Standard Methods for Examination of Water & Wastewater_9 potx

Part III

Unit Processes of Water
and Wastewater Treatment

Part III covers the various unit processes employed in water and wastewater treatment
including water softening; water stabilization; coagulation; removal of iron and man-
ganese by chemical precipitation; removal of nitrogen by nitrification–denitrification;
removal of phosphorus by chemical precipitation; ion exchange; and disinfection.

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Water Softening

Softening

is the term given to the process of removing ions that interfere with the
use of soap. These ions are called

hardness ions

due to the presence of multivalent
cations, mostly calcium and magnesium. In natural waters, other ions that may be
present to cause hardness but not in significant amounts are iron (Fe

2

+

), manganese
(Mn

2

+

), strontium (Sr

2

+

), and aluminum (Al

3

+

).
In the process of cleansing using soap, lather is formed causing the surface
tension of water to decrease. This decrease in surface tension makes water molecules
partially lose their mutual attraction toward each other, allowing them to wet ‘‘foreign”
solids, thereby, suspending the solids in water. As the water is rinsed out, the solids
are removed from the soiled material. In the presence of hardness ions, however,
soap does not form the lather immediately but reacts with the ions, preventing the
formation of lather and forming scum. Lather will only form when all the hardness
ions are consumed. This means that hard waters are hard to lather.

Hard waters

are
those waters that contain these hardness ions in excessive amounts. Softening using
chemicals is discussed in this chapter. Other topics related to softening are discussed
as necessary.

10.1 HARD WATERS

The following lists the general classification of hard waters:
Soft

<

50 mg/L as CaCO

3

Moderately hard 50–150 mg/L as CaCO

3

Hard 150–300 mg/L as CaCO

3

Very hard

>

300 mg/L as CaCO

3

A very soft water has a slimy feel. For example, rainwater, which is exceedingly
soft, is slimy when used with soap. For this reason, hardness in water used for
domestic purposes is not completely removed. Hardness is normally removed to the
level of 75 to 120 mg/L as CaCO

3

.

10.2 TYPES OF HARDNESS

Two basic types of hardness are associated with the ions causing hardness:

carbon-
ate

and

noncarbonate



hardness

. When the hardness ions are associated with the
ions in water, the type of hardness is called

carbonate hardness

; otherwise,
the type of hardness is called

noncarbonate hardness

. An example of carbonate
hardness is Ca(HCO

3

)

2

, and an example of noncarbonate hardness is MgCl

2

.
10
HCO
3
−

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468

In practice, when one addresses hardness removal, it means the removal of the
calcium and magnesium ions associated with the two types of hardness. Our dis-
cussion of hardness removal therefore is divided into the categories of calcium and
magnesium. Corresponding chemical reactions are developed. If other specific hard-
ness ions are also present and to be removed, such as iron, manganese, strontium,
and aluminum, corresponding specific reactions have to be developed for them. The
removal of iron and manganese will be discussed in a separate chapter.
In general, water is softened in three ways: chemical precipitation, ion exchange,
and reverse osmosis. Only the chemical precipitation method is discussed in this chapter.

10.3 PLANT TYPES FOR HARDNESS REMOVAL

In practice, two types of plants are generally used for chemical precipitation hardness
removal: One type uses a sludge blanket contact mechanism to facilitate the precip-
itation reaction. The second type consists of a flash mix, a flocculation basin, and
a sedimentation basin. The former is called a

solids-contact

clarifier. The latter
arrangement of flash mix, flocculation, and sedimentation were discussed in previous
chapters on unit operations.
A solids-contact clarifier is shown in Figure 10.1. The chemicals are introduced
into the primary mixing and reaction zone. Here, the fresh reactants are mixed by
the swirling action generated by the rotor impeller and also mixed with a return sludge
that are introduced under the hood from the clarification zone. The purpose of the
return sludge is to provide nuclei that are important for the initiation of the chemical
reaction. The mixture then flows up through the sludge blanket where secondary
reaction and mixing occur. The reaction products then overflow into the clarification

zone, where the clarified water is separated out by sedimentation of the reaction
product solids. The clarified water finally overflows into the effluent discharge. The
settled sludge from the clarification is drawn off through the sludge discharge pipe.

FIGURE 10.1

Solids-contact clarifier. (Courtesy of Infilco Degremont, Inc.)
Clarified
water
Motor drive
Secondary
mixing
and
Impeller
Chemical feed
cat walk
Sludge blanket
Primary mixing and reaction zone
Hood
Effluent
Return sludge
Influent
Concentrated sludge
Sludge discharge
Bleed off and drain
reaction
zone
Return sludge
Hood
Clarified

water

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469

10.4 THE EQUIVALENT CaCO

3

CONCENTRATION

In the literature, hardness is frequently expressed in terms of CaCO

3

. Expressing
concentrations in terms of CaCO

3

can be confusing. For example, when the hardness
of water is 60 mg/L as CaCO

3

, what does this really mean? Related to this question
is a second question: Given the concentration of a hardness substance, how is this
converted to the equivalent CaCO

3

hardness concentration?
To obtain the mass of any substance, the number of equivalents of that substance
is multiplied by its equivalent mass. Because the number of equivalents of all sub-
stances participating in a given chemical reaction are equal, what differentiates the
various species in this chemical reaction are their respective equivalent masses. Thus,
to obtain the concentration of 60 mg/L as CaCO

3

, this equal number of equivalents
must have been multiplied by the equivalent mass of calcium carbonate. This section
will determine this equivalent mass. This is actually 50 and converting the concentration
of any hardness substance to the equivalent CaCO

3

hardness concentration is obtained
by multiplying the number of equivalents of the substance by 50.
Let [

C

T

,

hard

]

eq

represent the total concentration of hardness in equivalents, where
the symbol [ ] is read as “the concentration of ” and the subscript

eq

means that the
concentration is expressed in terms of equivalents. If the only hardness ions present
are calcium and magnesium,
[

C

T

,

hard

]

eq



=

[Ca

2

+

]

eq



+

[Mg

2

+

]

eq

(10.1)
The number of equivalents of one substance in a given chemical reaction is equal
to the same number of equivalents of any other substance in this reaction, so it is
entirely correct to arbitrarily express the total concentration of hardness in terms of
only one of the ions that participates in the chemical reaction. The concentrations

of the other ions must then be subsumed in the concentration of this one ion being
chosen. For example, if the total hardness is to be expressed in terms of the mag-
nesium ion only, the previous equation will become
[

C

T

,

hard

]

eq



=

(10.2)
In this equation, the part of the total hardness contained in calcium is subsumed
in magnesium, . Note the prime. The term means that it is a
concentration in terms of magnesium but that the calcium ion concentration is
subsumed in it and expressed in terms of magnesium equivalents. Moreover, since
an equivalent of one is equal to the equivalent of another, it is really immaterial
under what substance the total equivalents of hardness is expressed. Then, in a
similar manner, the total concentration of hardness may also be expressed in terms
of the calcium ion alone as follows

[

C

T

,

hard

]

eq



=

(10.3)
Mg
2+
[]
eq
′
Mg
2+
[]
eq
′
Mg

2+
[]
eq
′
Ca
2+
[]
eq
′

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470

In this equation, the part of the total hardness contained in magnesium is now
subsumed in calcium, . Again, note the prime. Similar to the term ,



means that it is a concentration in terms of calcium but the magnesium ion
is subsumed in it and expressed in terms of calcium equivalents. If other hardness
ions are present as well, then the concentrations of these ions will also be subsumed
in the one particular ion chosen to express the hardness.
When any ion participates in a chemical reaction, it will react to the satisfaction
of its ionic charge. For example, when the calcium and magnesium ions participate
in a softening reaction, they will react to the satisfaction of their ionic charges of
two. Thus, if the chemical reaction is written out, the number of reference species
for Ca

2

+



and Mg

2

+

will be found to be two and the respective equivalent masses are
then Ca

/

2 and Mg

/

2. In terms of molar concentrations, [Ca

2

+

]

eq

is then equal to
[Ca

2

+

](Ca

/

Ca

/

2)

=

2[Ca

2

+

]. By analogy,

=

2 , where [Ca

2

+

]

′

is now
an equivalent molar concentration in terms of the calcium ion that subsumes all
concentrations. Also, [Mg

2

+

]

eq



=

[Mg

2

+

](Mg

/

Mg

/

2)

=

2[Mg

2

+

] and, again, by analogy,


=

2[Mg

2

+

]

′

. [Mg

2

+

]

′

is the equivalent molar concentration in terms of the
magnesium ion that also subsumes all concentrations. Considering Eqs. (10.2) and
(10.3) simultaneously and expressing the molar concentrations of the calcium and
magnesium hardness in terms of calcium only,
[

C

T

,

hard

]

eq



=



=



=

(10.4)
The term 2[Ca

2

+

]

′

can be converted to [CaCO

3

]

′

. This is done as follows: In
CaCO

3

, one mole of [Ca

2

+

] is equal to one mole of [CaCO

3

]. Hence,

2



=

2[CaCO

3

]

′

and Equation (10.4) becomes
[

C

T

,

hard

]

eq



=

= = 2 = 2[CaCo
3
]′ or (10.5)
[CaCO
3
]′ = = = (10.6)

[CaCO
3
]′ is an equivalent molar concentration expressed in terms of moles of
CaCO
3
. Thus, to get the equivalent calcium carbonate mass concentration, it must
be multiplied by CaCO
3
= 100. Therefore,
Ca
2+
[]
eq
′
Mg
2+
[]
eq
′
Ca
2+
[]
eq
′
Ca
2+
[]
eq
′
Ca

2+
[]
′
Mg
2+
[]′
eq
Mg
2+
[]
eq
′
Ca
2+
[]
eq
′
2Ca
2+
[]′
Ca
2+
[]′
Mg
2+
[]
eq
′
Ca
2+

[]
eq
′
Ca
2+
[]′
C
T,hard
[]
eq
2

Mg
2+
[]
eq
′
2

Ca
2+
[]
eq
′
2

CaCO
3
[]′100()
C

T,hard
[]
eq
2



100()50 C
T,hard
[]
eq
==
Mg
2+
[]
eq
′
2




100()50 Mg
2+
[]
eq
′
==
Ca
2+

[]
eq
′
2




100()50 Ca
2+
[]
eq
′
==
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Or,
= 50[C
T,hard
]
eq
= 50 = 50 (10.7)
Thus, the concentration of hardness expressed in terms of the mass of CaCO
3
is equal to the number of equivalents of hardness ([C
T,hard
]
eq
, , or )
divided 2 times the molecular mass of calcium carbonate. Or, simply put, the

concentration of hardness expressed in terms of the mass of CaCO
3
is equal to the
number of equivalents of hardness ([C
T,hard
]
eq
, , or ) times 50.
Note that Equation (10.7) merely states the concentration of hardness in terms
of the mass of CaCO
3
. Although the symbol for calcium carbonate is being used, it
does not state anything about the actual concentration of the CaCO
3
species present;
it is even possible that no species of calcium carbonate exists, but MgCO
3
or any
other species where the concentrations are simply being expressed in terms of CaCO
3
.
To apply Equation (10.7), suppose that the concentration of Mg
2+
in a sample
of water is given as 60 mg/L as CaCO
3
. The meq/L of Mg
2+
is then equal to [Mg
2+

]
eq
=
= 60/50 = 1.2 and the concentration of magnesium in mg/L is 1.2(Mg/2) =
1.2(24.3/2) = 14.58 mg/L.
Take the following reaction and the example of the magnesium ion:
In this reaction, the equivalent mass of CaCO
3
is 2(CaCO
3
)/2 = 100. Again, in any
given chemical reaction, the number of equivalents of all the participating species
are equal. Thus, if the number of equivalents of the magnesium ion is 1.2 meq/L,
the number of equivalents of CaCO
3
also must be 1.2 meq/L. Thus, from this reaction,
the calcium carbonate concentration corresponding to the 1.2 meq/L of Mg
2+
would be
1.2(100) = 120 mg/L as CaCO
3
. Or, because the species is really calcium carbonate,
it is simply 120 mg/L CaCO
3
––no more ‘‘as.”
How is the concentration of 120 mg/L CaCO
3
related to the concentration of
60 mg/L as CaCO
3

for the magnesium ion? The 60 mg/L is not a concentration of
calcium carbonate but a concentration of the magnesium expressed as CaCO
3
. These
two are very different. In the “120,” there is really calcium carbonate present, while
in the “60,” there is none.
Again, expressing the magnesium concentration as 60 mg/L CaCO
3
does not
mean that there are 60 mg/L of the CaCO
3
but that there are 60 mg/L of the ion of
magnesium expressed as CaCO
3
. Expressing the concentration of one substance in
terms of another is a normalization. This is analogous to expressing other currencies
in terms of the dollar. Go to the Philippines and you can make purchases with the
dollar, because there is a normalization (conversion) between the dollar and the peso.
Example 10.1 The concentration of total hardness in a given raw water is found
to be 300 mg/L as CaCO
3
. Calculate the concentration in milligram equivalents per
liter.
C
T,hard
[]
asCaCO
3
Mg
2+

[]
eq
′
Ca
2+
[]
eq
′
Mg
2+
[]
eq
′
Ca
2+
[]
eq
′
Mg
2+
[]
eq
′
Ca
2+
[]
eq
′
Mg
2+

[]
eq
′
Mg HCO
3
()
2
2Ca OH()
2
+MgOH()
2
↓
2CaCO
3
↓
2HOH++→
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Solution:
10.5 SOFTENING OF CALCIUM HARDNESS
Calcium hardness may either be carbonate or noncarbonate. The solubility product
constant of CaCO
3
is K
sp
= [Ca
2+
][ ] = 5(10
−9
) at 25°C. A low value of the K

sp
means that the substance has a low solubility; a value of 5(10
−9
) is very low. Because
of this very low solubility, calcium hardness is removed through precipitation of
CaCO
3
. Because there are two types of calcium hardness, there corresponds two
general methods of removing it. When calcium is associated with the bicarbonate
ion, the hardness metal ion can be easily removed by providing the hydroxide radical.
The H
+
of the bicarbonate becomes neutralized by the OH
−
provided forming water
and the ion necessary to precipitate calcium carbonate. The softening reaction
is as follows:
(10.8)
The precipitate, CaCO
3
, is indicated by a downward pointing arrow,↓. As shown,
the two bicarbonate species in the reactant side are converted to the two carbonate
species in the product side. The carbonate ion shown unpaired will pair with whatever
cation the OH
−
was with in the reactant side of the reaction.
Once all the bicarbonate ions have been destroyed by the provision of OH
−
, they
convert to the carbonate ion, as indicated. If the OH

−
source contains the calcium ion,
the carbonates all precipitate as CaCO
3
according to the following softening reaction:
(10.9)
In this case, no carbonate ion is left unpaired, because the calcium ion is present to
cause precipitation.
The other method of calcium removal involves the case when the hardness is in
the form of the noncarbonate, such as in the form of CaCl
2
. In this case, a carbonate
ion must be provided, whereupon Equation (10.9) will apply, precipitating calcium
as calcium carbonate. The usual source of the carbonate ion is soda ash.
10.6 SOFTENING OF MAGNESIUM HARDNESS
As in the case of calcium hardness, magnesium can also be present in the form of
carbonate and noncarbonate hardness. The K
sp
of Mg(OH)
2
is a low value of 9(10
−12
).
Thus, the hardness is removed in the form of Mg(OH)
2
. To remove the carbonate
hardness of magnesium, a source of the OH
−
ion is therefore added to precipitate
the Mg(OH)

2
as shown in the following softening chemical reaction:
(10.10)
C
T,hard
[]
meq
300
50
6.0 meq/L Ans==
CO
3
2−
CO
3
2−
Ca HCO
3
()
2
2OH
−
+ CaCO
3
↓ CO
3
2−
2HOH++→
Ca
2+

CO
3
2−
+ CaCO
3
↓→
Mg HCO
3
()
2
4OH
−
+ Mg OH()
2
↓ 2CO
3
2−
2HOH++→
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The carbonate ions in the product side will pair with whatever cation the OH
−
was
with in the reactant side of the reaction. If this cation is calcium, in the form of
Ca(OH)
2
, then the product will again be the calcium carbonate precipitate.
In natural waters, the form of noncarbonate hardness normally encountered is
the one associated with the sulfate anion; although, occasionally, large quantities of
the chloride and nitrate anions may also be found. The softening reactions for the

removal of the noncarbonate hardness of magnesium associated with the possible
anions are as follows:
(10.11)
(10.12)
(10.13)
As shown in all the previous reactions, the removal of the magnesium hardness,
both carbonate and noncarbonate, can use just one chemical. This chemical is normally
lime, CaO.
10.7 LIME–SODA PROCESS
As shown by the various chemical reactions above, the chemicals soda ash and lime
may be used for the removal of hardness caused by calcium and magnesium. Thus,
the lime–soda process is used. This process, as mentioned, uses lime (CaO) and
soda ash (Na
2
CO
3
). As the name of the process implies, two possible sets of chemical
reactions are involved: the reactions of lime and the reactions of soda ash. To
understand more fully what really is happening in the process, it is important to
discuss these chemical reactions. Let us begin by discussing the lime reactions.
CaO first reacts with water to form slaked lime, before reacting with the bicar-
bonate. The slaking reaction is
(10.14)
After slaking, the bicarbonates are neutralized according to the following reac-
tions:
(10.15)
(10.16)
Note that in Equation (10.16) two types of solids are produced: Mg(OH)
2
and

CaCO
3
and that the added calcium ion from the lime that would have produced an
added hardness to the water has been removed as CaCO
3
. Although the hardness
ions have been precipitated out, the resulting solids, however, pose a problem of
disposal in water softening plants.
MgSO
4
2OH
−
+ Mg OH()
2
↓ SO
4
2−
+→
MgCl
2
2OH
−
+ Mg OH()
2
↓ 2Cl+→
Mg NO
3
()
2
2OH

−
+ Mg OH()
2
↓ 2NO
3
+→
CaO HOH+ Ca OH()
2
→
Ca HCO
3
()
2
Ca OH()
2
+ 2CaCO
3
↓
2HOH+→
Mg HCO
3
()
2
2Ca OH()
2
+ Mg OH()
2
↓
2CaCO
3

↓
2 HOH++→
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Magnesium, whether in the form of the carbonate or noncarbonate hardness, is
always removed in the form of the hydroxide. Thus, to remove the total magnesium
hardness, more lime is added to satisfy the overall stoichiometric requirements for
both the carbonates and noncarbonates. (Later, we will also discuss the requirement
of adding more lime to raise the pH.) The pertinent softening reactions for the
removal of the noncarbonate hardness of magnesium follow.
(10.17)
(10.18)
(10.19)
As shown from the previous reactions, there is really no net mole removal of
hardness that results from the addition of lime. For every mole of magnesium hardness
removed [MgSO
4
, MgCl
2
, or Mg(NO
3
)
2
], there is a corresponding mole of by-product
noncarbonate calcium hardness produced [CaSO
4
, CaCl
2
, or Ca(NO
3

)
2
]. Despite the
fact that these reactions are useless, they still must be considered, because they will
always transpire if the noncarbonate hardness is removed using lime. For this reason,
the implementation of the lime-soda process should be such that as much magnesium
as possible is left unremoved and rely only on the removal of calcium to meet the
desired treated water hardness. If the desired hardness level is not met by the removal
of only the calcium ions, then removal of the magnesium may be initiated. This will
entail the use of lime followed by the possible addition of soda ash to remove the
resulting noncarbonate hardness of calcium. As shown by Eqs. (10.17), (10.18), and
(10.19), soda ash is needed to remove the by-product calcium hardness formed from
the use of the lime.
As noted before, the calcium ion is removed in the form of CaCO
3
. This is the
reason for the use of the second chemical known as soda ash for the removal of the
noncarbonate hardness of calcium. Thus, another set of chemical reactions involving
soda ash and calcium would have to written. The pertinent softening reactions are
as follows:
(10.20)
(10.21)
(10.22)
Some of the calcium hardness of these reactions would be coming from the by-product
noncarbonate hardness of calcium that results if lime were added to remove the
noncarbonate hardness of magnesium.
It is worth repeating that soda ash is used for two purposes only: to remove the
original calcium noncarbonate hardness in the raw water and to remove the by-product
calcium noncarbonate hardness that results from the precipitation of the noncarbonate
MgSO

4
Ca OH
2
()+ Mg OH()
2
↓
CaSO
4
+→
MgCl
2
Ca OH()
2
+ Mg OH()
2
↓
CaCl
2
+→
Mg NO
3
()
2
Ca OH()
2
+ Mg OH()
2
↓
Ca NO
3

()
2
+→
CaSO
4
Na
2
CO
3
+ CaCO
3
↓
Na
2
SO
4
+→
CaCl
2
Na
2
CO
3
+ CaCO
3
↓
2NaCl+→
Ca NO
3
()

2
Na
2
CO
3
+ CaCO
3
↓
2NaNO
3
+→
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© 2003 by A. P. Sincero and G. A. Sincero
hardness of magnesium. It is also important to remember that by using lime the
carbonate hardness of magnesium does not produce any calcium noncarbonate hard-
ness. It is only the noncarbonate magnesium that is capable of producing the by-
product calcium noncarbonate hardness when lime is used.
10.7.1 CALCULATION OF STOICHIOMETRIC LIME REQUIRED
IN THE LIME–SODA PROCESS
Because the raw water is exposed to the atmosphere, there will always be some CO
2
dissolved in it. As will be shown later, carbon dioxide also consumes lime. We will,
however, ignore this requirement for the moment, and discuss it in the latter part of this
chapter. Ignoring carbon dioxide, the amount of lime needed comes from the require-
ment to remove the carbonate hardness of calcium and the requirements to remove both
the carbonate and the noncarbonate hardness of magnesium. We will first calculate the
amount of lime required for the removal of the carbonate hardness of calcium.
Let be the mass of in the carbonate hardness of calcium
to be removed. From Equation (10.15), taking the positive oxidation state, the number
of reference species is 2 moles of positive charges. Thus, the equivalent mass of

calcium bicarbonate is /2 and the number of equivalents of the
mass of calcium bicarbonate is /( /2). This is also the same num-
ber of equivalents of CaO (or slaked lime) used to neutralize the bicarbonate.
From Eqs. (10.14) and (10.15), the equivalent mass of CaO is CaO/2. Thus, the
mass of CaO, , needed to neutralize the calcium bicarbonate is /
(CaO/2). Let be the total calcium bicarbonate in the raw
water and be its fractional removal. Thus, = and
(10.23)
Now, calculate the amount of lime needed to remove . Let
be the mass of in the carbonate hardness of magnesium to be removed.
From Equation (10.16), again taking the positive oxidation state, the number of ref-
erence species is 2 moles of positive charges. Thus, the equivalent mass of magnesium
bicarbonate is /2 and the number of equivalents of the mass of
magnesium bicarbonate is /( /2). This is also the same number
of equivalents of CaO (or slaked lime) used to neutralize the magnesium bicarbonate.
From Eqs. (10.14) and (10.16), the equivalent mass of CaO is 2CaO/2. Thus,
the mass of CaO, , needed to neutralize the magnesium bicarbonate is
/( /2)CaO. Let be the total magnesium bicarbonate
in the raw water and be its fractional removal. Thus, =
and
(10.24)
M
CaHCO
3
Ca HCO
3
()
2
Ca HCO
3

()
2
M
CaHCO
3
M
CaHCO
3
Ca HCO
3
()
2
M
CaOCaHCO
3
M
CaHCO
3
(Ca(HCO
3
)
2
/2) M
T CaHCO
3
f
CaHCO
3
M
CaHCO

3
f
CaHCO
3
M
T CaHCO
3
M
CaOCaHCO
3
f
CaHCO
3
M
T CaHCO
3
Ca HCO
3
()
2
/2

CaO/2()0.35 f
CaHCO
3
M
T CaHCO
3
==
Mg HCO

3
()
2
M
MgHCO
3
Mg HCO
3
()
2
Mg HCO
3
()
2
M
MgHCO
3
M
MgHCO
3
Mg(HCO
3
)
2
M
CaOMgHCO
3
M
MgHCO
3

Mg HCO
3
()
2
M
T MgHCO
3
f
MgHCO
3
M
MgHCO
3
f
MgHCO
3
M
T MgHCO
3
M
CaOMgHCO
3
f
MgHCO
3
M
T MgHCO
3
Mg HCO
3

()
2
/2

CaO 0.77 f
MgHCO
3
M
T MgHCO
3
()==
TX249_frame_C10.fm Page 475 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero
The removal of the noncarbonate hardness of magnesium using lime as the
precipitant produces a corresponding amount of noncarbonate hardness of calcium
as a by-product. This by-product requires the use of soda ash for its removal. Thus,
the amount of lime needed for the removal of this magnesium noncarbonate hardness
will be discussed in conjunction with the determination of the stoichiometric soda
ash required to remove the by-product calcium noncarbonate hardness that results.
10.7.2 KEY TO UNDERSTANDING SUBSCRIPTS
So far, we have used mass variables such as , , and
and fractional variables such as and . Later on, we will
also be using other mass variables such as M
solidsNonCarb
, and and fractional
variables such as f
MgCa
and f
Ca
. As we continue to develop equations, it would be

found that ascertaining the meaning of the subscripts of the variables would become
difficult. Thus, a technique must be developed to aid in remembering.
First, for the mass variables, M refers to the mass of the type of species involved.
The subscript of M is a string of words. In this string, T may or may not be present.
If the M of a given species can be both a partial mass and a total mass, the T would
be present as the first one in the subscipt to represent the total mass. Then if it is
not present, the mass is a partial mass. Partial here means a part of the total. For
example, consider and . As shown, the M’s in these symbols can
be both a partial mass and a total mass. Thus, the first one refers to the total mass
and the second refers to the partial mass of the type of species involved.
The next part of the string subscript is CaHCO
3
. This refers to the type of
species involved. This is not really the correct formula for calcium bicarbonate, but
because it is used as a subscript, it can be easily taken to mean calcium bicarbonate.
Thus, the symbols and refer to the total mass of calcium bicar-
bonate and the partial mass of calcium bicarbonate, respectively. Calcium bicar-
bonate, , is said to be the type of species involved.
If M can only have a total mass, the T is not used. For example, the total solids,
M
solids
, can come from the solids produced from carbonate hardness, from
the solids produced from noncarbonate hardness, M
solidsNonCarb
, and from the calcium
carbonate solids produced from carbon dioxide, . In this case, since the
designation for the components of total solids use different subscripts, the M of total
solids can only have the total mass and therefore M
solids
is used instead of M

Tsolids
.
No confusion occurs, however, if M
Tsolids
is used, because this would definitely refer
to the total.
Not counting T, a second portion of the string subscript (the third portion, if T
is present) may also be present. This will then refer to the reason for the existence
of the type of mass involved. For example, consider , mentioned previ-
ously. In this symbol, CO
2
is the reason for the existence of the mass of CaCO
3
.
Thus, is the mass of calcium carbonate produced from carbon dioxide.
Another example where the second portion of a string subscript is the reason
for existence of the type of mass involved is . In this symbol,
is the reason for the existence of CaO. In the pertinent chemical reaction, calcium
M
T CaHCO
3
M
CaHCO
3
M
CaOMgHCO
3
M
MgHCO
3

f
MgHCO
3
f
CaHCO
3
M
CaCO
3
MgCa
M
T CaHCO
3
M
CaHCO
3
M
T CaHCO
3
M
CaHCO
3
Ca HCO
3
()
2
M
solidsCarb
,
M

CaCO
3
CO
2
M
CaCO
3
CO
2
M
CaCO
3
CO
2
M
CaOCaHCO
3
Ca HCO
3
()
2
TX249_frame_C10.fm Page 476 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero
oxide is not produced from calcium bicarbonate, but, rather, calcium oxide is used
because of the bicarbonate. Nevertheless, the bicarbonate is still the reason for the
existence of calcium oxide.
The portions of a string subscript can be easily identified, since they conspicu-
ously stand out. For example, portions CaCO
3
and CO

2
conspicuously stand out in
. CaCO
3
and CO
2
are said to be portions of the string subscript of
.
In some situations, a string subscript may appear to have several portions con-
spicuously standing out. For example, consider M
solidsNonCarb
. This symbol seems to
have three portions standing out; however, solidsNonCarb does not refer to three
different entities but to only one and this is the noncarbonate solids. Thus, M
solidsNonCarb
does not have three subscript portions.
As far as the masses are concerned, consider this last example, . This
symbol indicates that the reason for the existence of the CaCO
3
solid is Mg, however,
a next symbol Ca also appears. The appearance of Ca simply indicates that Mg, as
the reason for the existence of CaCO
3
, bears a relationship to calcium. This rela-
tionship can only happen when the noncarbonate hardness of magnesium is involved,
for this will produce the corresponding noncarbonate hardness of calcium––the
relationship. An important thing to remember, as far as our method of subscripting
is concerned, is that whenever Mg and Ca are used as subscripts, they refer to the
noncarbonate form of their respective hardness. For example, M
T Ca

refers to the
total noncarbonate calcium hardness and M
Ca
refers to a partial noncarbonate calcium
hardness. As another example, consider . In this symbol, noncarbonate
Mg is the reason for the existence of the calcium carbonate. This magnesium, of
course, produced the noncarbonate Ca (the relationship) for the eventual production
of the calcium carbonate. In other words, to repeat, whenever Ca and Mg are used
as subscripts, they are intended to refer only to their respective noncarbonate form
of hardness.
Now, consider the fractional variables. For example, consider f
MgCa
and f
Ca
. As
in the mass variables, the first subscript of the fractional variable refers to the type
of fraction of the mass. Thus, in f
MgCa
, the type of fraction is the fraction of Mg;
and, considering the fact that when Mg and Ca are used as subscripts, they refer to
the noncarbonate forms of hardness, f
MgCa
stands for the fraction of the noncarbonate
form of magnesium hardness, with Ca reminding about the relationship. f
MgCa
could
just simply be written as f
Mg
and they would be the same, but Ca is there, again,
just as a reminder that it is also involved in the chemistry of the reaction. By the

convention we have just discussed, f
Ca
stands for the fraction of the noncarbonate
form of calcium hardness. As a further example, to what does refer? The
is the fraction of the calcium carbonate (bicarbonate) hardness.
10.7.3 CALCULATION OF STOICHIOMETRIC SODA ASH REQUIRED
The amount of soda ash needed comes from the requirement to remove the noncar-
bonate hardness of calcium. In addition, if the noncarbonate hardness of magnesium
was precipitated using lime, additional soda ash will also be required to remove the
M
CaCO
3
CO
2
M
CaCO
3
CO
2
M
CaCO
3
MgCa
M
CaCO
3
MgCa
f
CaHCO
3

TX249_frame_C10.fm Page 477 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero
by-product noncarbonate hardness of calcium that results from the magnesium
precipitation (if additional hardness removal is desired).
Let M
Ca
be the mass of the calcium species to be removed from the noncarbonate
hardness of calcium. From Eqs. (10.20), (10.21), or (10.22), taking the positive
oxidation state, the number of reference species is 2 moles of positive charges for
each of these reactions. Thus, the equivalent mass of calcium is Ca/2 and the number
of equivalents of the M
Ca
mass of calcium is M
Ca
/(Ca/2). This is also the same
number of equivalents of Na
2
CO
3
needed to precipitate the noncarbonate hardness
of calcium to CaCO
3
. Again, from Eqs. (10.20), (10.21), or (10.22), the equivalent
mass of Na
2
CO
3
is Na
2
CO

3
/2. Thus, the mass of soda ash, M
sodAshCa
, needed to
precipitate the calcium species from the noncarbonate hardness of calcium is
M
Ca
/(Ca/2)(Na
2
CO
3
/2). Let M
T Ca
be the total calcium in the noncarbonate hardness
of calcium in the raw water and f
Ca
be its fractional removal. Thus, M
Ca
= f
Ca
M
T Ca
and
(10.25)
Note that in M
sodAshCa
, Ca is the reason for the use of soda ash.
To calculate the soda ash requirement for the calcium noncarbonate hardness
by-product, let M
MgCa

represent the mass of the magnesium species that precipitates
and results in the production of the additional calcium hardness cation, where the
Ca, again, is written as a reminder. Refer to Eqs. (10.17), (10.18), and (10.19) to
see how the calcium hardness is produced from the precipitation of noncarbonate
magnesium. The number of equivalents of the calcium hardness produced is equal
to the number of equivalents of the noncarbonate hardness of magnesium precipitated
[which is equal to M
MgCa
/(Mg/2), where Mg/2 is the equivalent mass of Mg as
obtained from Eqs. (10.17), (10.18), and (10.19)]. Because this is calcium hardness,
we already have the method of determining the amount of soda ash needed to remove
it as shown in Equation (10.25). Letting this amount be M
sodAshMgCa
, M
sodAshMgCa
is
equal to M
MgCa
/(Mg/2)(Na
2
CO
3
/2). Let M
T Mg
be the total magnesium in the noncar-
bonate hardness of magnesium and f
MgCa
be its fractional removal. Thus, M
MgCa
=

f
MgCa
M
T Mg
, with this and Equation (10.25),
(10.26)
The mass of magnesium species from the noncarbonate hardness of magnesium
that precipitates also needs additional lime for its precipitation as magnesium
hydroxide. Note that this is the same mass of magnesium that produced the additional
by-product calcium hardness. As mentioned above, this mass is M
MgCa
. Again, from
Eqs. (10.17), (10.18), and (10.19), the number of equivalents of the M
MgCa
mass of
magnesium is M
MgCa
/(Mg/2). This is the same number of equivalents of the lime
needed to precipitate Mg(OH)
2
. From the preceding equations, the equivalent mass
of CaO is CaO/2. Let the mass of lime needed to precipitate the Mg(OH)
2
resulting
M
sodAshCa
f
Ca
M
T Ca

Ca/2

Na
2
CO
3
/2()2.65 f
Ca
M
TCa
==
M
sodAsh MgCa
f
MgCa
M
T Mg
Mg/2

Na
2
CO
3
/2()4.36 f
MgCa
M
T Mg
==
TX249_frame_C10.fm Page 478 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero

from M
MgCa
mass of magnesium be M
CaOMgCa
. Thus, M
CaOMgCa
is M
MgCa
/(Mg/2)(CaO/2),
but because M
MgCa
= f
MgCa
M
T Mg
,
(10.27)
10.7.4 CALCULATION OF SOLIDS PRODUCED
The solids produced in water softening plants, if not put to use, pose a disposal
problem. Conceptually, because of their basic nature, they can be used in absorption
towers that use alkaline solutions to scrub acidic gas effluents. These solids, as far
as the lime-soda process is concerned, come from the solids produced in the removal
of (1) the carbonate hardness and (2) the noncarbonate hardness.
Let us first derived the equations for solids production from the removal of the
carbonate hardness. The sources of these solids are the CaCO
3
produced from the
removal of the carbonate hardness of calcium [Equation (10.15)], the CaCO
3
pro-

duced from the removal of the carbonate hardness of magnesium [Equation (10.16)],
and the Mg(OH)
2
produced also from the removal of the carbonate hardness of
magnesium [Equation (10.16)].
The amount of calcium bicarbonate removed is = .
Thus, using Equation (10.15), the corresponding production of CaCO
3
is
(10.28)
is the calcium carbonate produced. Note that the second subscript,
CaHCO
3
, is the reason for the existence of the carbonate.
Now, the amount of magnesium bicarbonate removed is =
From Equation (10.16), the corresponding production of CaCO
3
is
(10.29)
is the calcium carbonate produced. Note that, similarly with M
CaCO
3
CaHCO
3
,
the second subscript, MgHCO
3
, is the reason for the existence of the carbonate.
Also, from the same Equation (10.16), the amount of Mg(OH)
2

solids produced is
(10.30)
M
CaOMgCa
f
MgCa
M
T Mg
Mg/2

CaO/2()2.30 f
MgCa
M
T Mg
==
M
CaHCO
3
f
CaHCO
3
M
T CaHCO
3
M
CaCO
3
CaHCO
3
2CaCO

3
Ca HCO
3
()
2

f
CaHCO
3
M
T CaHCO
3
1.23 f
CaHCO
3
M
T CaHCO
3
==
M
CaCO
3
CaHCO
3
M
MgHCO
3
f
MgHCO
3

M
T MgHCO
3
.
M
CaCO
3
MgHCO
3
2CaCO
3
Mg HCO
3
()
2

f
MgHCO
3
M
T MgHCO
3
1.37 f
MgHCO
3
M
T MgHCO
3
==
M

CaCO
3
MgHCO
3
M
MgOHMgHCO
3
Mg OH()
2
Mg HCO
3
()
2

f
MgHCO
3
M
T MgHCO
3
0.40 f
MgHCO
3
M
T MgHCO
3
==
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© 2003 by A. P. Sincero and G. A. Sincero
is the amount of the Mg(OH)

2
solids produced. Also, note the reason
for its existence, which is MgHCO
3
.
Combining Eqs. (10.28), (10.29), and (10.30), the total mass of solids, M
soildsCarb
,
produced from the removal of the carbonate hardness using the lime–soda ash
process is therefore
(10.31)
Now, let us derive the equations for solids production from the removal of the
noncarbonate hardness. The sources of these solids are the Mg(OH)
2
produced from
the removal of the M
MgCa
mass of magnesium [Eqs. (10.17), (10.18), and (10.19)]; the
CaCO
3
produced from the removal of the M
Ca
mass of calcium [Eqs. (10.20), (10.21),
and (10.22)]; and the CaCO
3
produced from the M
MgCa
mass of magnesium [Eqs.
(10.17), (10.18), and (10.19)] and [Eqs. (10.20), (10.21), and (10.22)].
Equations (10.17), (10.18), and (10.19) show the production of the Mg(OH)

2
solids. From these equations, the amount of hydroxide solids, M
MgOHMgCa
, produced
from the M
MgCa
mass of magnesium, is
(10.32)
The amount of noncarbonate calcium removed is M
Ca
= f
Ca
M
T Ca
. Thus, Eqs. (10.20),
(10.21), and (10.22), the corresponding production of CaCO
3
is
(10.33)
is the amount of the CaCO
3
solids produced. Note that Ca is the reason
for its existence.
The mass of CaCO
3
solids, , produced from the removal of the
M
MgCa
= f
MgCa

M
T Mg
mass of magnesium is obtained as follows: From Eqs. (10.17),
(10.18), and (10.19), one mole of Mg produces one mole of Ca. From Eqs. (10.20),
(10.21), and (10.22), one mole of Ca also produces the same one mole of CaCO
3
.
Thus,
(10.34)
From Eqs. (10.32), (10.33), and (10.34), the total mass of solids, M
soildsNonCarb
,
produced from the removal of the noncarbonate hardness is
(10.35)
M
MgOHMgHCO
3
M
soildsCarb
M
CaCO
3
CaHCO
3
M
CaCO
3
MgHCO
3
M

MgOHMgHCO
3
++=
M
MgOHMgCa
Mg OH()
2
Mg

f
MgCa
M
T Mg
()2.40 f
MgCa
M
T Mg
==
M
CaCO
3
Ca
CaCO
3
Ca

f
Ca
M
T Ca

2.49 f
Ca
M
T Ca
==
M
CaCO
3
Ca
M
CaCO
3
MgCa
M
CaCO
3
MgCa
CaCO
3
Mg

f
MgCa
M
T Mg
4.11 f
MgCa
M
T Mg
==

M
solidsNonCarb
M
MgOHMgCa
M
CaCO
3
Ca
M
CaCO
3
MgCa
++=
TX249_frame_C10.fm Page 480 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero
10.8 ORDER OF REMOVAL
Suppose that there are four gram-equivalents of Ca
2+
, one gram-equivalent of Mg
2+
,
and 2.5 gram-equivalents of . Suppose further that the decision has been made
to use lime. The question is which reaction takes precedence, Equation (10.15) or
Equation (10.16)? Assuming the former takes precedence, the bicarbonate species
present will be composed of 2.5 gram-equivalent of Ca(HCO
3
)
2
and none of the
Mg(HCO

3
)
2
. The amount of lime needed to remove this bicarbonate is 2.5(CaO/2) =
70 g. Assuming the latter takes precedence, the bicarbonate species will now be com-
posed of one gram-equivalent of Mg(HCO
3
)
2
and 1.5 gram-equivalent of Ca(HCO
3
)
2
.
The corresponding amount of lime needed to remove the bicarbonates is 1.0(2CaO/
2) + 1.5(CaO/2) = 98 g. These two simple calculations produce two very different
results, so knowledge of the order of precedence of the reactions is very important.
The order of precedence can be judged from the solubility of the precipitates, namely:
CaCO
3
and Mg(OH)
2
.
The K
sp
for CaCO
3
is 5(10
−9
) at 25°C, while that for Mg(OH)

2
is 9(10
−12
), also
at 25°C. Consider calcium carbonate, first. Before the solid is formed, the product
of the concentrations of the calcium ions and the carbonate ions must be, at least,
equal to 5(10
−9
). Let x be the concentration of the calcium ion; thus, x will also be
the concentration of the carbonate ion. Therefore,
x
2
= 5(10
−9
) ⇒ x = 0.000071 gmole/L
Now, consider the magnesium hydroxide. Again, before the solid is formed, the
product of the concentrations of the magnesium ions and the hydroxide ions must
be, at least, equal to 9(10
−12
). In magnesium hydroxide, there are two hydroxide ions
to one of the magnesium ion. Thus, letting y be the concentration of the magnesium
ions, the concentration of the hydroxide ions is 2y. Therefore,
y(y
2
) = 9(10
−12
) ⇒ y = 0.00021 gmole/L
Now, comparing the two answers, CaCO
3
is less soluble than Mg(OH)

2
, because
it only needs 0.000071 gmole/L to precipitate it compared to 0.00021 gmole/L for
the case of Mg(OH)
2
. Therefore, calcium carbonate will precipitate first and Equation
(10.15) takes precedence over Equation (10.16). In this example, the answer would
be 70 g instead of 98 g.
10.9 ROLE OF CO
2
IN REMOVAL
For the removal of the hardness of calcium and magnesium, the effect of carbon
dioxide on the amounts of sludge produced and chemicals used were not considered;
but because the type of raw water treated is exposed to the atmosphere, CO
2
always
dissolves in it. Therefore, in addition to the chemical reactions that have been
portrayed, the precipitant chemicals must also react with carbon dioxide. This means
HCO
3
−
TX249_frame_C10.fm Page 481 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero
that more lime than indicated in previous reactions would be needed. The reaction
that portrays the consumption of lime by carbon dioxide is as follows:
(10.36)
As shown in Equation (10.36), solids of CaCO
3
are produced. Let be the
mass of carbon dioxide in the raw water. Therefore, letting be the car-

bonate solids produced from the dissolved carbon dioxide,
(10.37)
Also, from Equation (10.36), the lime needed, , to react with the carbon
dioxide in the raw water is
(10.38)
10.10 EXCESS LIME TREATMENT AND OPTIMUM
OPERATIONAL pH
Refer to Figure 10.2. As shown, the lowest solubility of Ca
2+
in equilibrium with
CaCO
3
is at a pH of approximately 9.3. This represents the optimum pH for adding
lime or soda ash to precipitate calcium as CaCO
3
. Raising the pH to about 11
practically dissolves all the CaCO
3
. This is illustrated in the reaction below. The
anion species responsible for raising the pH to 11 is the OH
−
. Thus, the chemical
reaction is
(10.39)
This means that, in order to precipitate the calcium carbonate optimally, the pH
should be maintained at the vicinity of 9.3 and that it should not be raised to pH 11.
As shown in the chemical reaction, the calcium ion is converted into the complex
CaOH
+
, which has a relatively high K

sp
of 10
−1.49
, indicating the calcium carbonate
is practically dissolved when the pH is raised to 11.
Also, from the figure, the lowest solubility of Mg
2+
in equilibrium with Mg(OH)
2
occurs at about pH 10.4; raising the pH above this value does not affect the solubility,
since the solubility is already zero at pH 10.4. At this value of pH, the solubility of
Ca
2+
in equilibrium with CaCO
3
is not affected very much. Also, lowering the pH
to below 9.8 practically dissolves the Mg(OH)
2
. Thus, these two limiting conditions
leave only a very small window of pH 9.8 to 10.4 for an optimal concurrent
precipitation of Ca
2+
and Mg
2+
. Above pH 10.4, the precipitation of CaCO
3
suffers
and below pH 9.8, the precipitation of Mg(OH) suffers.
CO
2

Ca OH()
2
+ CaCO
3
↓
H
2
O+→
M
CO
2
M
CaCO
3
CO
2
M
CaCO
3
CO
2
CaCO
3
CO
2

M
CO
2
()2.27 M

CO
2
()==
M
CaOCO
2
M
CaOCO
2
CaO
CO
2

M
CO
2
()1.27 M
CO
2
()==
CaCO
3
OH
−
+ Ca OH()
2
CO
3
2−
+→

Ca
2+
OH
−
 CaOH
+
K
sp
+
Ca
2+
{}OH
−
{}
CaOH
+

10
1.49–
==
TX249_frame_C10.fm Page 482 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero
It has been found experimentally that adding lime to satisfy the stoichiometric
amount for precipitating Mg(OH)
2
does not raise the pH value to 10.4. Lime dissolves
in water according to the following reaction: CaO + HOH → Ca(OH)
2
 Ca
2+

+
2OH
−
. Empirically, based on this reaction (equivalent mass = CaO/2), 1.0 milligram-
equivalent per liter of water of excess alkalinity over the computed stoichiometric
amount is needed to raise the pH to 10.4. This means that this amount must be added
if the pH is to be maintained at this level. Letting M
CaOExcess
be the total amount of
excess lime and be the volume of water treated in cubic meters,
(10.40)
FIGURE 10.2 Concentrations of calcium and magnesium ions in equilibrium with the
calcium carbonate and magnesium hydroxide solids, respectively. (From Powell, S. T. (1954).
Water Conditioning for Industry. McGraw-Hill, New York. With permission.)
Concentration of calcium ion and magnesium ion. ppm
200
180
160
140
120
100
80
60
40
20
0
7 8 9 10 11 12 13 14
pH value
Concentration of magnesium
ion in equilibrium with

magnesium hydroxide
Concentration of calcium
ion in equilibrium with
calcium carbonate
V
M
CaOExcess
1.0 CaO/2()1000()
V
1000 1000()

0.028
V
kg==
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© 2003 by A. P. Sincero and G. A. Sincero
10.11 SUMMARY OF CHEMICAL REQUIREMENTS
AND SOLIDS PRODUCED
To summarize, let M
CaO
be the total lime requirement; M
sodAsh
be the total soda ash
requirement; and M
solids
be the total solids produced. M
CaO
is equal to the amount of
lime used for the removals of the carbonate hardness of calcium , the
carbonate hardness of magnesium ( ), the noncarbonate hardness of mag-

nesium (M
CaOMgCa
), the requirement to neutralize the dissolved carbon dioxide
( ), and the requirement to raise the pH to 10.4 for the precipitation of
Mg(OH)
2
. Thus,
(10.41)
Note: The unit of M
CaOExcess
is kilograms. Thus, in order for the terms of this
equation to be dimensionally equivalent to M
CaOExcess
, they must all be
expressed in terms of kilograms.
M
sodAsh
is equal to the amount of soda ash needed to precipitate the noncarbonate
hardness of calcium and the amount of soda ash needed to precipitate the noncar-
bonate hardness of calcium produced from the precipitation of the noncarbonate
hardness of magnesium. Thus,
(10.42)
M
solids
is equal to the solids produced from the removals of the carbonate and
the noncarbonate hardness (M
solidsCarb
and M
solidsNonCarb
, respectively) and the solids

produced from the neutralization of the dissolved carbon dioxide using lime
( ). Thus,
M
solids
= M
solidsCarb
+ M
solidsNonCarb
+ (10.43)
10.12 SLUDGE VOLUME PRODUCTION
The mass of solids precipitated enmeshes in an enormous amount of water. As these
solids are flocculated and settled, they retain extraordinary amounts of water resulting
in a huge volume of sludge to be disposed of. The amount of the solids is negligible
compared to the total amount of sludge. In fact, sludges contains approximately 99%
water. Let f
solids
be the fraction of solids in the sludge. Then the mass of sludge M
sludge
is M
solids
/f
solids
.

If S
sl
is the specific gravity of the sludge, its volume in cubic meters is
(10.44)
The 1,000 in the denominator is the mass density of water in kg/m
3

at the temperature
of 4°C used as the reference temperature in the definition of specific gravity. All
units in this equation must be in the mks (meter-kilogram-second) system.
M
CaOCaHCO
3
()
M
CaOMgHCO
3
M
CaOCO
2
M
CaO
M
CaOCaHCO
3
M
CaOMgHCO
3
M
CaOMgCa
M
CaOCO
2
M
CaOExcess
++++=
M

sodAsh
M
sodAshCa
M
sodAshMgCa
+=
M
CaCO
3
CO
2
M
CaCO
3
CO
2
V
sl
V
sl
M
solids
f
solids
S
sl
1000()

M
sludge

S
sl
1000()

==
TX249_frame_C10.fm Page 484 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero
10.13 CHEMICAL SPECIES IN THE TREATED WATER
In addition to the H
+
and OH
−
species, a great majority of ions present in the treated
water are the cations Ca
2+
, Mg
2+
, and Na
+
and the anions and . Occa-
sionally, the anions Cl
−
and may also be present when the hardness cations in
natural waters are associated with these anions.
The effluent water from the hardness removal reactor is basic and tends to deposit
scales in distribution pipes. For this reason, this water should be stabilized. Stabili-
zation is normally done using carbon dioxide, a process called recarbonation. Sta-
bilization using carbon dioxide affects the concentration of the bicarbonate ion in
the treated water. The concentrations of the , Cl
−

and ions are not affected,
however, because they do not react with carbon dioxide. Their concentrations remain
the same as when they were in the influent to the treatment plant. The original cation
Na
+
from the influent raw water is also not affected for the same reason that it does
not react with carbon dioxide. Na
+
is, however, introduced with the soda ash.
Let , , and be the concentrations of the
indicated ions in milligram equivalents per liter in the influent to the reactor. Also, let
, , and be the concentrations of the ions in the treated water.
Thus,
10.13.1 LIMITS OF TECHNOLOGY
The precipitation of CaCO
3
and Mg(OH)
2
is never complete. At 0°C, the solubility
of CaCO
3
is 15 mg/L and at 25°C, it is 14 mg/L. These solubility values imply that
no matter how much precipitant is applied to the water, there will always remain
some calcium and carbonate ions which, in the aggregate, amounts to 14 to 15 mg/L
of CaCO
3
dissolved in the treated water.
For Mg(OH)
2
, its solubility at 0°C is 17 mg/L as CaCO

3
; at 18°C it is 15.5 mg/L
as CaCO
3
. As in the case of CaCO
3
, some ions of magnesium and, of course, the
hydroxide will always remain dissolved in the treated effluent no matter how much
precipitant is employed. Operationally, let us adopt the following figures as the limit
of technology: CaCO
3
= 15 mg/L; Mg(OH)
2
= 16 mg/L as CaCO
3
. This brings a
total to 15 + 16 = 31 mg/L as CaCO
3
.
10.13.2 CONCENTRATION OF Ca
2++
++
The concentration of the calcium ion in the treated water comes from the calcium
bicarbonate not precipitated, the calcium from the noncarbonate hardness of cal-
cium not precipitated, and the calcium that results from the precipitation of the
SO
4
2−
HCO
3

−
NO
3
−
SO
4
2−
NO
3
−
SO
4
2−
[]
meq,inf
Cl
−
[]
meq,inf
NO
3
−
[]
meq,inf
SO
4
2−
[]
meq
Cl

−
[]
meq
NO
3
−
[]
meq
SO
4
2−
[]
meq
SO
4
2−
[]
meq,inf
=
Cl
−
[]
meq
Cl
−
[]
meq,inf
=
NO
3

−
[]
meq
NO
3
−
[]
meq,inf
=
TX249_frame_C10.fm Page 485 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero

noncarbonate hardness of magnesium. Note that, as the treated water is recarbonated,
the 15 mg/L of CaCO

3

from the limit of technology converts to the bicarbonate form,
(10.45)
From this reaction, the number of milligram equivalents per liter of Ca

2

+

and
are 15

/

(CaCO

3

/

2)

=

0.3, respectively.
Recall that mass of calcium bicarbonate from the
carbonate hardness of calcium was precipitated. Let this mass be measured in kilo-
grams. Also, let , in mg/L, be the concentration of calcium bicar-
bonate not precipitated. Thus,
(10.46)
is the volume of water treated in cubic meters.
Ca(HCO

3

)

2

ionizes to ; thus, the equivalent mass
of calcium bicarbonate is Ca(HCO

3

)

2

/

2. The calcium bicarbonate not precipitated there-
fore produces

/(

)

=

0.0123( )

=

0.0123[1000

/

]

=

milligram
equivalents per liter of Ca

2

+

and , respectively.
Now, also recall that

M

Ca

(

=

f

Ca

M

T

Ca

) mass of calcium from the noncarbonate
hardness of calcium precipitated. As before, let this mass be measured in kilograms.
Also, let , in mg/L, be the concentration of calcium not precipitated. Thus,
(10.47)
Consider CaSO

4

as representing the noncarbonate hardness of calcium species.
This ionizes as CaSO

4



→

Ca

2

+



+

; thus, the equivalent mass of calcium in the
noncarbonate hardness of calcium is Ca

/

2. The calcium not precipitated therefore
produces [Ca

2

+

]

mgnot

/

(Ca

/

2)

=

0.050[Ca

2

+

]

mgnot

=

50.0 [

M

TCa

(1

−

f

Ca

)

/

] milligram
equivalents of Ca

2

+

.
Summing all the calcium ion concentrations, the total concentration [Ca

2

+

]

meq

is
(10.48)
Again, take note that and

M

T

Ca

must be in kilograms and must be in
cubic meters. [Ca

2

+

]

meq

is then in milligram equivalents per liter.

CaCO
3
H
2
CO
3
CO
2
HOH H
2
CO
3
→+()+ Ca HCO
3
()
2
→
HCO
3
−
M
CaHCO
3
= f
CaHCO
3
M
T CaHCO
3
)(

Ca HCO
3
()
2
[]
mgnot
Ca HCO
3
()
2
[]
mgnot
M
T CaHCO
3
f
CaHCO
3
M
T CaHCO
3
–
V

1000 1000()
1000

=
1000
M

T CaHCO
3
1 f
CaHCO
3
–()
V

=
V
Ca HCO
3
()
2
Ca
2+
2HCO
3
−
+→
Ca HCO
3
()
2
[]
mgnot
Ca HCO
3
()
2

/2 Ca HCO
3
()
2
[]
mgnot
M
T CaHCO
3
1 f
CaHCO
3
–()
V
12.3 M
T CaHCO
3
1 f
CaHCO
3
–()/
V
[]
HCO
3
−
Ca
2+
[]
mgnot

Ca
2+
[]
mgnot
M
T Ca
f
Ca
M
T Ca
–
V

1000 1000()
1000

1000
M
T Ca
1 f
Ca
–()
V

==
SO
4
2−
V
Ca

2+
[]
meq
12.3
M
T CaHCO
3
1 f
CaHCO
3
–()
V

50.0
M
T Ca
1 f
Ca
–()f
MgCa
M
TMg
+
V

+=
M
T CaHCO
3
V

TX249_frame_C10.fm Page 486 Wednesday, June 19, 2002 10:43 AM
© 2003 by A. P. Sincero and G. A. Sincero
10.13.3 CONCENTRATION OF Mg
2++
++
The concentration of the magnesium ion in the treated water comes from the mag-
nesium from the magnesium bicarbonate not precipitated, and the magnesium from
the noncarbonate hardness of magnesium not precipitated.
As the treated water is recarbonated, the 16 mg/L of Mg(OH)
2
as CaCO
3
from
the limit of technology converts to the bicarbonate form,
(10.49)
Equation (10.7) has been written as
(10.50)
Thus, the 16 mg/L of Mg(OH)
2
as CaCO
3
is equivalent to 16/50 = 0.32 meq/L of
Mg
2+
= 0.32 meq/L of .
The possibility exists that the 16 mg/L of Mg(OH)
2
as CaCO
3
will react in the

presence of excess bicarbonates to form magnesium carbonate; however, the K
sp
of
magnesium carbonate is around 10
−5
compared to that of the K
sp
of magnesium
hydroxide, which is 9(10
−12
). Thus, Mg(OH)
2
will stay as Mg(OH)
2
until recarbonation.
Recall that (= ) mass of magnesium bicarbonate from
the carbonate hardness of magnesium was precipitated. Again, let all this mass be
measured in kilograms. Also, let [Mg(HCO
3
)
2
]
mgnot
, in mg/L, be the concentration
of magnesium bicarbonate not precipitated. Thus,
(10.51)
is the volume of water treated in cubic meters.
Mg(HCO
3
)

2
ionizes to Mg(HCO
3
)
2
→ Mg
2+
+ ; thus, the equivalent mass
of magnesium bicarbonate is Mg(HCO
3
)
2

/2. The magnesium bicarbonate not pre-
cipitated therefore produces:
milligram equivalents of Mg
2+
and , respectively.
Now, also recall that M
MgCa
(= f
MgCa
M
T Mg
) mass of magnesium from the noncar-
bonate hardness of magnesium precipitated. As before, let this mass be measured
Mg OH()
2
2H
2

CO
3
Mg HCO
3
()
2
2HOH+→+
C
T,hard
[]
asCaCO
3
Mg
2+
[]
eq
′
2




100()50 Mg
2+
[]
eq
′
==
HCO
3

−
M
MgHCO
3
f
MgHCO
3
M
T MgHCO
3
Mg HCO
3
()
2
[]
mgnot
M
T Mg HCO
3
f
MgHCO
3
M
T Mg HCO
3
–
V

1000 1000()
1000

=
M
T Mg HCO
3
1 f
MgHCO
3
–()
V

1000[]=
V
2HCO
3
−
Mg HCO
3
()
2
[]
mgnot
Mg HCO
3
()
2
/2

0.0137 Mg HCO
3

()
2
[]
mgnot
()=
13.7
M
T MgHCO
3
(1 f
MgHCO
3
)–
V

=
HCO
3
−
TX249_frame_C10.fm Page 487 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero
in kilograms. Also, let [Mg
2+
]
mgnot
, in mg/L, be the concentration of magnesium not
precipitated. Thus,
(10.52)
Consider MgSO
4

as representing the noncarbonate hardness of calcium species.
This ionizes as MgSO
4
→ Mg
2+
+ ; thus, the equivalent mass of magnesium
in the noncarbonate hardness of magnesium is Mg/2. The magnesium not precipi-
tated therefore produces:
milligram equivalents of Mg
2+
.
Summing all the magnesium ion concentrations, the total concentration [Mg
2+
]
meq
is
(10.53)
Again, take note that and must be in kilograms and must be in
cubic meters. [Mg
2+
]
meq
is then in milligram equivalents per liter.
10.13.4 CONCENTRATION OF
Two sources of the bicarbonate ion in the treated water are those coming from the
bicarbonates of the carbonate hardness of calcium and magnesium not removed
[12.3( − )/ + 13.7( − )/ ]. In addition, bicar-
bonates may also result from the recarbonation of the magnesium hydroxide from
the limits of technology. Call this concentration [HCO
3

]
OHmeq
. If no carbonate hard-
ness of calcium is present but noncarbonate, bicarbonates may also result from the
recarbonation of the calcium carbonate that precipitates when soda ash is used to
remove the noncarbonate hardness of calcium. Call this concentration
and let equal [HCO
3
]
OHmeq
plus . Letting the total con-
centration of bicarbonate be designated as , in milligram equivalents
per liter, we obtain
(10.54)
Mg
2+
[]
mgnot
M
T Mg
f
MgCa
M
T Mg
–
V

1000 1000()
1000

M
T Mg
1 f
MgCa
–()
V

1000[]==
SO
4
2−
Mg
2+
[]
mgnot
Mg/2

0.0823 Mg
2+
[]
mgnot
82.3
M
T Mg
(1 f
MgCa
)–
V




==
Mg
2+
[]
meq
13.7
M
T MgHCO
3
1 f
MgHCO
3
–()
V

82.3
M
T Mg
1 f
MgCa
–()
V

+=
M
T MgHCO
3
M
T Mg

V
H
CO
3
−
−
−
−
M
T CaHCO
3
M
CaHCO
3
V
M
T MgHCO
3
M
MgHCO
3
V
[HCO
3
]
CO
3
meq
HCO
3

[]
OHCO
3
me
q
[HCO
3
]
CO
3
meq
[HCO
3
−
]
meqtent
HCO
3
−
[]
meqtent
HCO
3
[]
OHCO
3
meq
12.3
M
T CaHCO

3
1 f
CaHCO
3
–()
V

+=
13.7
M
T MgHCO
3
1 f
MgHCO
3
–()
V

+
TX249_frame_C10.fm Page 488 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero
One of the subscripts is tent. This stands for tentative, because the concentration
will be modified as discussed below. [HCO
3
]
OHmeq
is equal to 0.32 meq/L, if mag-
nesium hydroxide is recarbonated, and it is equal to zero, if no magnesium hydroxide
is recarbonated. is equal to zero, if the carbonate hardness of calcium
is present, and it is equal to 3.0 meq/L, if no carbonate hardness of calcium is present

but noncarbonate and the noncarbonate hardness of calcium is removed using soda
ash.
Because of equilibrium, however, dissociates into and as
follows:
(10.55)
where the activities must be in gram moles per liter and is the K
sp
of the
bicarbonate. Thus, the concentration of bicarbonate at the treated effuent will be
less than that portrayed by Equation (10.54). In addition, some is also present
in the effluent.
The geq/L of is equal to the gmol/L of . Now, let us find the gram
mole per liter of the bicarbonate ion at equilibrium that results from an original one
gram mole per liter of the ion. Let x be the concentration of and H
+
, respec-
tively, produced from the dissociation. Thus, at equilibrium, = 1 − x.
Substituting into the K
sp
equation,
(10.56)
Solving for x,
(10.57)
Therefore, one milligram equivalent per liter of original bicarbonate produces
1 − mgeq/L of bicarbonate at equilibrium, and
thus,the of original bicarbonate produces {[ ]
meqtent
}{1 −
+ } mgeq/L. From Equation (10.54) and letting
meq

be the milligram equivalent per liter of the bicarbonate in the effluent,
(10.58)
HCO
3
[]
CO
3
meq
HCO
3
−
H
+
CO
3
2−
HCO
3
−
 H
+
CO
3
2−
K
sp,HCO
3
+
H
+

{}CO
3
2−
{}
HCO
3
−
{}

=
K
sp,HCO
3
CO
3
2−
HCO
3
−
HCO
3
−
CO
3
2−
HCO
3
−
{}
K

sp,HCO
3
H
+
{}CO
3
2−
{}
HCO
3
−
{}

x
2
1 x–

==
x
K–
sp,HCO
3
K
sp,HCO
3
2
4K
sp,HCO
3
++

2

gmol/L or geq/L()=
( K–
sp,HCO
3
K
sp,HCO
3
2
4K
sp,HCO
3
+
)+/2
[HCO
3
−
]
meqtent
HCO
3
−
( K–
sp,HCO
3
K
sp,HCO
3
2

4K
sp,HCO
3
+
)/2
[HCO
3
−
]
HCO
3
−
[]
meq
HCO
3
[]
OHCO
3
meq
12.3
M
T CaHCO
3
1 f
CaHCO
3
–()
V

+


=
13.7
M
T Mg HCO
3
1 f
MgHCO
3
–()
V



+
1
K
sp,HCO
3
– K
sp,HCO
3
2
4K
sp,HCO
3
++
2

–



TX249_frame_C10.fm Page 489 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero
10.13.5 CONCENTRATION OF CO
The one milligram equivalent per liter of original bicarbonate also produces
( + )/2 gram moles per liter of the carbonate ion. The
equivalent mass of carbonate is CO
3
/2. Thus, the milligram equivalent per liter of
the carbonate, , appearing in the effluent is
(10.59)
10.13.6 CONCENTRATION OF Na
++
++
The sources of the sodium ion are the original sodium from the influent and the sodium
added with the soda ash. Recall that there were M
sodAsh
mass of Na
2
CO
3
used. Let this
be measured in terms of kilograms. Its reaction with the Ca
2+
is Na
2
CO

3
+ Ca
2+
→
CaCO
3
+ 2Na
+
. Thus, the equivalent mass of Na
2
CO
3
is Na
2
CO
3
/2 and the number of
kilogram equivalents of the M
sodAsh
mass of Na
2
CO
3
is M
sodAsh
/(Na
2
CO
3
/2) = 0.01887

M
sodAsh
. For the cubic meters of water treated, the equivalent concentration in meq/L
is 0.01887M
sodAsh
/ [1000(1000)/1000] = 18.9(M
sodAsh
/ ). Let the concentration in the
influent be [Na
+
]
meq,in f
milligram equivalents per liter. Then, if [Na
+
]
meq
is the concen-
tration of the sodium ion in the treated water,
(10.60)
10.14 RELATIONSHIPS OF THE FRACTIONAL REMOVALS
The fractional removals , , f
Ca
, and f
MgCa
have been used in the fore-
going derivations. Let us formalize the definitions of these parameters in this heading
as follows:
(10.61)
(10.62)
(10.63)

(10.64)
In addition, define f
1
as the overall removal of the total calcium hardness, f
2
as
the overall removal of the total magnesium hardness, and f as the overall removal
of total hardness. Recall that is the total volume of water treated and that the
3
2−−
−−
K–
sp,HCO
3
K
sp,HCO
3
2
4K
sp,HCO
3
+
CO
3
−
[]
meq
CO
3
−

[]
meq
2
K–
sp,HCO
3
K
sp,HCO
3
2
4K
sp,HCO
3
++
2




=
V
V
V
Na
+
[]
meq
Na
+
[]

meq,in f
18.9
M
sodAsh
V



+=
f
CaHCO
3
f
MgHCO
3
f
CaHCO
3
M
CaHCO
3
M
T CaHCO
3

=
f
MgHCO
3
M

MgHCO
3
M
T Mg HCO
3

=
f
Ca
M
Ca
M
T Ca

=
f
MgCa
M
MgCa
M
T Mg

=
V
TX249_frame_C10.fm Page 490 Friday, June 14, 2002 4:40 PM
© 2003 by A. P. Sincero and G. A. Sincero