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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
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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
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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 )
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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.
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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
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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.
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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
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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.
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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,
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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
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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.
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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
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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 ]
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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
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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
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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
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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
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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.
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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 }
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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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