Background Chemistry and Fluid Mechanics

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Trong tính toán, một số yếu tố có thể tham gia vào một thuật ngữ và điều quan trọng là để giữ cho đường của các đơn vị của mỗi trong những yếu tố để xác định đơn vị cuối cùng của thời hạn. Ví dụ, hãy xem xét chuyển đổi 88 kg để microgram. Để thực hiện chuyển đổi này, một số yếu tố có mặt trong thời gian tính. Giả sử chúng ta làm cho việc chuyển đổi như sau: 88 kg = 88 (1000) (1000) (1000...

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TX249_Frame_C00 Page 25 Friday, June 14, 2002 1:47 PM Background Chemistry and Fluid Mechanics As mentioned, this chapter discusses the background knowledge needed in order to understand the subsequent chapters of this book. The student must have already gained this knowledge, but it is presented here as a refresher. Again, the background knowledge to be reviewed includes chemistry and ﬂuid mechanics. Before they are discussed, however, units used in calculations need to be addressed, ﬁrst. This is important because confusion may arise if there is no technique used to decipher the units used in a calculation. UNITS USED IN CALCULATIONS In calculations, several factors may be involved in a term and it is important to keep tract of the units of each of the factors in order to ascertain the ﬁnal unit of the term. For example, consider converting 88 kilograms to micrograms. To make this con- version, several factors are present in the term for the calculation. Suppose we make the conversion as follows: 88 kg = 88 ( 1000 ) ( 1000 ) ( 1000 ) = 88,000,000,000 µ g (1) where 88(1000)(1000)(1000) is called the term of the calculation and 88 and the 1000s are the factors of the term. As can be seen, it is quite confusing what each of the 1000s refers to. To make the calculation more tractable, it may be made by putting the units in each of the factors. Thus, µg 88 kg = 88 kg  1000 -----  1000 ------   1000 ------  = 88,000,000,000 µ g g mg - - - (2)  kg  g  mg This second method makes the calculations more tractable, but it makes the writing long, cumbersome, and impractical when several pages of calculations are done. For example, in designs, the length of the calculations can add up to the thickness of a book. Thus, in design calculations, the ﬁrst method is preferable with its attendant possible confusion of the units. Realizing its simplicity, however, we must create some method to make it tractable. This is how it is done. Focus on the right-hand side of the equation of the ﬁrst calculation. It is known that the unit of 88 is kg, and we want it converted to µg. Remember that conversions follow a sequence of units. For example, to convert © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 26 Friday, June 14, 2002 1:47 PM 26 Physical–Chemical Treatment of Water and Wastewater kg to µg, the sequence might be any one of the following: kg ⇒ g ⇒ mg ⇒ µ g (3) kg ⇒ µ g (4) In Sequence (3), the conversion follows the detailed steps: ﬁrst, from kg to g, then from g to mg and, ﬁnally, from mg to µg. Looking back to Equation (1), this is the sequence followed in the conversion. The ﬁrst 1000 then refers to the g; the second 1000 refers to the mg; and the last 1000 refers to the µg. Note that in this scheme, the value of a given unit is exactly the equivalent of the previous unit. For example, the value 1000 for the g unit is exactly the equivalent of the previous unit, which is the kg. Also, the value 1000 for the mg unit is exactly the equivalent of its previous unit, which is the g, and so on with the µg. In Sequence (4), the method of conversion is a short cut. This is done if the 6 number of micrograms in a kilogram is known. Of course, we know that there are 10 micrograms in a kilogram. Thus, using this sequence to convert 88 kg to micrograms, we proceed as follows: 88 kg = 88 ( 10 ) = 88,000,000,000 µ g 6 Example 1 Convert 88,000,000,000 µg to kg using the detailed-step and the short-cut methods. Solution: Detailed step: 88,000,000,000 µ g = 88,000,000,000 ( 10 ) ( 10 ) ( 10 ) = 88 kg –3 –3 –3 Ans −3 −3 Note that the ﬁrst (10 ) refers to the mg; the second (10 ) refers to the g; and the −3 last (10 ) refers to the kg. There is no need to write the units speciﬁcally. Short cut: 88,000,000,000 µ g = 88,000,000,000 ( 10 ) = 88 kg –6 Ans −6 Note that there are 10 kg in one µg. 3 Example 2 Convert 10 cfs to m /d. Solution: cfs is cubic feet per second. 1 ft = 1 / 3.281 m 1 s = 1/60 min; 1 min = 1/60 hr; 1 hr = 1/24 d Therefore, 13 10 cfs = 10 ft /s = 10  ------------  ---------------------------- = 24,462 m /d 1 3 3 - Ans  3.281  1   1   1  ----- ----- ----- - - -  60  60  24 © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 27 Friday, June 14, 2002 1:47 PM Background Chemistry and Fluid Mechanics 27 Now, let us turn to a more elaborate problem of substituting into an equation. Of course, to use the equation, all the units of its parameters must be known. Accordingly, when the substitution is done, these units must be satisﬁed. For exam- ple, take the following equation: 96,494 [ C o ] Q o η ( m / 2 ) I = ----------------------------------------------------- - (5) M It is impossible to use the above equation if the units of the factors are not known. Thus, any equation, whether empirically or analytically derived, must always have its units known. In the above equation, the following are the units of the factors: I amperes, A 3 [Co] gram equivalents per liter, geq/m 3 Qo cubic meters per second, m /s η dimensionless m dimensionless Now, with all the units known, it is easy to substitute the values into the equation; the proper unit of the answer will simply fall into place. The values can be substituted into the equation in two ways: direct substitution and indirect substitution. Direct substitution means substituting the values directly into the equation and making the conversion into proper units while already substi- tuted. Indirect substitution, on the other hand, means converting the values into their proper units outside the equation before inserting them into the equation. These methods will be elaborated in the next example. Example 3 In the equation I = 96,494[ C oMQo η ( m / 2) , the following values for the ] ------------------------------------------------------ - factors are given: [Co] = 4000 mg/L of NaCl; Qo = 378.51 m /d; η = 0.77, m = 400, 3 and M = 0.90. Calculate the value of I by indirect substitution and by direct substi- tution. Solution: Indirect substitution: In indirect substitution, all terms must be in their proper units before making the substitution, thus: NaCl = 23 + 35.45 = 58.45, molecular mass of sodium chloride Therefore, –3 4000 ( 10 ) [ C o ] = 4000 mg/L = -------------------------- = 0.068 gmols/L - 58.45 3 = 0.068 geq/L = 68 geq/m 1 3 3 Q o = 378.51 m /d = 378.51 --------------------------- = 0.0044 m /s - 24 ( 60 ) ( 60 ) © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 28 Friday, June 14, 2002 1:47 PM 28 Physical–Chemical Treatment of Water and Wastewater And, now, substituting, therefore, 96,494 [ C o ] Q o η ( m / 2 ) 96,494 ( 68 ) ( 0.0044 ) ( 0.77 ) ( 400 / 2 ) I = ----------------------------------------------------- = ------------------------------------------------------------------------------------ - - M 0.90 = 122.84 A Ans Direct substitution: 96,494 [ C o ] Q o η ( m / 2 ) I = ----------------------------------------------------- - M    –3 96,494  4000 ( 10 - ( 1000 )  378.51 ---------------------------- ( 0.77 )/ ( 400 / 2 ) ) 1 -------------------------- 24 ( 60 ) ( 60 )   58.45  = ----------------------------------------------------------------------------------------------------------------------------------------------------- - 0.90 = 122.84 A Ans Note that the conversions into proper units are done inside the equation, and that the conversions for [Co] and Qo are inside pairs of braces {}. Once accustomed to viewing these conversions, you may not need these braces anymore. One last method of ascertaining units in a calculation is the use of consistent units. If a system of units is used consistently, then it is not necessary to keep track of the units in a given calculation. The proper unit of the answer will automatically fall into place. The system of units is based upon the general dimensions of space, mass, and time. Space may be in terms of displacement or volume and mass may be in terms of absolute mass or relative mass. An example of absolute mass is the gram, and an example of relative mass is the mole. The mole is a relative mass, because it expresses the ratio of the absolute mass to the molecular mass of the substance. When the word mass is used without qualiﬁcation, absolute mass is intended. The following are examples of systems of units: meter-kilogram-second (mks), meter-gram-second (mgs), liter-gram-second (lgs), centimeter-gram-second (cgs), liter-grammoles-second (lgmols), meter-kilogrammoles-second (mkmols), centi- meter-grammoles-second (cgmols), etc. Any equation that is derived analytically does not need to have its units speciﬁed, because the units will automatically conform to the general dimension of space, mass, and time. In other words, the units are automatically speciﬁed by the system of units chosen. For example, if the mks system of units is chosen, then the measurement of distance is in units of meters, the measurement of mass is in units of kilograms, and the measurement of time is in seconds. Also, if the lgs system of units is used, then the volume is in liters, the mass is in grams, and the time is in seconds. To repeat, if consistent units are used, it is not necessary to keep track of the units of the various factors, because these units will automatically fall into place by virtue of the choice of the system of units. The use of a consistent system of units is illustrated in the next example. © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 29 Friday, June 14, 2002 1:47 PM Background Chemistry and Fluid Mechanics 29 Example 4 The formula used to calculate the amount of acid needed to lower the pH of water is – pH – pH cur 10 to – 10 [ A cadd ] geq = [ A cur ] geq + -------------------------------------- - φ Calculate the amount of acid needed using the lgmols system of units. Solution: Of course, to intelligently use the above equation, all the factors should be explained. We do not need to do it here, however, because we only need to make substitutions. Because the lgmols units is to be used, volume is in liters, mass is in gram moles, and time is in seconds. Therefore, the corresponding con- centration is in gram moles per liter (gmols/L). Another unit of measurement of concentration is also used in this equation, and this is equivalents per liter. For the lgmols system, this will be gram equivalents per liter (geq/L). Now, values for the factors need to be given. These are shown below and note that no units are given. Because the lgmols system is used, they are understood to be either geq/L or gmols/L. Again, it is not necessary to keep track of the units; they are understood from the system of units used. –3 [ A cur ] geq = 2.74 ( 10 ) pH to = 8.7 pH cur = 10.0 Therefore, – 8.7 – 10 10 – 10 –3 –3 [ A cadd ] geq = 2.74 ( 10 ) + ------------------------------ = 2.74 ( 10 ) geq/L Ans 0.00422 Note that the values are just freely substituted without worrying about the units. By the system of units used, the unit for [Acadd]geq is automatically geq/L. GENERAL CHEMISTRY Chemistry is a very wide ﬁeld; however, only a very small portion, indeed, of this seemingly complex subject is used in this book. These include equivalents and equivalent mass, methods of expressing concentrations, activity and active concen- tration, equilibrium and solubility product constants, and acids and bases. This knowledge of chemistry will be used under the unit processes part of this book. EQUIVALENTS EQUIVALENT MASSES AND The literature shows confused deﬁnitions of equivalents and equivalent masses and no universal deﬁnition exists. They are deﬁned based on speciﬁc situations and are never uniﬁed. For example, in water chemistry, three methods of deﬁning equivalent mass are used: equivalent mass based on ionic charge, equivalent mass based on © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 30 Friday, June 14, 2002 1:47 PM 30 Physical–Chemical Treatment of Water and Wastewater acid–base reactions, and equivalent mass based on oxidation–reduction reactions (Snoeyink and Jenkins, 1980). This section will unify the deﬁnition of these terms by utilizing the concept of reference species; but, before the deﬁnition is uniﬁed, the aforementioned three methods will be discussed ﬁrst. The result of the discussion, then, will form the basis of the uniﬁcation. Equivalent mass based on ionic charge. In this method, the equivalent mass is deﬁned as (Snoeyink and Jenkins, 1980): molecular weight Equivalent mass = ----------------------------------------- (6) ionic charge For example, consider the reaction, Fe ( HCO 3 ) 2 + 2Ca ( OH ) 2 → Fe ( OH ) 2 + 2CaCO 3 + 2H 2 O Calculate the equivalent mass of Fe(HCO3)2. When this species ionizes, the Fe will form a charge of plus 2 and the bicarbonate ion will form a charge of minus 1 but, because it has a subscript of 2, the total ionic charge is minus 2. Thus, from the previous formula, the equivalent mass is Fe(HCO3)2/2, where Fe(HCO3)2 must be evaluated from the respective atomic masses. The positive ionic charge for calcium hydroxide is 2. The negative ionic − charge of OH is 1; but, because the subscript is 2, the total negative ionic charge for calcium hydroxide is also 2. Thus, if the equivalent mass of Ca(OH)2 were to be found, it would be Ca(OH)2/2. It will be mentioned later that Ca(OH)2/2 is not compatible with Fe(HCO3)2/2 and, therefore, Equation (6) is not of universal appli- cation, because it ought to apply to all species participating in a chemical reaction. Instead, it only applies to Fe(HCO3)2 but not to Ca(OH)2, as will be shown later. Equivalent mass based on acid–base reactions. In this method, the equivalent mass is deﬁned as (Snoeyink and Jenkins, 1980): molecular weight Equivalent mass = ----------------------------------------- (7) n where n is the number of hydrogen or hydroxyl ions that react in a molecule. For example, consider the reaction, + 2− H 3 PO 4 + 2NaOH → 2Na + HPO 4 + 2H 2 O Now, calculate the equivalent mass of H3PO4. It can be observed that H3PO4 converts 2− to HPO 4 . Thus, two hydrogen ions are in the molecule of H3PO4 that react and the equivalent mass of H3PO4 is H3PO4/2, using the previous equation. The number of hydroxyl ions that react in NaOH is one; thus, using the previous equation, the equivalent mass of NaOH is NaOH/1. © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 31 Friday, June 14, 2002 1:47 PM Background Chemistry and Fluid Mechanics 31 Equivalent mass based on oxidation–reduction reactions. For oxidation– reduction reactions, the equivalent mass is deﬁned as the mass of substance per mole of electrons involved (Snoeyink and Jenkins, 1980). For example, consider the reaction, 4Fe ( OH ) 2 + O 2 + 2H 2 O → 4Fe ( OH ) 3 The ferrous is oxidized to the ferric form from an oxidation state of +2 to +3. The difference between 2 and 3 is 1, and because there are 4 atoms of Fe, the amount of electrons involved is 1 × 4 = 4. The equivalent mass of Fe(OH)2 is then 4Fe(OH)2/4. Note that the coefﬁcient 4 has been included in the calculation. This is so, because in order to get the total number of electrons involved, the coefﬁcient must be included. The electrons involved are not only for the electrons in a molecule but for the electrons in all the molecules of the balanced chemical reaction, and, therefore must account for the coefﬁcient of the term. For oxygen, the number of moles electrons involved will also be found to be 4; thus, the equivalent mass of oxygen is O2/4. Now, we are going to unify this equivalence using the concept of the reference species. The positive or negative charges, the hydrogen or hydroxyl ions, and the moles of electrons used in the above methods of calculation are actually references species. They are used as references in calculating the equivalent mass. Note that the hydrogen ion is actually a positive charge and the hydroxyl ion is actually a negative charge. From the results of the previous three methods of calculating equivalent mass, we can make the following generalizations: 1. The mass of any substance participating in a reaction per unit of the number of reference species is called the equivalent mass of the substance, and, it follows that 2. The mass of the substance divided by this equivalent mass is the number of equivalents of the substance. The expression molecular weight/ionic charge is actually mass of the substance per unit of the reference species, where ionic charge is the reference species. The expression molecular weight n is also actually mass of the substance per unit of the reference species. In the case of the method based on the oxidation–reduction reaction, no equation was developed; however, the ratios used in the example are ratios of the masses of the respective substances to the reference species, where 4, the number of electrons, is the number of reference species. From the discussion above, the reference species can only be one of two possi- bilities: the electrons involved in an oxidation–reduction reaction and the positive (or, alternatively, the negative) charges in all the other reactions. These species (electrons and the positive or negative charges) express the combining capacity or valence of the substance. The various examples that follow will embody the concept of the reference species. Again, take the following reactions, which were used for the illustrations above: 4Fe(OH)2 + O2 + 2H2O → 4Fe(OH)3 Fe(HCO2)3 + 2Ca(OH)2 → Fe(OH)2 + 2CaCO3 + 2H2O © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 32 Friday, June 14, 2002 1:47 PM In the ﬁrst reaction, the ferrous form is oxidized to the ferric form from an oxidation state of +2 to +3. Thus, in this reaction, electrons are involved, making them the reference species. The difference between 2 and 3 is 1, and since there are 4 atoms of Fe, the amount of electrons involved is 1 × 4 = 4. For the oxygen molecule, its atom has been reduced from 0 to −2 per atom. Since there are 2 oxygen atoms in the molecule, the total number of electrons involved is also equal to 4 (i.e., 2 × 2 = 4). In both these cases, the number of electrons involved is 4. This number is called the number of reference species, combining capacity, or valence. (Number of reference species will be used in this book.) Thus, to obtain the equivalent masses of all the participating substances in the reaction, each term must be divided by 4: 4Fe(OH)2/4, O2/4, 2H2O/4, and 4Fe(OH)3/4. We had the same results for Fe(OH)2 and O2 obtained before. If the total number of electrons involved in the case of the oxygen atom were different, a problem would have arisen. Thus, if this situation occurs, take the convention of using the smaller of the number of electrons involved as the number of reference species. For example if the number of electrons involved in the case of oxygen were 2, then all the participating substances in the chemical reaction would have been divided by 2 rather than 4. For any given chemical reaction or series of related chemical reactions, however, whatever value of the reference species is chosen, the answer will still be the same, provided this number is used consistently. This situation of two competing values to choose from is illustrated in the second reaction to be addressed below. Also, take note that the reference species is to be taken from the reactants only, not from the products. This is so, because the reactants are the ones responsible for the initiation of the interaction and, thus, the initiation of the equivalence of the species in the chemical reaction. In the second reaction, no electrons are involved. In this case, take the convention that if no electrons are involved, either consider the positive or, alternatively, the negative oxidation states as the reference species. For this reaction, initially consider the positive oxidation state. Since the ferrous iron has a charge of +2, ferrous bicarbonate has 2 for its number of reference species. Alternatively, consider the negative charge of bicarbonate. The charge of the bicarbonate ion is −1 and because two bicarbonates are in ferrous bicarbonate, the number of reference species is, again, 1 × 2 = 2. From these analyses, we adopt 2 as the number of reference species for the reaction, subject to a possible modiﬁcation as shown in the paragraph below. (Notice that this is the number of reference species for the whole reaction, not only for the individual term in the reaction. In other words, all terms and each term in a chemical reaction must use the same number for the reference species.) In the case of the calcium hydroxide, since calcium has a charge of +2 and the coefﬁcient of the term is 2, the number of reference species is 4. Thus, we have now two possible values for the same reaction. In this situation, there are two alternatives: the 2 or the 4 as the number of reference species. As mentioned previously, either can be used provided, when one is chosen, all subsequent calculations are based on the one particular choice; however, adopt the convention wherein the number of reference species to be chosen should be the smallest value. Thus, the number of reference species in the second reaction is 2, not 4—and all the equivalent masses of the participating substances are obtained by dividing each balanced term of the reaction by 2: Fe(HCO3)2/2, 2Ca(OH)2/2, Fe(OH)2/2, 2CaCO3/2 and 2H2O/2. © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 33 Friday, June 14, 2002 1:47 PM Note that Ca(OH)2 has now an equivalent mass of 2Ca(OH)2/2 which is different from Ca(OH)2/2 obtained before. This means that the deﬁnition of equivalent mass in Equation (6) is not accurate, because it does not apply to Ca(OH)2. The equivalent mass of Ca(OH)2/2 is not compatible with Fe(HCO3)2/2, Fe(OH)2/2, 2CaCO3/2, or 2H2O/2. Compatibility means that the species in the chemical reaction can all be related to each other in a calculation; but, because the equivalent mass of Ca(OH)2 is now made incompatible, it could no longer be related to the other species in the reaction in any chemical calculation. In contrast, the method of reference species that is developed here applies in a uniﬁed fashion to all species in the chemical reaction: Fe(HCO3)2, Ca(OH)2, Fe(OH)2, CaCO3, and H2O and the resulting equiv- alent masses are therefore compatible to each other. This is so, because all the species are using the same number of reference species. Take the two reactions of phosphoric acid with sodium hydroxide that follow. These reactions will illustrate that the equivalent mass of a given substance depends upon the chemical reaction in which the substance is involved. + 2− H 3 PO 4 + 2NaOH → 2Na + HPO 4 + 2H 2 O + 3− H 3 PO 4 + 3NaOH → 3Na + PO 4 + 3H 2 O + Consider the positive electric charge and the ﬁrst reaction. Because Na of NaOH (remember that only the reactants are to be considered in choosing the reference species) has a charge of +1 and the coefﬁcient of the term is 2, two positive charges (2 × 1 = 2) are involved. In the case of H3PO4, the equation shows that the acid 2− breaks up into HPO 4 and other substances with one H still “clinging” to the PO4 + on the right-hand side of the equation. This indicates that two H ’s are involved in + the breakup. Because the charge of H is +1, two positive charges are accordingly + + involved. In both the cases of Na and H , the reference species are the two positive charges and the number of reference species is 2. Therefore, in the ﬁrst reaction, the equivalent mass of a participating substance is obtained by dividing the term (including the coefﬁcient) by 2. Thus, for the acid, the equivalent mass is H3PO4/2; for the base, the equivalent mass is 2NaOH/2, etc. In the second reaction, again, basing on the positive electric charges and per- forming similar analysis, the number of reference species would be found to be +3. Thus, in this reaction, for the acid, the equivalent mass is H3PO4/3; for the base, the equivalent mass is 3NaOH/3, etc., indicating differences in equivalent masses with the ﬁrst reaction for the same substances of H3PO4 and NaOH. Thus, the equivalent mass of any substance depends upon the chemical reaction in which it participates. In the previous discussions, the unit of the number of reference species was not established. A convenient unit would be the mole (i.e., mole of electrons or mole of positive or negative charges). The mole can be a milligram-mole, gram-mole, etc. The mass unit of measurement of the equivalent mass would then correspond to the type of mole used for the reference species. For example, if the mole used is the gram-mole, the mass of the equivalent mass would be expressed in grams of the substance per gram-mole of the reference species; and, if the mole used is the milligram-mole, © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 34 Friday, June 14, 2002 1:47 PM the equivalent mass would be expressed in milligrams of the substance per milligram- mole of the reference species and so on. Because the reference species is used as the standard of reference, its unit, the mole, can be said to have a unit of one equivalent. From this, the equivalent mass of a participating substance may be expressed as the mass of the substance per equivalent of the reference species; but, because the substance is equivalent to the reference species, the expression “per equivalent of the reference species” is the same as the expression “per equivalent of the substance.” Thus, the equivalent mass of a substance may also be expressed as the mass of the substance per equivalent of the substance. Each term of a balanced chemical reaction, represents the mass of a participating substance. Thus, the general formula for ﬁnding the equivalent mass of a substance is mass of substance Equiv. mass = ------------------------------------------------------------------------------------ - number of equivalents of substance term in balanced reaction = ----------------------------------------------------------------------------------------- number of moles of reference species Example 5 Water containing 2.5 moles of calcium bicarbonate and 1.5 moles of calcium sulfate is softened using lime and soda ash. How many grams of calcium carbonate solids are produced (a) using the method of equivalent masses and (b) using the balanced chemical reaction? Pertinent reactions are as follows: Ca ( HCO 3 ) 2 + Ca ( OH ) 2 → 2CaCO 3 ↓ + 2H 2 O CaSO 4 + Na 2 CO 3 → CaCO 3 + Na 2 SO 4 Solution: (a) Ca ( HCO 3 ) 2 + Ca ( OH ) 2 → 2CaCO 3 ↓ + 2H 2 O number of reference species = 2 2CaCO 3 Therefore, eq. mass of CaCO 3 = ------------------- = 100 - 2 Ca ( HCO 3 ) 40 + 2 [ 1 + 12 + 3 ( 16 ) ] 162 eq. mass of Ca ( HCO 3 ) 2 = ---------------------------2 = ------------------------------------------------------- -------- = 81 - -- 2 2 2 2.5 ( 162 ) qeq of Ca ( HCO 3 ) 2 = -------------------- = 5 = geq of CaCO 3 - 81 g of CaCO 3 solids = 5 ( 100 ) = 500 g Ans CaSO 4 + Na 2 CO 3 → CaCO 3 + Na 2 SO 4 © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 35 Friday, June 14, 2002 1:47 PM Background Chemistry and Fluid Mechanics 35 number of reference species = 2 CaCO 3 Therefore, eq. mass of CaCO 3 = ---------------- = 50 - 2 40 + 32 + 4 ( 16 ) CaSO 4 136 eq. mass of CaSO 4 = ---------------- = -------------------------------------- = -------- = 68 - 2 2 2 1.5 ( 136 ) qeq of CaSO 4 = -------------------- = 3 = geq of CaCO 3 - 68 g of CaCO 3 solids = 3 ( 50 ) = 150 g Ans Total grams of CaCO 3 = 500 + 150 = 650 Ans (b) Ca ( HCO 3 ) 2 + Ca ( OH ) 2 → 2CaCO 3 ↓ + 2H 2 O 2CaCO 3 g of CaCO 3 solids = --------------------------- ( 2.5 ) ( 162 ) = 500 - Ca ( HCO 3 ) 2 CaSO 4 + Na 2 CO 3 → CaCO 3 + Na 2 SO 4 CaCO 3 g of CaCO 3 solids = ---------------- ( 1.5 ) ( 136 ) = 150 - CaSO 4 Total grams of CaCO 3 = 500 + 150 = 650 Ans METHODS EXPRESSING CONCENTRATIONS OF Several methods are used to express concentrations in water and wastewater and it is appropriate to present some of them here. They are molarity, molality, mole fraction, mass concentration, equivalents concentration, and normality. Molarity. Molarity is the number of gram moles of solute per liter of solution, where from the general knowledge of chemistry, gram moles is the mass in grams divided by the molecular mass of the substance in question. Take note that the statement says “per liter of solution.” This means the solute and the solvent are taken together as a mixture in the liter of solution. The symbol used for molarity is M. For example, consider calcium carbonate. This substance has a mass density ρ CaCO 3 equals 2.6 g/cc. Suppose, 35 mg is dissolved in a liter of water, ﬁnd the corresponding molarity. The mass of 35 mg is 0.035 g. Calcium carbonate has a molecular weight of 100 g/mol; thus, 0.035 g is 0.035/100 = 0.00035 gmol. The volume corresponding to 0.35 g is 0.035/2.6 = 0.0135 cc = 0.0000135 L and the total volume of the mixture then is 1.0000135 L. Therefore, by the deﬁnition of molarity, the corre- sponding molarity of 35 mg dissolved in one liter of water is 0.00035/1.0000135 = 0.00035 M. Molality. Molarity is the number of gram moles of solute per 1000 g of solvent. Take note of the drastic difference between this deﬁnition of molality and the diﬁnition of molarity. The solvent is now “separate” from the solute. The symbol used for molality is m. © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 36 Friday, June 14, 2002 1:47 PM Now, consider the calcium carbonate example above, again, and ﬁnd the corre- sponding molality. The only other calculation we need to do is to ﬁnd the number of grams of the liter of water. To do this, an assumption of the water temperature must be made. Assuming it is 5°C, its mass density is 1.0 g/cc and the mass of one liter is then 1000 g. Thus, the 0.00035 gmol of calcium carbonate is dissolved in 1000 gm of water. This is the very deﬁnition of molality and the corresponding molality is therefore 0.00035m. Note that there is really no practical difference between molarity and molality in this instance. Note that the mass density of water does not vary much from the 5°C to 100°C. Mole fraction. Mole fraction is a method of expressing the molar fractional part of a certain species relative to the total number of moles of species in the mixture. Letting ni be the number of moles of a particular species i, the mole fraction of this species xi is ni x i = ----------- (8) N ∑1 ni N is the total number of species in the mixture. Example 6 The results of an analysis in a sample of water are shown in the table below. Calculate the mole fractions of the respective species. Ions Conc (mg/L) Ca(HCO3)2 150 Mg(HCO3)2 12.0 Na2SO4 216.0 Solution: The calculations are shown in the table, which should be self- explanatory. Conc Mole Ions (mg/L) Molecular Mass Moles/Liter Fraction a b 40.1(2) + 2{1 + 12 + 3(16)} = 202.2 Ca(HCO3)2 150 0.74 0.32 24.3(2) + 2{1 + 12 + 3(16)} = 170.6 Mg(HCO3)2 12.0 0.07 0.03 23(2) + 32.1 + 4(16) = 142.1 Na2SO4 216.0 1.52 0.65 Sum = ∑ 2.33 1.00 150 a ------------ = 0.74 - 202.2 0.74 b --------- = 0.32 - 2.33 Mass concentration. Generally, two methods are used to express mass concen- tration: mass of solute per unit volume of the mixture (m/v basis) and mass of the © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 37 Friday, June 14, 2002 1:47 PM solute per unit mass of the mixture (m/m basis). In environmental engineering, the most common expression in the m/v basis is the mg/L. The most common in the m/m basis is the ppm, which means parts per million. In other words, in a ppm, there is 6 one mass of the solute in 10 mass of the mixture. One ppm for a solute dissolved in water can be shown to be equal to one mg/L. 6 3 This is shown as follows: 1 ppm = (1 mg)/(10 mg) = (1 mg)/(10 g). The mass density of water at 5°C is 1.0 g/cc. Therefore, 1 mg 1 mg 1 mg 1 mg 1 ppm = ---------------- = ----------- = ------------- = ----------- - - - 6 3 3 L 10 mg 10 g 10 cc The mass density of water decreases from 1.0 g/cc at 5°C to 0.96 g/cc at 100°C. Thus, for practical purposes, 1 ppm = 1 mg/L Molar concentration. In concept, molar concentrations can also be expressed on the m/v basis and the m/m basis; however, the most prevalent practice in envi- ronmental engineering is the m/v basis. Molar concentration, then, is the number of moles of the solute per unit volume of the mixture. There are several types of moles: milligram-moles, gram-moles, tonne-moles, and so on corresponding to the unit of mass used. In chemistry, the gram moles is almost exclusively used. When the type of moles is not speciﬁed, it is understood to be gram moles. So, normally, molar concentration is expressed in gram moles per liter (gmmols/L). An important application of molar concentration is in a molar mass balance. For example, in the removal of phosphorus using alum, the following series of reactions occurs: 3+ 3− Al + PO 4 AlPO 4 ( s ) ↓ 3+ 3− AlPO 4 ( s ) ↓ Al + PO 4 3− + 2− PO 4 + H HPO 4 2− + − HPO 4 + H H 2 PO 4 − + H 2 PO 4 + H H 3 PO 4 An indicator of the efﬁciency of removal is that the concentration of phosphorus in solution must be the minimum. To ﬁnd this phosphorus in solution, it is necessary to perform a phosphate (PO4) molar mass balance in solution using the above reactions. AlPO4(s) is a solid; thus, not in solution and will not participate in the balance. The other species, however, are in solution and contain the PO4 radical. 3− 2− − They are [PO 4 ], [HPO 4 ], [H 2 PO 4 ], and [H 3 PO 4 ]. Note the PO4 embedded in the respective formulas; thus, the formulas are said to contain the PO4 radical. Because they contain this PO4 radical, they will participate in the molar mass balance. Let [ sp PO4 Al ] © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 38 Friday, June 14, 2002 1:47 PM represent the total molar concentration of the phospate radical contained in all the species. Then, the molar mass balance on PO4 becomes 3− 2− − [ sp PO4 Al ] = [ PO 4 ] + [ HPO 4 ] + [ H 2 PO 4 ] + [ H 3 PO 4 ] The symbol [ sp PO4 Al ] needs to be explained further, since this mode of subscript- ing is used in the unit processes part of this book. First, [ ] is read as “the concentration of.” The symbol sp stands for species and its ﬁrst subscript, PO4, stands for the type of species, which is the PO4 radical. The second subscript, Al, stands for the “reason” for the existence of the type of species. In other words, the use of alum (the Al) is the reason for the existence of the phosphate radical, the type of species. Example 7 In removing the phosphorus using a ferric salt, the following series of reactions occurs: 3+ 3− Fe + PO 4 FePO 4 ↓ 3+ 3− FePO 4 ↓ Fe + PO 4 3− + 2− PO 4 + H HPO 4 2− + − HPO 4 + H H 2 PO 4 − + H 2 PO 4 + H H 3 PO 4 Write the molar mass balance for the total molar concentration of the phosphate radical. Solution: The solution to this problem is similar to the one discussed in the text, except that [ sp PO4 Al ] will be replaced by [ sp PO4 FeIII ] . The second subscript, FeIII, stands for the ferric salt. The molar mass balance is 3− 2− − [ sp PO4 FeIII ] = [ PO 4 ] + [ HPO 4 ] + [ H 2 PO 4 ] + [ H 3 PO 4 ] Ans Example 8 State the symbol [ sp PO4 FeIII ] in words. 3− Solution: [ sp PO4 FeIII ] is the total molar concentration of the PO 4 radical as a result of using a ferric salt. Ans Example 9 The complexation reaction of the calcium ion with the carbonate − 2− species, OH , and SO 4 are given below: 2+ 2− o Ca + CO 3 CaCO 3 + 2+ − Ca + HCO 3 CaHCO 3 + 2+ − Ca + OH CaOH 2+ 2− o Ca + SO 4 CaSO 4 © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 39 Friday, June 14, 2002 1:47 PM Write the molar mass balance for the total molar concentration of Ca. Solution: Let [CaT] represent the total molar concentration of Ca. Therefore, 2+ + + o o [ Ca T ] = [ Ca ] + [ CaCO 3 ] + [ CaHCO 3 ] + [ CaOH ] + [ CaSO 4 ] Ans Equivalent concentration and normality. This method of expressing concen- trations is analogous to molar concentrations, with the solute expressed in terms of equivalents. One problem that is often encountered is the conversion of a molar concentration to equivalent concentration. Let [C] be the molar concentration of any substance, where the symbol [ ] is read as “the concentration of.” Convert this to equivalent concentration. To do the conversion, ﬁrst convert the molar concentration to mass concentration by multiplying it by the molecular mass (MM). Thus, [C] (MM) is the corresponding mass concentration. By deﬁnition, the number of equivalents is equal to the mass divided by the equivalent mass (eq. mass). Therefore, the equivalent concentration, [C]eq, is [ C ] ( MM ) [ C ] eq = ----------------------- - eq. mass The concentration expressed as geq/L is the normality. The symbol for normality is N. Example 10 The concentration of Ca(HCO3)2 is 0.74 gmol/L. Convert this concentration to geq/L. Solution: [ C ] ( MM ) [ C ] eq = ----------------------- - eq. mass [ C ] = 0.74 gmol/L MM = 40.1 ( 2 ) + 2 { 1 + 12 + 3 ( 16 ) } = 202.2 No. of reference species = 2 Ca ( HCO 3 ) 202.2 Therefore, eq. mass = ---------------------------2 = ------------ - - 2 2 0.74 ( 202.2 ) [ C ] eq = ---------------------------- = 1.48 geq/L Ans - 202.2 ------------ - 2 Another problem often encountered in practice is the conversion of equivalent concentration to molar concentration. Let us illustrate this situation using the car- 2− − bonate system. The carbonate system is composed of the species CO 3 , HCO 3 , OH , + 2+ + and H . In addition, Ca may also be a part of this system. Note that OH and H are always part of the system, because a water solution will always contain these species. To perform the conversion, the respective equivalent masses of the species should ﬁrst be found. In order to ﬁnd the number of reference species, the pertinent © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 40 Friday, June 14, 2002 1:47 PM chemical reactions must all be referred to a common end point. For the carbonate + system, this end point is the methyl orange end point when H is added to the system to form H2CO3. − The reaction to the end point of HCO 3 is − + HCO 3 + H H 2 CO 3 Because the number of reference species from this reaction is 1, the equivalent mass − 2− of HCO 3 is HCO 3 / 1 . For CO 3 , the reaction to the end point is 2− + CO 3 = 2H H 2 CO 3 This reaction gives the number of reference species equal to 2; thus, the equivalent + − 2− mass of CO 3 is CO3/2. The reaction of H and OH to the end point gives their 2+ respective equivalent masses as H/1 and OH/1. The reaction of Ca in the carbonate system is 2+ 2− Ca + CO 3 CaCO 3 2+ This gives the equivalent of Ca as Ca/2. The conversion from equivalent to molar concentrations is illustrated in the following example. Example 11 An ionic charge balance in terms of equivalents for a carbonate system is shown as follows: 2− − + 2+ [ CO 3 ] eq + [ HCO 3 ] eq + [ OH ] eq = [ H ] eq + [ Ca ] eq Convert the balance in terms of molar concentrations. Solution: [ C ] ( MM ) [ C ] eq = ----------------------- - eq. mass CO 3 HCO OH 2− − ( CO 3 ) eq. mass = --------- ; ( HCO 3 ) eq. mass = -------------3 ( OH ) eq. mass = ------- - -; - 2 1 1 H Ca + 2+ ( H ) eq. mass = --- ; ( Ca ) eq. mass = ----- - - 1 2 Therefore, substituting, 2− − [ CO 3 ] ( CO 3 ) [ HCO 3 ] ( HCO 3 ) [ OH − ] ( OH ) + 2+ [ H ] ( H ) [ Ca ] ( Ca ) -------------------------------- + --------------------------------------- + ---------------------------- = -------------------- + --------------------------- - - - - CO OH H Ca HCO 3 3 ------- - --- - ----- - --------- - -------------- - 1 1 2 2 1 2− − + 2+ 2 [ CO 3 ] + [ HCO 3 ] + [ OH ] = [ H ] + 2 [ Ca ] Ans © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 41 Friday, June 14, 2002 1:47 PM ACTIVITY ACTIVE CONCENTRATION AND In simple language, activity is a measure of the effectiveness of a given species in a chemical reaction. It is an effective or active concentration and is proportional to concentration. It has the units of concentration. Since activity bears a relationship to concentration, its value may be obtained using the value of the corresponding concentration. This relationship is expressed as follows: { sp } = γ [ sp ] (9) 2+ 2− − where sp represents any species involved in the equilibria such as Ca , CO 3 , HCO 3 and so on, in the case of the carbonate equilibria. The pair of braces, { }, is read as the “activity of ” and the pair of brackets, [ ], is read as “the concentration of;” γ is called activity coefﬁcient. The activity coefﬁcient expresses the effect of ions on the reactive ability of a species. As the species become crowded, the reactive ability or effectiveness of the species to react per unit individual of the species is diminished. As they become less concentrated, the effectiveness per unit individual is improved. Thus, at very dilute solutions, the activity coefﬁcient approaches unity; at a more concentrated solution, the activity coefﬁcient departs from unity. Because of the action of the charges upon each other, the activity of the ionized particles is smaller than those of the unionized particles. Ionized particles tend to maintain “relationship” between counterparts, slowing down somewhat their inter- 2+ 2− actions with other particles. Thus, Ca and CO 3 , which are ionization products of CaCO3, have activity coefﬁcients less than unity; they tend to maintain relationship with each other rather than with other particles. Thus, their activity with respect to other particles is diminished. On the other hand, a solid that is not ionized such as CaCO3 before ionization, has an activity coefﬁcient of unity; a liquid or solvent such as water (not ionized) has an activity coefﬁcient of unity. Gases that are not disso- ciated have activity coefﬁcients of unity. (Because the activity coefﬁcients are unity, all molar concentrations of these unionized species have a unit of activity.) In other words, unionized particles are free to interact with any other particles, thus having the magnitude of the highest possible value of the activity coefﬁcient. They are not restricted to maintain any counterpart interaction, because they do not have any. EQUILIBRIUM SOLUBILITY PRODUCT CONSTANTS AND Let a reactant be represented by the solid molecule AaBb. As this reactant is mixed with water, it dissolves into its constituent solute ions. The equilibrium dissolution reaction is a A + bB Aa Bb (10) a b { A} {B} The ratio ------------------------- is called the reaction quotient. Note that in the deﬁnition of { Aa Bb } reaction quotients, the activities of the reactants and products are raised to their © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 42 Friday, June 14, 2002 1:47 PM respective coefﬁcients. At equilibrium, the reaction quotient becomes the equilibrium constant. Thus, the equilibrium constant is a b { A} {B} K = ------------------------ - (11) { Aa Bb } Note the pair of braces denoting activity. Because AaBb is a solid, its activity is unity. For this reason, the product K{AaBb} is a constant and is designated as Ksp. Ksp is called the solubility product constant of the equilibrium dissolution reaction. Equation (11) now transforms to a b K sp = { A } { B } (12) At equilibrium, neither reactants nor products increase or decrease with time. Thus, Ksp’s, being constants, can be used as indicators whether or not a given solid will form or dissolve in solution. −9 For example, CaCO3 has a Ksp of 4.8(10 ) at 25°C. This value decreases to −9 2.84(10 ) at 60°C. The equilibrium reaction for this solid is 2+ 2− 2+ 2− –9 Ca + CO 3 . K sp = { Ca ) ( CO 3 } = 4.8 ( 10 ) at 25 ° C CaCO 3 ( s ) 2+ 2− –9 = { Ca ) ( CO 3 } = 2.84 ( 10 ) at 60 ° C Assuming the reaction is at equilibrium at 25°C, the ions at the right side of the reaction will neither cause CaCO3 to form nor dissolve. At 25°C, by the deﬁnition of the solubility product constant and, since the reaction is assumed to be in equilibrium, −9 the product of the activities will equal 4.8(10 ). As the temperature increases from −9 −9 25°C to 60°C, the product of the activities decreases from 4.8(10 ) to 2.84(10 ). This means that there are fewer particles of the ions existing than required to maintain equilibrium at this higher temperature. As a consequence, some of the particles will combine to form a precipitate, the CaCO3 solid. Table 1 shows values of Ksp’s of solids that are of importance in ascertaining whether or not a certain water sample will form or dissolve the respective solids. The subscripts s and aq refers to “solid” and “aqueous,” respectively. In writing the chemical reactions for the discussions in this book, these subscripts will not be indicated unless necessary for clarity. Of all the Ksp’s discussed previously, the one involving the carbonate system equilibria is of utmost importance in water and wastewater treatment. This is because carbon dioxide in the atmosphere affects any water body. As carbon dissolves in ∗ − 2− water, the carbonate system species H 2 CO 3 , HCO 3 , and CO 3 are formed. Cations + − will then interact with these species and, along with the H and OH that always ∗ exist in water solution, complete the equilibrium of the system. H 2 CO 3 is a mixture of CO2 in water and H2CO3 (note the absence of the asterisk in H2CO3). The CO2 ∗ in water is written as CO2(aq) and H 2 CO 3 is carbonic acid. © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 43 Friday, June 14, 2002 1:47 PM TABLE 1 Solubility Product Constants of Respective Solids at 25°C Ksp reaction of solid Ksp Signiﬁcance + − −10 Ag ( aq ) + Cl ( aq ) AgCl ( s ) 1.8(10 ) Chloride analysis 3+ − −33 Al ( OH ) 3 ( s ) + 3OH Al 1.9(10 ) Coagulation ( aq ) ( aq ) 2+ 2− −10 + SO BaSO 4 ( s ) Ba 10 Sulfate analysis ( aq ) 4 ( aq ) 2+ 2− −9 + CO CaCO 3 ( s ) Ca 4.8(10 ) Hardness removal, scales ( aq ) 3 ( aq ) 2+ − −6 Ca ( OH ) 2 ( s ) + 2OH Ca 7.9(10 ) Hardness removal ( aq ) ( aq ) 2+ 2− −5 + SO CaSO 4 ( s ) Ca 2.4(10 ) Flue gas desulfurization ( aq ) 4 ( aq ) 2− 2− −25 Ca 3 ( PO 4 ) 2 ( s ) + 2PO 3Ca 10 Phosphate removal ( aq ) ( aq ) 2− −6 Ca ( aq ) + HPO CaHPO 4 ( s ) 5.0(10 ) Phosphate removal 4 ( aq ) 2+ − −11 + 2F CaF 2 ( s ) Ca 3.9(10 ) Fluoridation ( aq ) ( aq ) 3+ − −31 Cr ( OH ) 3 ( s ) + 3OH Cr 6.7(10 ) Heavy metal removal ( aq ) ( aq ) 2+ − −20 Cu ( OH ) 2 ( s ) + 2OH Cu 5.6(10 ) Heavy metal removal ( aq ) ( aq ) 3+ − −36 Fe ( OH ) 3 ( s ) + 3OH Fe 1.1(10 ) Coagulation, iron removal, ( aq ) ( aq ) corrosion 2+ − −15 Fe ( OH ) 2 Fe ( aq ) + 2OH ( aq ) 7.9(10 ) Coagulation, iron removal, corrosion 2+ 2− −5 Mg ( aq ) + CO 3 ( aq ) MgCO 3 ( s ) 10 Hardness removal, scales 2+ − −11 Mg ( OH ) 2 + 2OH Mg 1.5(10 ) Hardness removal, scales ( aq ) ( aq ) 2+ − −14 Mn ( OH ) 2 + 2OH Mn 4.5(10 ) Manganese removal ( aq ) ( aq ) − −14 2+ Ni ( OH ) 2 + 2OH Ni 1.6(10 ) Heavy metal removal ( aq ) ( aq ) − −17 2+ Zn ( OH ) 2 + 2OH Zn 4.5(10 ) Heavy metal removal ( aq ) ( aq ) From A. P. Sincero and G. A. Sincero (1996). Enviromental Engineering: A Design Approach. Prentice Hall, Upper Saddle River, NJ, 42. Another important application of the equilibrium constant K in general, [Equation (11)], is in the coagulation treatment of water using alum, Al2(SO4)3 ·14 H2O. (The “14” actually varies from 13 to 18.) In coagulating a raw water using alum, a number 3+ of complex reactions are formed by the Al ion. These reactions are as follows: 3+ 2+ + Al + H 2 O Al ( OH ) + Η 2+ + (13) { Al ( OH ) } { H } −5 K Al ( OH ) c = ------------------------------------------- = 10 - 3+ { Al } © 2003 by A. P. Sincero and G. A. Sincero
TX249_Frame_C00 Page 44 Friday, June 14, 2002 1:47 PM 3+ 4+ + 7Al + 17H 2 O Al 7 ( OH ) 17 + 17 Η 4+ + 17 (14) { Al 7 ( OH ) 17 } { H } − 48.8 = ------------------------------------------------- = 10 - K Al7 ( OH )17 c 3+ 7 { Al } 3+ 5+ + 13Al + 34H 2 O Al 13 ( OH ) 34 + 34 Η 5+ + 34 (15) { Al 13 ( OH ) 34 } { H } − 97.4 K Al13 ( OH )34 c = --------------------------------------------------- = 10 - 3 + 13 { Al } 3+ − Al ( OH ) 3 ( s ) ( fresh precipitate ) Al + 3OH (16) 3+ −3 – 33 K sp ,Al ( OH )3 = { Al } { OH } = 10 − − Al ( OH ) 3 ( s ) + OH Al ( OH ) 4 − (17) { Al ( OH ) 4 } +1.3 K Al ( OH )4 c = --------------------------- = 10 - − OH 3+ 4+ + 2Al + 2H 2 O Al 2 ( OH ) 2 + 2H 4+ +2 (18) { Al 2 ( OH ) 2 } { H } – 6.3 K Al2 ( OH )2 c = ----------------------------------------------- = 10 - 3+ 2 { Al } The equilibrium constants K Al(OH) c , K Al7 (OH)17 c , K Al13 (OH)34 c , K sp ,Al(OH)3 , K Al(OH)4 c , and K Al2 (OH)2 c apply at 25°C. Note that the subscript c is used for the equilibrium constants of the complexes; it stands for “complex.” Al(OH)3(s) is not a complex; thus, K sp ,Al(OH)3 does not contain the subscript c. The previous equations can be used to determine the conditions that will allow maximum precipitation of the solid represented by Al(OH)3. The maximum precip- itation of Al(OH)3 will produce the utmost clarity of the treated water. To allow for 2+ this maximum precipitation, the concentrations of the complex ions Al(OH) , 3+ 4+ − 4+ Al 7 (OH) 17 , Al13(OH)34, Al (OH) 4 , and Al 2 (OH) 2 and Al must be held to a minimum. This will involve ﬁnding the optimum pH of coagulation. This optimum pH may be determined as follows: It is obvious that the complex ions contain the Al atom. Thus, the technique is to sum up their molar concentrations in terms of their Al atom content. Once they have been summed up, they are then eliminated using the previous Κ equations, Eqs. (13) to (18), with the objective of expressing the resulting equation in terms of the constants and the hydrogen ion concentration. Because the K constants are constants, the equations would simply be expressed in terms of one variable, the hydrogen ion concentration, and the equation can then be easily differentiated to obtain the optimum pH of coagulation. © 2003 by A. P. Sincero and G. A. Sincero

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