it would refer to the mass of alum, which could be MAl or MAlkeq, if the masses are expressed as absolute mass or equivalent mass, respectively. The second subscript, Ca(HCO3)2, is the reason for the existence of Al, which means that calcium bicarbonate is the one reacting with the alum in this fraction. Thus, in other words, fAlCa(HCO ) is the fraction of the alum that is reacted by calcium bicarbonate. By the same token, fAlCaO is the fraction of alum that is reacted by lime. Other symbols of fractional variables can be interpreted similarly.

12.9.3 Chemical Requirements in Ferrous Coagulation Treatment

The chemical reactions for the ferrous coagulant FeSO2 • 7H2O are:

FeSO4 • 7H2O + Ca(HCO3)2 + 2Ca(OH)2 ^ Fe(OH)2i + CaSO4 + 2CaCO3i + 9H2O

FeSO4 • 7H2O + Ca(OH)2 ^ Fe(OH)2i + CaSO4 + 7H2O (12.55b)

These reactions are optimum at pH of 11.95 when the dissolved solids concentration is 140 mg/L and when the temperature is 25°C.

In the first reaction, because calcium bicarbonate is always present in natural waters, it is a participant. Only when all the natural alkalinities are expended will the second reaction proceed. The second reaction assumes that lime is the one used. In the presence of dissolved oxygen, Fe(OH)2 oxidizes to Fe(OH)3. Ferric hydroxide is the intended precipitate in the coagulation process using copperas, since it is a more insoluble precipitate than the ferrous form. In addition, the reaction with oxygen cannot be avoided, because oxygen will always be present in water. The oxidation reaction is

Note that to have equivalence among the species of Reactions (12.55a), (12.55b), and (12.56), Fe(OH)2 is used as the "tying species" between the three reactions. To satisfy the equivalence, one mole of Fe(OH)2 is used in each of the reactions. The number of reference species is then equal to 2. Thus, the equivalent masses are as follows: copperas = FeSO4 • 7H2O/2 = 138.95; Ca(HCO3)2 = Ca(HCO3)2/2 = 81.05; lime = 2CaO/2 = 56.1 [Reaction (12.55a)]; lime = CaO/2 = 28.05 [Reaction (12.55b)]; and oxygen = (1/4)O2/2 = 4.

In Equation (12.56), the equivalent masses could have been based on the number of electrons involved. The ferrous ion is oxidized to the ferric ion; thus, the number of reference species is 1 mole of electrons, for which the equivalent masses would have been divided by 1. If this were done, however, Equation (12.56) would stand unrelated to Eqs. (12.55a), (12.55b).

As in the case of alum, it is impossible to determine the optimum dose using chemical reaction. This value must be obtained through the jar test. Let [FeIIopt]mg and [FeIIopt]geq be the milligrams per liter and gram equivalents per liter of optimum copperas dose, respectively, and let V be the cubic meters of water or wastewater treated. Also, let MCaOkgeqPeII and MCaOFeII be the kilogram equivalents and kilogram mass of lime, respectively, used at a fractional purity of PCaO. In the case of Ca(HCO3)2, the respective symbols are Ma(HCO3)2kgeqFeII and Ma^^en at a fractional purity of PCa(HCO ) . PCa(HCO;,)2 = 1 in natural waters; and, in the case of oxygen, the respective symbols are MO2kgeq and MO2. The fractional purity will be 1.

The number of kilogram equivalents of copperas needed is ([FeIIopt]geq(1000) / 1000) V = [FeIIopt]geqV = ([FeIIopt]mg/1000(138.95))V. Let MFeIIkeq and MFeII be the kilogram equivalents and kilograms of copperas used, respectively, at a fractional purity of PMI. Thus,

The number of kilogram equivalents of pure copperas is the same number of equivalents of pure calcium calcium bicarbonate, lime, and dissolved oxygen. Let/FeIICaO be the fraction of MFeIIkgeq reacted by lime alone without the presence of bicarbonate and f FeiiCa(HCO3 )2 be the fraction reacted in the presence of calcium bicarbonate and lime. f FeiiCaO + f FeiiCa(HCO3)2 = 1. Therefore, considering fractional purities, f FeiiCa(HCO3)2[ FeIIopt ]mg MCaOakgeqFeII = 1000( 13895)pc V [Reaction ( 12.55a)] (12.59)

M CaOaFeii = -(138"-V [ Reaction ( 12.55a )] (12.60)

f FeiiCa(HCO3)2[ F^ 0pt ]mg M Ca(HC°3)2kgeqFeE = 1000( 138.95) V (12.63)

0.081 f FeIICa(HCO3)2[FeIIOpt]mg _

7I3895)

M 0.004[FeIIopt]mg V (1266)

Example 12.5 A raw water containing 140 mg/L of dissolved solids is subjected to coagulation treatment using copperas. If the optimum dose as determined by a jar test is 50 mg/L, calculate the alkalinity of calcium bicarbonate, lime, and dissolved oxygen required. If the natural alkalinity is 100 mg/L as CaCO3, is there enough alkalinity to neutralize the coagulant dose? If the dissolved oxygen is 6.0 mg/L, is this enough to satisfy the DO requirement?

Solution: First, convert 100 mg/L as CaCO3 to kgeq/L. Thus,

100 3 3 100 mg/L as CaCO = J^- = 2.0 ( 10-3 ) kgeq/m3

Assume that the optimum dose is reacted in the presence of both the bicarbonate and lime. Therefore, f FeIICa(HCO3)2[ FeIIopt ]mg M Ca(HC°3)2kgeqFeH = 1000(138.95) V

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