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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:
(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,
(2)
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

µ

g

.

Remember that conversions follow a sequence of units. For example, to convert
88 kg 88 1000()1000()1000()88,000,000,000
µ
g==
88 kg 88 kg 1000
g
kg



1000
mg
g



1000
µ
g
mg



88,000,000,000
µ
g==

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26

Physical–Chemical Treatment of Water and Wastewater

kg to

µ

g, the sequence might be any one of the following:
(3)
(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

µ

g. 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

µ

g. 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

µ

g.
In Sequence (4), the method of conversion is a

short cut

. This is done if the
number of micrograms in a kilogram is known. Of course, we know that there are 10

6

micrograms in a kilogram. Thus, using this sequence to convert 88 kg to micrograms,
we proceed as follows:

Example 1

Convert 88,000,000,000

µ

g to kg using the detailed-step and the
short-cut methods.

Solution:

Detailed step:
Note that the ﬁrst (10

−

3

) refers to the mg; the second (10

−

3

) refers to the g; and the
last (10

−

3

) refers to the kg. There is no need to write the units speciﬁcally.
Short cut:
Note that there are 10

−

6

kg in one

µ

g.

Example 2

Convert 10 cfs to m

3

/d.

Solution:

cfs is cubic feet per second.

1 s

=

1

/

60 min; 1 min

=

1

/

60 hr; 1 hr

=

1

/

24 d
Therefore,

kggmg
µ
g⇒⇒ ⇒

kg
µ
g⇒
88 kg 88 10
6
()88,000,000,000
µ
g==
88,000,000,000
µ
g 88,000,000,000 10
3–
()10
3–
()10
3–
()88 kg Ans==
88,000,000,000
µ
g 88,000,000,000 10
6–
()88 kg Ans==
1ft 1/3.281 m=
10 cfs 10 ft
3
/s 10
1
3.281




3
1
1
60



1
60



1
24




24,462 m
3
/d Ans== =

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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:
(5)
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

[

C

o

] gram equivalents per liter, geq/m

3

Q

o

cubic meters per second, m

3

/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

=

, the following values for the
factors are given: [

C

o

]

=

4000 mg/L of NaCl;

Q

o

=

378.51 m

3

/

d

;

η

=

0.77,

m

=

400,
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,
I
96,494 C
o

[]Q
o
η
m/2()
M
=
96,494[C
o
]Q
o
η
(m/2)
M

C
o
[] 4000 mg/L
4000 10
3–
()
58.45

0.068 gmols/L===
0.068 geq/L 68 geq/m
3
==
Q
o
378.51 m
3

/d 378.51
1
24 60()60()

0.0044 m
3
/s== =

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28 Physical–Chemical Treatment of Water and Wastewater
And, now, substituting, therefore,
Direct substitution:
Note that the conversions into proper units are done inside the equation, and
that the conversions for [C
o
] and Q
o
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.
I
96,494 C
o
[]Q
o
η
m/2()
M

96,494 68()0.0044()0.77()400/2()
0.90

==
122.84 A Ans=

I
96,494 C
o
[]Q
o
η
m/2()
M

=
96,494
4000 10
3–
()
58.45

1000()



378.51
1
24 60()60()




0.77()/400/2()
0.90

=
122.84 A Ans=
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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
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.
Therefore,
Note that the values are just freely substituted without worrying about the units. By
the system of units used, the unit for [A
cadd
]
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 AND EQUIVALENT MASSES
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
A
cadd
[]
geq
A
cur
[]
geq
10
pH–
to
10
pHcur–
–
φ

+=
A
cur
[]
geq
2.74 10
3–

() pH
to
8.7 pH
cur
10.0===
A
cadd
[]
geq
2.74 10
3–
()
10
8.7–
10
10–
–
0.00422

+ 2.74 10
3–
() geq/L Ans==
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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):
(6)
For example, consider the reaction,
Calculate the equivalent mass of Fe(HCO
3
)
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(HCO
3
)
2
/2, where Fe(HCO
3
)
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(HCO
3
)
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(HCO
3
)
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):
(7)
where n is the number of hydrogen or hydroxyl ions that react in a molecule.
For example, consider the reaction,
Now, calculate the equivalent mass of H
3
PO
4
. It can be observed that H
3
PO
4
converts

to . Thus, two hydrogen ions are in the molecule of H
3
PO
4
that react and the
equivalent mass of H
3
PO
4
is H
3
PO
4
/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.
Equivalent mass
molecular weight
ionic charge

=
Fe HCO
3
()
2
2Ca OH()
2
+ Fe OH()
2
2CaCO

3
2H
2
O++→
Equivalent mass
molecular weight
n

=
H
3
PO
4
2NaOH 2Na
+
HPO
4
2−
2H
2
O++→+
HPO
4
2−
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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,
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 O
2
/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
+ O
2
+ 2H
2
O → 4Fe(OH)
3
Fe(HCO
2
)
3
+ 2Ca(OH)
2
→ Fe(OH)
2
+ 2CaCO
3

+ 2H
2
O
4Fe OH()
2
O
2
2H
2
O 4Fe OH()
3
→++
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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, O
2
/4, 2H
2
O/4,

and 4Fe(OH)
3
/4. We had the same results for Fe(OH)
2
and O
2
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(HCO
3
)
2
/2, 2Ca(OH)
2
/2, Fe(OH)
2
/2, 2CaCO
3
/2 and 2H
2
O/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(HCO
3
)
2
/2, Fe(OH)
2
/2, 2CaCO
3
/2, or
2H
2
O/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(HCO
3
)
2

, Ca(OH)
2
, Fe(OH)
2
, CaCO
3
, and H
2
O 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.
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 H
3
PO
4
, the equation shows that the acid
breaks up into and other substances with one H still “clinging” to the PO
4
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 H
3
PO
4
/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 H
3
PO
4
/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 H
3
PO
4
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,
H
3
PO
4
2NaOH 2Na
+
HPO
4
2−
2H
2
O++→+
H
3
PO
4
3NaOH 3Na
+
PO
4
3−
3H
2
O++→+

HPO
4
2−
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© 2003 by A. P. Sincero and G. A. Sincero
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
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:
Solution:
number of reference species = 2
Equiv. mass
mass of substance
number of equivalents of substance

=

term in balanced reaction
number of moles of reference species

=
Ca HCO
3
()
2
Ca OH()
2
2CaCO
3
↓
2H
2
O+→+
CaSO
4
Na
2
CO
3

CaCO
3
Na
2
SO
4
+→+

a() Ca HCO
3
()
2
Ca OH()
2
2CaCO
3
↓
2H
2
O+→+
Therefore, eq. mass of CaCO
3
2CaCO
3
2

100==
eq. mass of Ca HCO
3
()
2
Ca HCO
3
()
2
2

40 2 1 12 3 16()++[]+

2

162
2

81== =
qeq of Ca HCO
3
()
2
2.5 162()
81

5 geq of CaCO
3
===
g of CaCO
3
solids 5 100()500 g Ans==
CaSO
4
Na
2
CO
3
CaCO
3
Na
2
SO

4
+→+
TX249_Frame_C00 Page 34 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
Background Chemistry and Fluid Mechanics 35
number of reference species = 2
METHODS OF EXPRESSING CONCENTRATIONS
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
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.
Therefore, eq. mass of CaCO
3

CaCO
3
2

50==
eq. mass of CaSO
4
CaSO
4
2

40 32 4 16()++
2

136
2

68== ==
qeq of CaSO
4
1.5 136()
68

3 geq of CaCO
3
===
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+→+
g of CaCO
3
solids
2CaCO
3
Ca HCO
3
()
2

2.5()162()500==
CaSO
4
Na
2
CO

3
CaCO
3
Na
2
SO
4
+→+
g of CaCO
3
solids
CaCO
3
CaSO
4

1.5()136()150==
Total grams of CaCO
3
500 150+ 650 Ans==
ρ
CaCO
3
TX249_Frame_C00 Page 35 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
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 n
i
be the number of moles of a particular species i, the mole fraction
of this species x
i
is
(8)
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.
Solution: The calculations are shown in the table, which should be self-
explanatory.
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
Ions Conc (mg/L)
Ca(HCO
3
)
2
150
Mg(HCO
3
)
2

12.0
Na
2
SO
4
216.0
Ions
Conc
(mg/L) Molecular Mass Moles/Liter
Mole
Fraction
Ca(HCO
3
)
2
150 40.1(2) + 2{1 + 12 + 3(16)} = 202.2 0.74
a
0.32
b
Mg(HCO
3
)
2
12.0 24.3(2) + 2{1 + 12 + 3(16)} = 170.6 0.07 0.03
Na
2
SO
4
216.0 23(2) + 32.1 + 4(16) = 142.1 1.52 0.65
Sum = ∑ 2.33 1.00

a

b

x
i
n
i
∑
1
N
n
i

=
150
202.2

0.74=
0.74
2.33

0.32=
TX249_Frame_C00 Page 36 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
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
one mass of the solute in 10
6

mass of the mixture.
One ppm for a solute dissolved in water can be shown to be equal to one mg/L.
This is shown as follows: 1 ppm = (1 mg)/(10
6
mg) = (1 mg)/(10
3
g). The mass
density of water at 5°C is 1.0 g/cc. Therefore,
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,
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:
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 (PO
4
) molar mass balance in solution using the above
reactions. AlPO
4(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 PO

4
radical.
They are . Note the PO
4
embedded in the
respective formulas; thus, the formulas are said to contain the PO
4
radical. Because they
contain this PO
4
radical, they will participate in the molar mass balance. Let
1 ppm
1mg
10
6
mg

1mg
10
3
g

1mg
10
3
cc

1mg
L

====
1 ppm 1 mg/L=
Al
3+
PO
4
3−
 AlPO
4 s()
↓+
AlPO
4 s()
↓  Al
3+
PO
4
3−
+
PO
4
3−
H
+
 HPO
4
2−
+
HPO
4
2−

H
+
 H
2
PO
4
−
+
H
2
PO
4
−
H
+
 H
3
PO
4
+
[PO
4
3−
], [HPO
4
2−
], [H
2
PO
4

−
], and [H
3
PO
4
]
[sp
PO
4
Al
]
TX249_Frame_C00 Page 37 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
represent the total molar concentration of the phospate radical contained in all the
species. Then, the molar mass balance on PO
4
becomes
The symbol 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, PO
4
, stands for the type
of species, which is the PO
4
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:
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 will be replaced by . The second subscript,
FeIII, stands for the ferric salt. The molar mass balance is
Example 8 State the symbol in words.
Solution: is the total molar concentration of the radical as a
result of using a ferric salt. Ans
Example 9 The complexation reaction of the calcium ion with the carbonate
species, OH
−
, and are given below:
sp
PO
4
Al
[]PO
4
3−
[]= HPO
4
2−
[]H
2
PO
4
−
[]H
3
PO
4

[]+++
[sp
PO
4
Al
]
Fe
3+
PO
4
3−
 FePO
4
↓+
FePO
4
↓  Fe
3+
PO
4
3−
+
PO
4
3−
H
+
 HPO
4
2−

+
HPO
4
2−
H
+
 H
2
PO
4
−
+
H
2
PO
4
−
H
+
 H
3
PO
4
+
[sp
PO
4
Al
][sp
PO

4
FeIII
]
sp
PO
4
FeIII
[]PO
4
3−
[]HPO
4
2−
[]H
2
PO
4
−
[]H
3
PO
4
[]Ans+++=
[sp
PO
4
FeIII
]
[sp
PO

4
FeIII
]PO
4
3−
SO
4
2−
CaCO
3
o
 Ca
2+
CO
3
2−
+
CaHCO
3
+
 Ca
2+
HCO
3
−
+
CaOH
+
 Ca
2+

OH
−
+
CaSO
4
o
 Ca
2+
SO
4
2−
+
TX249_Frame_C00 Page 38 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
Write the molar mass balance for the total molar concentration of Ca.
Solution: Let [Ca
T
] represent the total molar concentration of Ca. Therefore,
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
The concentration expressed as geq/L is the normality. The symbol for normality is N.
Example 10 The concentration of Ca(HCO
3
)
2
is 0.74 gmol/L. Convert this
concentration to geq/L.
Solution:
No. of reference species = 2
Another problem often encountered in practice is the conversion of equivalent
concentration to molar concentration. Let us illustrate this situation using the car-
bonate system. The carbonate system is composed of the species ,
and H
+
. In addition, Ca
2+
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
Ca
T
[] Ca
2+
[]CaCO
3
o
[]CaHCO

3
+
[]CaOH
+
[]CaSO
4
o
[]++ ++ Ans=
C[]
eq
C[]MM()
eq. mass

=
C[]
eq
C[]MM()
eq. mass

=
C[] 0.74 gmol/L=
MM 40.1 2() 2 1 12 3 16()++{}+ 202.2==
Therefore, eq. mass
Ca HCO
3
()
2
2

202.2

2

==
C[]
eq
0.74 202.2()
202.2
2

1.48 geq/L== Ans
CO
3
2−
, HCO
3
−
, OH
TX249_Frame_C00 Page 39 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
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 H
2
CO
3
.
The reaction to the end point of is

Because the number of reference species from this reaction is 1, the equivalent mass
of is . For , the reaction to the end point is
This reaction gives the number of reference species equal to 2; thus, the equivalent
mass of is CO
3
/2. The reaction of H
+
and OH
−
to the end point gives their
respective equivalent masses as H/1 and OH/1. The reaction of Ca
2+
in the carbonate
system is
This gives the equivalent of Ca
2+
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:
Convert the balance in terms of molar concentrations.
Solution:
Therefore, substituting,
HCO
3
−
HCO
3
−
H

+
 H
2
CO
3
+
HCO
3
−
HCO
3
/1CO
3
2−
CO
3
2−
2H
+
 H
2
CO
3
=
CO
3
2−
Ca
2+
CO

3
2−
 CaCO
3
+
CO
3
2−
[]
eq
HCO
3
−
[]
eq
OH[]
eq
++ H
+
[]
eq
Ca
2+
[]
eq
+=
C[]
eq
C[]MM()
eq. mass

=
CO
3
2−
()
eq. mass
CO
3
2

; HCO
3
−
()
eq. mass
HCO
3
1

;OH()
eq. mass
OH
1

===
H
+
()
eq. mass

H
1

;Ca
2+
()
eq. mass
Ca
2

==
CO
3
2−
[]CO
3
()
CO
3
2

HCO
3
−
[]HCO
3
()
HCO
3

1

OH
−
[]OH()
OH
1

++
H
+
[]H()
H
1

Ca
2+
[]Ca()
Ca
2

+=
2CO
3
2−
[]HCO

3
−
[]OH[]++ H
+
[]2Ca
2+
[]Ans+=
TX249_Frame_C00 Page 40 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
ACTIVITY AND ACTIVE CONCENTRATION
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:
(9)
where sp represents any species involved in the equilibria such as Ca
2+
, ,
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-
actions with other particles. Thus, Ca
2+
and , which are ionization products of
CaCO
3
, 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
CaCO
3
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 AND SOLUBILITY PRODUCT CONSTANTS
Let a reactant be represented by the solid molecule A
a
B
b
. As this reactant is mixed
with water, it dissolves into its constituent solute ions. The equilibrium dissolution
reaction is
(10)
The ratio is called the reaction quotient. Note that in the deﬁnition of

reaction quotients, the activities of the reactants and products are raised to their
sp{}
γ
= sp[]
CO
3
2−
HCO
3
−
CO
3
2−
A
a
B
b
 aA bB+
{ A}
a
{B}
b
{ A
a
B
b
}

TX249_Frame_C00 Page 41 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero

respective coefﬁcients. At equilibrium, the reaction quotient becomes the equilibrium
constant. Thus, the equilibrium constant is
(11)
Note the pair of braces denoting activity. Because A
a
B
b
is a solid, its activity is
unity. For this reason, the product K{A
a
B
b
} is a constant and is designated as K
sp
.
K
sp
is called the solubility product constant of the equilibrium dissolution reaction.
Equation (11) now transforms to
(12)
At equilibrium, neither reactants nor products increase or decrease with time. Thus,
K
sp
’s, being constants, can be used as indicators whether or not a given solid will
form or dissolve in solution.
For example, CaCO
3
has a K
sp
of 4.8(10

−9
) at 25°C. This value decreases to
2.84(10
−9
) at 60°C. The equilibrium reaction for this solid is
Assuming the reaction is at equilibrium at 25°C, the ions at the right side of the
reaction will neither cause CaCO
3
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,
the product of the activities will equal 4.8(10
−9
). As the temperature increases from
25°C to 60°C, the product of the activities decreases from 4.8(10
−9
) to 2.84(10
−9
).
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 CaCO
3
solid.
Table 1 shows values of K
sp
’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 K
sp
’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
water, the carbonate system species and 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. is a mixture
of CO
2
in water and H
2
CO
3
(note the absence of the asterisk in H
2
CO
3
). The CO
2
in water is written as CO
2(aq)
and is carbonic acid.
K
A{}
a

B{}
b
A
a
B
b
{}

=
K
sp
A{}
a
= B{}
b
CaCO
3s()
 Ca
2+
CO
3
2−
. K
sp
+ Ca
2+
) CO
3
2−
({}4.8 10

9–
() at 25°C==
Ca
2+
) CO
3
2−
({}2.84 10
9–
() at 60°C==
H
2
CO
3
∗
, HCO
3
−
,CO
3
2−
H
2
CO
3
∗
H
2
CO
3

∗
TX249_Frame_C00 Page 42 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
Another important application of the equilibrium constant K in general, [Equation
(11)], is in the coagulation treatment of water using alum, Al
2
(SO
4
)
3
·14 H
2
O. (The
“14” actually varies from 13 to 18.) In coagulating a raw water using alum, a number
of complex reactions are formed by the Al
3+
ion. These reactions are as follows:
(13)
TABLE 1
Solubility Product Constants of Respective Solids at 25°C
K
sp
reaction of solid K
sp
Signiﬁcance
1.8(10
−10
) Chloride analysis
1.9(10
−33

) Coagulation
10
−10
Sulfate analysis
4.8(10
−9
) Hardness removal, scales
7.9(10
−6
) Hardness removal
2.4(10
−5
) Flue gas desulfurization
10
−25
Phosphate removal
5.0(10
−6
) Phosphate removal
3.9(10
−11
) Fluoridation
6.7(10
−31
) Heavy metal removal
5.6(10
−20
) Heavy metal removal
1.1(10
−36

) Coagulation, iron removal,
corrosion
7.9(10
−15
) Coagulation, iron removal,
corrosion
10
−5
Hardness removal, scales
1.5(10
−11
) Hardness removal, scales
4.5(10
−14
) Manganese removal
1.6(10
−14
) Heavy metal removal
4.5(10
−17
) Heavy metal removal
From A. P. Sincero and G. A. Sincero (1996). Enviromental Engineering: A Design
Approach. Prentice Hall, Upper Saddle River, NJ, 42.
AgCl
s()
 Ag
aq()
+
Cl
aq()

−
+
Al OH()
3 s()
 Al
aq()
3+
3OH
aq()
−
+
BaSO
4 s()
 Ba
aq()
2+
SO
4 aq()
2−
+
CaCO
3 s()
 Ca
aq()
2+
CO
3 aq()
2−
+
Ca OH()

2 s()
 Ca
aq()
2+
2OH
aq()
−
+
CaSO
4 s()
 Ca
aq()
2+
SO
4 aq()
2−
+
Ca
3
PO
4
()
2 s()
 3Ca
aq()
2−
2PO
aq()
2−
+

CaHPO
4 s()
 Ca
aq()
HPO
4 aq()
2−
+
CaF
2 s()
 Ca
aq()
2+
2F
aq()
−
+
Cr OH()
3 s()
 Cr
aq()
3+
3OH
aq()
−
+
Cu OH()
2 s()
 Cu
aq()

2+
2OH
aq()
−
+
Fe OH()
3 s()
 Fe
aq()
3+
3OH
aq()
−
+
Fe OH()
2
 Fe
aq()
2+
2OH
aq()
−
+
MgCO
3 s()
 Mg
aq()
2+
CO
3 aq()

2−
+
Mg OH()
2
 Mg
aq()
2+
2OH
aq()
−
+
Mn OH()
2
 Mn
aq()
2+
2OH
aq()
−
+
Ni OH()
2
 Ni
aq()
2+
2OH
aq()
−
+
Zn OH()

2
 Zn
aq()
2+
2OH
aq()
−
+
Al
3+
H
2
O  Al OH()
2+
Η
+
++
K
Al OH()c
Al OH()
2+
{}H
+
{}
Al
3+
{}

10
−5

==
TX249_Frame_C00 Page 43 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
(14)
(15)
(16)
(17)
(18)
The equilibrium constants ,
and 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, 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
this maximum precipitation, the concentrations of the complex ions Al(OH)
2+
,
, Al
13
(OH)
34
, Al , and and Al
3+

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.
7Al
3+
17H
2
O  Al
7
OH()
17
4+
17Η
+
++
K
Al
7
OH()
17

c
Al
7
OH()
17
4+
{}H
+
{}
17
Al
3+
{}
7

10
−48.8
==
13Al
3+
34H
2
O  Al
13
OH()
34
5+
34Η
+
++

K
Al
13
OH()
34
c
Al
13
OH()
34
5+
{}H
+
{}
34
Al
3+
{}
13

10
−97.4
==
Al OH()
3 s()
fresh precipitate() Al
3+
3OH
−
+

K
sp,Al OH()
3
Al
3+
{}OH
−
{}
3
10
33–
==
Al OH()
3 s()
OH
−
 Al OH()
4
−
+
K
Al OH()
4
c
Al OH()
4
−
{}
OH
−

10
+1.3
==
2Al
3+
2H
2
O  Al
2
OH()
2
4+
2H
+
++
K
Al
2
OH()
2
c
Al
2
OH()
2
4+
{}H
+
{}

2
Al
3+
{}
2

10
6.3–
==
K
Al(OH)c
, K
Al
7
(OH)
17
c
, K
Al
13
(OH)
34
c
, K
sp,
Al(OH)
3
, K
Al(OH)
4

c
K
Al
2
(OH)
2
c
K
sp,Al(OH)
3
Al
7
(OH)
17
4+
(OH)
4
−
Al
2
(OH)
2
4+
TX249_Frame_C00 Page 44 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
Gleaning from Eqs. (13) to (18), a molar mass balance on the aluminum atom
may be performed. Let sp
Al
represent all the species that contain the aluminum atom
standing in solution. Thus, the concentration of all the species containing the alu-

minum atom, is
(19)
Note that, because we are summing the species containing the aluminum atom,
the coefﬁcients of the terms of the previous equation contain the number of aluminum
atoms in the respective species. For example, contains 7 aluminum atoms;
thus, its coefﬁcient is 7. Similarly, contains 13 aluminum atoms; there-
fore, its coefﬁcient is 13. This explanation holds for the other species as well.
Using Eqs. (13) to (18) and the relations of molar concentration and activity,
Equation (9), the following equations are obtained [for the purpose of eliminating
the complex ions in Equation (19)]:
(20)
(21)
(22)
(23)
(24)
sp
Al
[] Al
3+
[]Al OH()
2+
[]7Al
7
OH()
17
4+
[]13 Al
13
OH()
34

5+
[] ++ +=
Al OH()
4
−
[]2Al
2
OH()
2
4+
[]++
Al
7
(OH)
17
4+
Al
13
(OH)
34
5+
Al
3+
[]
Al
3+
{}
γ
Al

K
sp,Al OH()
3
γ
Al
OH
−
{}
3

K
sp,Al OH()
3
H
+
{}
3
γ
Al
K
w
3

K
sp,Al OH()
3
γ
H
3
H

+
[]
3
γ
Al
K
w
3

== = =
Al OH()
2+
[]
Al OH()
2+
{}
γ
Al OH()c

K
Al OH()c
Al
3+
{}
γ
Al OH()c
H
+
{}

K
Al OH()c
K
sp,Al OH()
3
γ
H
2
H
+
[]
2
γ
Al OH()c
K
w
3

== =
Al
7
OH()
17
4+
[]
Al
7
OH()
17
{}

γ
Al
7
OH()
17
c

K
Al
7
OH()
17
c
Al
3+
{}
7
γ
Al
7
OH()
17
c
H
+
{}
17

K
Al

7
OH()
17
c
γ
Al
7
Al
3+
[]
7
γ
Al
7
OH()
17
c
γ
H
17
H
+
[]
17

== =

K
Al
7

OH()
17
c
K
sp,Al OH()
3
7
γ
H
4
H
+
[]
4
γ
Al
7
OH()
17
c
K
w
21

=
Al
13
OH()
34
5+

[]
Al
13
OH()
34
5+
{}
γ
Al
13
OH()
34
c

K
Al
13
OH()
34
c
Al
3+
{}
13
γ
Al
13
OH()
34
c

H
+
()
34

K
Al
13
OH()
34
c
γ
Al
13
Al
3+
[]
13
γ
Al
13
OH()
34
c
γ
H
34
H
+
[]

34

== =

K
Al
13
OH()
34
c
K
sp,Al OH()
3
13
γ
H
5
H
+
[]
5
γ
Al
13
OH()
34
c
K
w
39

=
Al OH()
4
−
[]
Al OH()
4
−
{}
γ
Al OH()
4
c

K
Al OH()
4
c
OH
−
{}
γ
Al OH()
4
c

K
Al OH()
4

c
K
w
γ
Al OH()
4
c
γ
H
H
+
[]

== =
TX249_Frame_C00 Page 45 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
(25)
are, respectively, the
activity coefﬁcients of the aluminum ion and the hydrogen ion and the complexes
Al(OH)
2+
, Al
7
Al
13
, and . is the
solubility product constant of the solid Al(OH)
3(s)
and K
w

is the ion product of water.
K
Al(OH)c
, , , , and are, respectively, the equi-
librium constants of the complexes Al(OH)
2+
, , ,
and .
Let us explain the derivation of one of the above equations. Pick Equation (20).
The ﬁrst part of this equation is
(26)
This equation is simply the application of Equation (9), where sp is Al
3+
and
γ
is
γ
Al
and where the molar concentration is solved in terms of the activity.
The next part of the equation, , is derived, in part, from
Equation (16), which reads
(16)
From the deﬁnition of K
sp
,
(27)
This equation may be solved for {Al
3+
} to produce
(28)

Substituting {Al
3+
} = in Equation (26) produces
(29)
Al
2
OH()
2
4+
[]
Al
2
OH()
2
4+
{}
γ
Al
2
OH()
2
c

K
Al
2
OH()
2
c
Al

3+
{}
2
γ
Al
2
OH()
2
c
H
+
{}
2

K
Al
2
OH()
2
c
γ
Al
2
Al
3+
[]
2
γ
Al
2

OH()
2
c
γ
H
2
H
+
[]
2

== =

K
Al
2
OH()
2
c
K
sp,Al OH()
3
2
γ
H
4
H
+
[]
4

γ
Al
2
OH()
2
c
K
w
6

=
γ
Al
,
γ
H
,
γ
Al(OH)c
,
γ
Al
7
(OH)
17
c
,
γ
Al
13

(OH)
34
c
,
γ
Al(OH)
4
c
,
γ
Al
2
(OH)
2
c
(OH)
17
4+
, (OH)
34
5+
, Al(OH)
4
−
Al
2
(OH)
2
4+
K

sp,Al(OH)
3
K
Al
7
(OH)
17
c
K
Al
13
(OH)
34
c
K
Al(OH)
4
c
K
Al
2
(OH)
2
c
Al
7
(OH)
17
4+
,

Al
13
(OH)
34
5+
Al(OH)
4
−
Al
2
(OH)
2
4+
Al
3+
[]
Al
3+
{}
γ
Al
=
K
sp,Al(OH)
3
/γ
Al
{OH
−
}

3
Al OH()
3 s()
fresh precipitate() Al
3+
3OH
−
K
sp,Al OH()
3
10
33–
=+
K
sp,Al OH()
3
Al
3+
{}OH
−
{}
3
=
Al
3+
{}
K
sp,Al OH()
3
OH

−
{}
3

=
K
sp,Al(OH)
3
/{OH
−
}
3
Al
3+
[]
Al
3+
{}
γ
Al

K
sp,Al OH()
3
γ
Al
OH
−
{}
3

==
TX249_Frame_C00 Page 46 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
The next part of the equation, , should be easy to derive
and will no longer be pursued. Note that K
w
came from the ion product of water
that reads:
(30)
This ion product is an equilbirium constant for the dissociation of water:
(31)
Equations (20) to (25) may now be substituted into Equation (19). This produces
Equation (32), where, now, the complexes are eliminated and the equation only
expressed in terms of the K’s and the hydrogen ion concentration. This equation
may then be differentiated and equated to zero to obtain the optimum pH. We will
not, however, complete this differentiation and equate to zero in this chapter, but
will do this in the unit processes part of this book.
(32)
Still another important application of the concept of K equilibrium constants is
the coprecipitation of FePO
4
and Fe(OH)
3
in the removal of phosphorus from water.
As in the case of coagulation using alum, it is desired to have a ﬁnal equation that
is expressed only in terms of the constants and the hydrogen ion. Once this is done,
the equation can then also be manipulated to obtain an optimum pH for the removal
of phosphorus.
In phosphorus removal, the phosphorus must be in the phosphate form and,

because the reaction occurs in water, the phosphate ion originates a series of reactions
with the hydrogen ion. The series is as follows:
(33)
K
sp,Al(OH)
3
{H
+
}
3
[]/
γ
Al
K
w
3
K
w
H
+
{}OH{}=
H
2
O  H
+
OH
−
+
sp
Al

[]
K
sp,Al OH()
3
γ
H
3
H
+
[]
3
γ
Al
K
w
3

K
Al OH()c
K
sp,Al OH()
3
γ
H
2
H
+
[]
2
γ

Al OH()c
K
w
3

+=

7K
Al
7
OH()
17
c
K
sp,Al OH()
3
7
γ
H
4
H
+
[]
4
γ
Al
7
OH()
17
c

K
w
21

+

13K
Al
13
OH()
34
c
K
sp,Al OH()
3
13
γ
H
5
H
+
[]
5
γ
Al
13
OH()
34
c
K

w
39

+
K
Al OH()
4
c
K
w
γ
Al OH()
4
c
γ
H
H
+
[]

+

2K
Al
2
OH()
2
c
K
sp,Al OH()

3
2
γ
H
4
H
+
[]
4
γ
Al
2
OH()
2
c
K
w
6

+
PO
4
3−
H
+
 HPO
4
2−
HPO
4

2−
 PO
4
3−
H
+
+⇒+
K
HPO
4
PO
4
3−
{}H
+
{}
HPO
4
2−
{}

10
12.3–
==
TX249_Frame_C00 Page 47 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
(34)
(35)
From the previous equations, a molar mass balance on the phosphate radical
may be performed. Let represent the species in solution containing PO

4
species, using a ferric salt as the precipitant. (Note the FeIII as the second subscript,
indicating that a ferric salt is used.) Therefore,
(36)
To indicate the removal of phosphate using the ferric salt, we use the following
precipitation reaction to incorporate the iron into the above equation:
(37)
(38)
, and [H
3
PO
4
] may be eliminated by expressing them in terms
of the K constants using Eqs. (33) to (35). The results are:
(39)
(40)
(41)
By using Equation (38), may also be eliminated in favor of to
produce
(42)
HPO
4
2−
H
+
 H
2
PO
4
−

H
2
PO
4
−
 HPO
4
2−
H
+
+⇒+
K
H
2
PO
4
HPO
4
2−
{}H
+
{}
H
2
PO
4
−
{}

10

7.2–
==
H
2
PO
4
−
H
+
 H
3
PO
4
H
3
PO
4
 H
2
PO
4
−
H
+
+⇒+
K
H
3
PO
4

H
2
PO
4
−
{}H
+
{}
H
3
PO
4
{}

10
2.1–
==
sp
PO
4
FeIII
sp
PO
4
FeIII
[]PO
4
3−
[]HPO
4

2−
[]H
2
PO
4
−
[]H
3
PO
4
[]+++=
FePO
4
↓  Fe
3+
PO
4
3−
K
sp,FePO
4
+ 10
21.9–
at 25°C=
K
sp,FePO
4
Fe
3+
{}PO

4
3−
{}=
[HPO
4
2−
], [H
2
PO
4
−
]
HPO
4
2−
[]
HPO
4
2−
{}
γ
HPO
4

PO
4
3−
{}H
+
{}

γ
HPO
4
K
HPO
4

γ
PO
4
γ
H
PO
4
3−
[]H
+
[]
γ
HPO
4
K
HPO
4

== =
H
2
PO
4

2−
[]
H
2
PO
4
−
{}
γ
H
2
PO
4

HPO
4
2−
{}H
+
{}
γ
H
2
PO
4
K
H
2
PO
4

γ
PO
4
γ
H
2
PO
4
3−
[]H
+
[]
2
γ
H
2
PO
4
K
H
2
PO
4
K
HPO
4

== =
H

3
PO
4
[]H
3
PO
4
{}
H
2
PO
4
−
{}H
+
{}
K
H
3
PO
4

γ
PO
4
γ
H
3
PO
4

3−
[]H
+
[]
3
K
H
3
PO
4
K
H
2
PO
4
K
HPO
4

== =
[PO
4
3−
] K
sp,FePO
4
PO
4
3−
[]

PO
4
3−
{}
γ
PO
4

K
sp,FePO
4
γ
PO
4
Fe
3+
{}

K
sp,FePO
4
γ
PO
4
γ
FeIII
Fe
3+
[]

== =
TX249_Frame_C00 Page 48 Friday, June 14, 2002 1:47 PM
© 2003 by A. P. Sincero and G. A. Sincero
This will incorporate the iron into Equation (36) to indicate the use of the ferric
salt. Finally, substituting the whole result into Equation (36), produces
(43)
are, respectively, the activity coefﬁcients of the
phosphate, aluminum, hydrogen, hydrogen phosphate, and dihydrogen phosphate ions.
and are, respectively, the equilibrium constants of the hydro-
gen phosphate and dihydrogen phosphate ions and phosphoric acid.
γ
FeIII
is the
activity coefﬁcient of the ferric ion.
As shown, Equation (43) still expresses in terms of two variables,
[Fe
3+
] and [H
+
]. We want it expressed only in terms of the K constants and the
hydrogen ion concentration. This is where an expression needs to be found to elim-
inate [Fe
3+
]. If this expression indicates that some compound of iron is coprecipitating
with FePO
4
, then [Fe
3+
] could be eliminated using the K constant of this compound.
When two compounds are coprecipitating, they are at equilibrium at this partic-

ular instance of coprecipitation. They are at equilibrium, so their equilibrium con-
stants can be used in a calculation. This instance of coprecipitating with FePO
4
is,
indeed, possible through the use of the compound ferric hydroxide which, of course,
must still be investigated whether, in fact, it is possible. This is investigated next.
The dissociation reaction of ferric hydroxide is
(44)
If the results of this investigation show that the molar concentration of iron needed
to precipitate Fe(OH)
3
is about the same as the molar concentration of the iron
needed to precipitate FePO
4
, then Fe(OH)
3
and FePO
4
must be coprecipitating.
Let x be the {Fe
3+
}. Because the coefﬁcient of OH
−
in the previous reaction is
3, {OH
−
} is therefore 3x. Thus, substituting into the K
sp
equation,

Now, from Equation (37), . Letting y equal
to {Fe
3+
}, is also equal to y. Substituting into the K
sp
equation,
sp
PO
4
FeIII
[]
K
sp,FePO
4
γ
PO
4
γ
FeIII
Fe
3+
[]

γ
H
K
sp,FePO
4
H
+