Standard Methods for Examination of Water & Wastewater_12 potx

Removal of Iron and
Manganese by Chemical
Precipitation

Iron concentrations as low as 0.3 mg/L and manganese concentrations as low as
0.05 mg/L can cause dirty water complaints. At these concentrations, the water may
appear clear but imparts brownish colors to laundered goods. Iron also affects the
taste of beverages such as tea and coffee. Manganese flavors tea and coffee with
medicinal tastes.
Some types of bacteria derive their energy by utilizing soluble forms of iron
and manganese. These organisms are usually found in waters that have high levels
of iron and manganese in solution. The reaction changes the species from soluble
forms into less soluble forms, thus causing precipitation and accumulation of black
or reddish brown gelatinous slimes. Masses of mucous iron and manganese can clog
plumbing and water treatment equipment. They also slough away in globs that
become iron or manganese stains on laundry.
Standards for iron and manganese are based on levels that cause taste and staining
problems and are set under the Environmental Protection Agency Secondary Drinking
Water Standards (EPA SDWA). They are, respectively, 0.3 mg/L for iron and 0.05 mg/L
for manganese. Iron and manganese are normally found in concentrations not exceed-
ing 10 mg/L and 3 mg/L, respectively, in natural waters. Iron and manganese can
be found at higher concentrations; however, these conditions are rare. Iron concen-
trations can go as high as 50 mg/L.
Iron and manganese may be removed by

reverse osmosis

and

ion exchange

. The
unit operation of reverse osmosis was discussed in a previous chapter; the unit
process of ion exchange is discussed in a later chapter. This chapter discusses the
removal of iron and manganese by the unit process of chemical precipitation.
One manufacturer claimed that these elements could also be removed by bio-
logical processes. It claimed that a process called the

bioferro

process encourages
the growth of naturally occurring iron assimilating bacteria, such as

Gallionella
ferruginea

, thus reducing iron concentration. An experimental result shows a reduc-
tion of iron from 6.0 mg/L to less than 0.1 mg/L. They also claimed that a companion
process called

bioman

could remove manganese down to 0.08 mg/L. This uses
naturally occurring manganese bacteria to consume manganese. Figure 13.1 is a
photograph showing growths of “bioferro” and “bioman” bacteria.
13

TX249_frame_C13.fm Page 593 Friday, June 14, 2002 2:32 PM
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594

13.1 NATURAL OCCURRENCES OF IRON
AND MANGANESE

Iron and manganese have the electronic configurations of [Ar]3

d

6

4

s

2

and [Ar]3

d

5

4

s

2

,

and are located in Groups

VIIIB

and

VIIB

of the Periodic Table, respectively. They
are both located in the fourth period. [Ar] means that these elements have the
electronic configuration of the noble gas argon. The letters

d

and

s

refer to the

d

and

s

orbitals; the superscripts indicate the number of electrons that the orbitals
contain. Thus, the

d

orbital of iron contains 6 electrons and that of manganese contains
5 electrons. Both elements contain 2 electrons in their

s

orbitals. This means that in
their most reduced positive state, they acquire oxidation states of 2

+

. Also, because
of the

d

orbitals, they can form a number of oxidation states. The multiplicity of
oxidation states give iron and manganese the property of imparting colors such as
the imparting of brownish colors to laundered goods.
Surface waters always contain dissolved oxygen in it. Thus, iron and manganese
would not exist in their most reduced positive state of 2

+

in these waters. The reason
is that they will simply be oxidized to higher states of oxidation by the dissolved
oxygen forming hydroxides and precipitate out. Groundwater is a source where these
elements could come from. Groundwaters occurring deep down in the earth can become
devoid of oxygen, thus, any iron or manganese present would have to be reduced.
Therefore, the waters where removal of iron and manganese could be undertaken are

groundwaters and the form of the elements are in the 2

+

oxidation states, Fe(II)



and
Mn(II), respectively

FIGURE 13.1

A photograph of “bioferro” and “bioman” bacteria.

TX249_frame_C13.fm Page 594 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero

595

13.2 MODES OF REMOVAL OF IRON AND
MANGANESE

The best place to investigate for determining the mode by which the elements can
be removed is the table of solubility products constants as shown in Table 13.1. In
general, a precipitation product that has the lowest

K

sp

means that the substance is
the most insoluble. As shown in the table, for iron, the lowest

K

sp

is that of Fe(OH)

3

,
an Fe(III) iron, and has the value of 1.1(10

−

38

). For manganese, the lowest

K

sp

is that
of MnS, an

Mn(II)



manganese, and has the value of 1.1(10

−

22

).
These

K

sp

’s indicate that the elements must be removed in the form of ferric
hydroxide and manganese sulfide, respectively; however, from the table, manganese
can also be removed as



Mn(OH)

2

at a

K

sp

=

4.5(10

−

14

). Of course, lime has many
uses, while sulfide has only few. Sodium sulfide is used in photographic film devel-
opment; however, lime is used in water and wastewater treatment, as an industrial
chemical, as well as being used in agriculture. Thus, because of its varied use, lime
is much cheaper. In addition, using a sulfide to remove iron and manganese would
be a new method. Its health effect when found in drinking water is not documented.
On the other hand, lime has been used for years. We will therefore use lime as the
precipitant for the removal of iron and manganese. The probable use of sulfide in
removing iron and manganese could be a topic for investigation in applied research.

13.3 CHEMICAL REACTIONS OF THE FERROUS
AND THE FERRIC IONS

The chemical reactions of the ferrous and the ferric ions were already discussed in
a previous chapter. From the topic in the preceding section, iron is more efficiently
removed as ferric hydroxide. The natural iron is in the form of Fe(II), so this ferrous
must therefore oxidize to the ferric form in order to precipitate as the ferric hydroxide,
if, in fact, the iron is to be removed in the ferric form. In Chapter 12, this was done
using the dissolved oxygen that is relatively abundant in natural waters. It must be

TABLE 13.1
Solubility Product Constant of Iron and
Manganese Precipitation Products

Precipitation Product Solubility Product,

K

sp

Fe(OH)

2

3.16(10

−15

)
FeCO

2

2.11(10

−11

)
FeS 8(10

−26

)
Fe(OH)

3

1.1(10

−38

)
Mn(OH)

2

4.5(10

−14

)
MnCO

3

8.8(10

−11

)
MnS 4.3(10

−22

)

TX249_frame_C13.fm Page 595 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero

596

noted, however, that based on the

K

sp

values, iron can also be removed in the ferrous
form as Fe(OH)

2

. For convenience, the reactions are reproduced next.
Ferrous:
(13.1)
(13.2)
(13.3)
(13.4)
Ferric:

(13.5)
(13.6)
(13.7)
(13.8)
(13.9)
Remember that the values of the solubility product constants, and
and all the other equilibrium constants for the complex ions apply only
at 25

°

C.
The presence of the complex ions increase the solubility of the iron species and
therefore increase the concentration of these species in solution. For the ferrous and
the ferric species, they are

sp

FeII



and

sp

FeIII

and were derived in Chapter 12, respec-
tively, as

(13.10)
(13.11)
Fe(OH)
2 s()
 Fe
2+
2OH
−
K
sp,Fe OH()
2
+ 10
−14.5
=
Fe(OH)
2 s()
 FeOH
+
OH
−
K
FeOHc
+ 10
−9.4
=
Fe(OH)
2 s()
OH
−
 Fe OH()

3
−
K
Fe OH()
3
c
+ 10
−5.1
=
Fe(OH)
2
1
4

O
2
1
2

H
2
O → Fe(OH)
3
↓ (conversion from ferrous to ferric)++
Fe(OH)
3 s()
 Fe
3+
3OH
−

K
sp,Fe OH()
3
+ 10
−38
=
Fe(OH)
3 s()
 FeOH
2+
2OH
−
K
FeOHc
+ 10
−26.16
=
Fe(OH)
3 s()
 Fe(OH)
2
+
OH
−
K
Fe OH()
2
c
+ 10
−16.74

=
Fe(OH)
3 s()
OH
−
 Fe OH()
−
4
K
Fe OH()
4
c
+ 10
−5
=
2Fe(OH)
3 s()
 Fe
2
OH()
2
4+
4OH
−
K
Fe
2
OH()
2
c

+ 10
−50.8
=
K
sp,Fe OH()
2
K
sp,Fe OH()
3
,
sp
FeII
[]Fe
2+
[]FeOH
+
[]Fe OH()
3
−
[]++=
K
sp Fe OH()
2
,
γ
H
2
H
+
[]

2
γ
FeII
K
w
2

K
FeOHc
γ
H
H
+
[]
γ
FeOHc
K
w

K
Fe OH()
3
c
K
w
γ
Fe OH()
3
c
γ

H
H
+
[]

++=
sp
FeIII
[]Fe
3+
[]FeOH
2+
[]Fe OH()
2
+
[]Fe OH()
4
−
[]2Fe
2
OH()
2
4+
[]++++=
K
sp Fe OH()
3
,
γ
H

3
H
+
[]
3
γ
FeIII
K
w
3

K
FeOHc
γ
H
2
H
+
[]
2
γ
FeOHc
K
w
2

K
Fe OH()
2
c

γ
H
H
+
[]
γ
Fe OH()
2
c
K
w

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

++ +=
2K
Fe
2

OH()
2
c
γ
H
4
H
+
[]
4
γ
Fe
2
OH()
2
c
K
w
4

+
TX249_frame_C13.fm Page 596 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
Example 13.1 From the respective optimum pH’s of 11.95 and 8.2 for sp
FeII
and sp
FeIII
, calculate the concentrations [sp
FeII
] and [sp

FeIII
], respectively. Assume the
water contains 140 mg/L of dissolved solids.
Solution:
Therefore,
sp
FeII
[]
K
sp,Fe OH()
2
γ
H
2
H
+
[]
2
γ
FeII
K
w
2

K
FeOHc
γ
H
H
+

[]
γ
FeOHc
K
w

K
Fe OH()
3
c
K
w
γ
Fe OH()
3
c
γ
H
H
+
[]

++=
K
sp Fe OH()
2
,
10
14.5–
=

µ
2.5 10
−5
()TDS
γ
10
0.5z
i
2
µ
()
1+1.14
µ
()
–
==
µ
2.5 10
−5
()140()3.5 10
3–
()==
γ
H
γ
FeOHc
γ
Fe OH()
3
c

10
0.5 1()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]
–
0.94== = =
γ
FeII
10
0.5 2()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]
–
0.77K
w
10
14–
()===
K
FeOHc

10
9.4–
K
Fe(OH )
3
c
10
5.1–
==
sp
FeII
[]
10
14.5–
0.94()
2
10
11.95–
[]
2
0.77()10
14–
()
2
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
10
9.4–
( ) 0.94()10
11.95–
[]

0.94()
10
14–
()
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
10
5.1–
()10
14–
()
0.94( ) 0.94()10
11.95–
[]
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−++
=
4.57 10
11–
()4.47 10
8–
()8.01 10
8–
()++ 1.0 10
7–
() gmol/L Ans==
0.0056 mg/L=
sp
FeIII
[]
K
sp,Fe OH()

3
γ
H
3
H
+
[]
3
γ
FeIII
K
w
3

K
FeOHc
γ
H
2
H
+
[]
2
γ
FeOHc
K
w
2

K

Fe OH()
2
c
γ
H
H
+
[]
γ
Fe OH()
2
c
K
w

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

++ +=

2K
Fe
2
OH()
2
c
γ
H
4
H
+
[]
4
γ
Fe
2
OH()
2
c
K
w
4

+
K
sp,Fe OH()
3
10
38–
γ

FeIII
10
0.5 3()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]
–
0.56== =K
FeOHc
10
26.16–
=
γ
FeOHc
10
0.5 2()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]
–
0.77 K
Fe OH()

2
c
10
16.74–
== =
γ
Fe OH()
2
c
10
0.5 1()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]
–
0.94==
TX249_frame_C13.fm Page 597 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
Therefore,
13.3.1 PRACTICAL OPTIMUM pH RANGE FOR THE REMOVAL
OF
FERROUS AND FERRIC
As shown in Chapter 12, at 25°C and at a solids concentration of 140 mg/L, the
optimum pH’s correspond to 11.95 and 8.2 (or around 12 and 8), respectively, for
ferrous and ferric. The respective concentrations for sp
FeII

and sp
FeII
at these condi-
tions as obtained in the previous example are [sp
FeII
] = 0.0056 mg/L and [sp
FeIII
] =
0.0000016 mg/L. A pH range exists, however, at which units used for the removal
of the elements can be operated and effect good results. This range is called the
practical optimum pH range.
Tables 13.2 through 13.5 show the respective concentrations of sp
FeII
and sp
FeIII
at other conditions of pH and total solids. The values for [sp
FeII
] were obtained using
Equation (13.10) and the values for [sp
FeIII
] were obtained using Equation (13.11).
Note that these equations require the values of the activity coefficients of the ions.
The activity coefficients are needed by the equations and, since activity coefficients
are functions of the dissolved solids, dissolved solids are used as parameters in the
tables, in addition to pH.
The previous tables indicate that the total solids (or equivalently, the activities
of the ions) do not have a significant effect on the optimum pH values, which for
sp
FeII
remain at about 12.0 and for sp

FeIII
remain at about 8.0. For practical purposes,
however, the practical optimum pH for sp
FeII
ranges from 11 to 13 and for sp
FeIII
, it
ranges from 5.0 to 13.0. Note that for the range of pH for sp
FeII
, it is assumed the
element is to be removed as Fe(OH)
2
. If it is to be removed as Fe(OH)
3
, the pH of
the solution during its oxidation by dissolved oxygen or any oxidizer need not be
adjusted since the practical optimum pH for the precipitation of ferric hydroxide
varies over a wide range from 5.0 to 13.0 and already includes the range for the
ferrous removal.
K
Fe(OH)
4
c
10
5–
γ
Fe(OH)
4
c
γ

Fe(OH)
2
c
0.94 K
Fe
2
(OH)
2
c
10
50.8–
=== =
γ
Fe
2
(OH)
2
c
10
0.5 4()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]
–
0.36==
sp

FeIII
[]
10
38–
()0.94()
3
10
8.2–
[]
3
0.56()10
14–
()
3
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
10
26.16–
( ) 0.94()
2
10
8.2–
[]
2
0.77()10
14–
()
2
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
10
16.74–

( ) 0.94()10
8.2–
[]
0.94()10
14–
()
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−++
=
10
5–
()10
14–
()
0.94()10
8.2–
[]

2 10
50.8–
()0.94()
4
10
8.2–
[]
4
0.36()10
14–
()
4

++
2.09 10
63–
()
5.6 10
43–
()

2.43 10
43–
()
7.7 10
29–
()

1.08 10
25–
()
9.4 10
15–
()

1.0 10
19–
()
5.93 10
9–
()

3.92 10

84–
()
3.6 10
57–
()

++++=
3.73 10
21–
()3.16 10
15–
()1.15 10
11–
()1.69 10
11–
()1.08 10
27–
()++++=
2.84 10
11–
() gmol/L 0.0000016 mg/L Ans==
TX249_frame_C13.fm Page 598 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
TABLE 13.2
Concentration of sp
FeII
as a
Function of pH at 25°°
°°
C

pH, Dissolved
Solids = 140 mg/L [sp
FeII
], mg/L
0 5.0(10
+18
)
1 2.0(10
+16
)
2 2.0(10
+14
)
3 2.0(10
+12
)
4 2.0(10
+10
)
5 2.0(10
+8
)
6 2.0(10
+6
)
7 2.0(10
+4
)
8 2.2(10
+2

)
9 4.2(10
+0
)
10 2.4(10
−1
)
11 2.3(10
−2
)
12 7.3(10
−3
)
13 5.1(10
−2
)
14 5.0(10
−1
)
15 5.0(10
+0
)
TABLE 13.3
Concentration of sp
FeII
as a
Function of pH at 25°°
°°
C
pH, Dissolved

Solids = 35,000 mg/L [sp
FeII
], mg/L
0 5.0(10
+18
)
1 5.0(10
+16
)
2 5.0(10
+14
)
3 5.0(10
+12
)
4 5.0(10
+10
)
5 5.0(10
+8
)
6 5.0(10
+6
)
7 5.0(10
+4
)
8 5.2(10
+2
)

9 7.2(10
+0
)
10 2.7(10
−1
)
11 2.4(10
−2
)
12 1.5(10
−2
)
13 1.3(10
−1
)
14 1.3(10
+0
)
15 1.3(10
+1
)
TX249_frame_C13.fm Page 599 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
TABLE 13.4
Concentration of sp
FeIII
as a
Function of pH at 25°°
°°
C

pH, Dissolved
Solids = 140 mg/L [sp
FeII
], mg/L
0 1.2(10
+10
)
1 8.6(10
+5
)
2 1.3(10
+3
)
3 5.3(10
+0
)
4 5.5(10
−2
)
5 1.4(10
−3
)
6 1.1(10
−4
)
7 1.0(10
−5
)
8 1.6(10
−6

)
9 6.0(10
−6
)
10 6.0(10
−5
)
11 6.0(10
−4
)
12 6.0(10
−3
)
13 6.0(10
−2
)
14 6.0(10
−1
)
15 6.0(10
+0
)
TABLE 13.5
Concentration of sp
FeIII
as a
Function of pH at 25°°
°°
C
pH, Dissolved

Solids = 35,000 mg/L [sp
FeIII
], mg/L
0 1.2(10
+10
)
1 1.2(10
+7
)
2 1.3(10
+4
)
3 2.3(10
+1
)
4 1.2(10
−1
)
5 1.1(10
−3
)
6 1.1(10
−5
)
7 2.0(10
−7
)
8 9.4(10
−7
)

9 9.4(10
−6
)
10 9.4(10
−5
)
11 9.4(10
−4
)
12 9.4(10
−3
)
13 9.4(10
−2
)
14 9.4(10
−1
)
15 9.4(10
+0
)
TX249_frame_C13.fm Page 600 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
The previous optimum values of pH apply only at 25°C. As indicated in the
formulas, equilibrium constants are being used to compute these values. The values
of the equilibrium constants vary with temperature, so the optimum range at other
temperatures would be different. Equilibrium constants at other temperatures can be
calculated using the Van’t Hoff equation which, however, requires the values of the
standard enthalpy changes. At present, these values are unavailable making values
of optimum pH range at other temperatures impossible to calculate. For this reason,

the pH range found above must be modified to a conservative range. Hence, adopt
the following: for ferrous removal as Fe(OH)
2
, 11.5 ≤ optimum pH ≤ 12.5 and for
ferrous removal as Fe(OH)
3
, 5.5 ≤ optimum pH ≤ 12.5.
13.4 CHEMICAL REACTIONS OF THE MANGANOUS
ION [Mn(II)]
Manganese can be removed as Mn(OH)
2
using a suitable source. Upon intro-
duction of the hydroxide source, however, it is not only this solid that is produced.
Manganese forms complex ions with the hydroxide. The complex equilibrium reac-
tions are as follows (Snoeyink and Jenkins, 1980):
(13.12)
(13.13)
(13.14)
(13.15)
The values of the equilibrium constants given above are at 25°C. The complexes
are Mn(OH)
+
, Mn and Mn . Also note that the ion is a participant
in these reactions. This means that the concentrations of each of these complex ions
are determined by the pH of the solution. In the application of the previous equations
in an actual treatment of water, conditions must be adjusted to allow maximum precip-
itation of the solid represented by Mn(OH)
2(s)
. To allow for this maximum precipita-
tion, the concentrations of the complex ions must be held to the minimum. The pH

corresponding to this condition is the optimum pH. From the previous reactions, the
equivalent mass of Mn
2+
is Mn/2 = 27.45.
13.4.1 DETERMINATION OF THE OPTIMUM pH
Let sp
Mn
represent the collection of species standing in solution containing the Mn(II)
species. Thus,
(13.16)
All the concentrations in the right-hand side of the above equation will now be
expressed in terms of the hydrogen ion concentration. This will result in expressing
OH
−
Mn(OH)
+
 Mn
2+
OH
−
K
MnOHc
+ 10
3.4–
=
Mn(OH)
2 s()
 Mn
2+
2OH

−
K
sp,Mn(OH)
2
+ 4.5 10
14–
()=
Mn(OH)
2
0
 Mn
2+
2OH
−
K
Mn(OH)
2
c
+ 10
6.8–
=
Mn(OH)
3
−
 Mn
2+
3OH
−
K
Mn(OH)

3
c
+ 10
7.8–
=
(OH)
2
0
, (OH)
3
−
OH
−
sp
Mn
[]Mn
2+
[]Mn(OH)
+
[]Mn(OH)
2
0
[]Mn(OH)
3
−
[]+++=
TX249_frame_C13.fm Page 601 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
[sp
Mn

] in terms of the hydrogen ion. Differentiating the resulting equation of [sp
Mn
]
with respect to [H
+
] and equating the result to zero will produce the minimum
concentration of sp
Mn
and, thus, the optimum pH determined.
Using the equations and equilibrium constants of Eqs. (13.12) through (13.15),
along with the ion product of water, we proceed as follows:
(13.17)
(13.18)
(13.19)
(13.20)
γ
Mn
,
γ
MnOHc
,
γ
Mn(OH)2c
, and
γ
Mn(OH)3c
are, respectively, the activity coefficients
of Mn(II) and the complexes Mn(OH)
+
, Mn , and Mn . is

the solubility product constant of the solid Mn(OH)
2(s)
. K
MnOHc
, and
are, respectively, the equilibrium constants of the complexes Mn(OH)
+
,
Mn and Mn .
Equations (13.17) through (13.20) may now be substituted into Equation (13.16)
to produce
(13.21)
Differentiating with respect to [H
+
], equating to zero, rearranging, and changing H
+
to , the concentration of the hydrogen ion at optimum conditions, the following
expression is produced:
(13.22)
The value of [H
opt
] may be solved by trial error.
Mn
2+
[]
Mn
2+
{}
γ
Mn

K
sp Mn, OH()
2
γ
Mn
OH
−
{}
2

K
sp Mn, OH()
2
H
+
{}
2
γ
Mn
K
w
2

K
sp Mn, OH()
2
γ
H
2

H
+
[]
2
γ
Mn
K
w
2

== = =
Mn OH()
+
[]
Mn OH()
+
{}
γ
Mn OH()c

Mn
2+
{}OH
−
{}
γ
Mn OH()c
K
MnOHc

K
sp Mn, OH()
2
γ
H
H
+
[]
γ
Mn OH()c
K
MnOHc
K
w

== =
Mn(OH)
2
0
[]
Mn(OH)
2
0
{}
γ
Mn(OH)
2
c=1

Mn(OH)

2
0
{}
Mn
2+
{}OH
−
{}
2
K
Mn OH()
2
c

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

=== =
Mn(OH)
3
−
[]
Mn(OH)
3
−

{}
γ
Mn OH()
3
c

Mn
2+
{}OH
−
{}
3
γ
Mn OH()
3
c
K
Mn OH()
3
c

K
sp Mn, OH()
2
K
w
γ
Mn OH()
3
c

K
Mn OH()
3
c
γ
H
H
+
[]

== =
(OH)
2
0
OH()
3
−
K
sp,Mn(OH)
2
K
Mn(OH)
2
c
,
K
Mn(OH)
3
c
(OH)

2
0
,OH()
3
−
sp
Mn
[]
K
sp Mn, OH()
2
γ
H
2
H
+
[]
2
γ
Mn
K
w
2

K
sp Mn, OH()
2
γ
H
H

+
[]
γ
Mn OH()c
K
MnOHc
K
w

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

++=
K
sp Mn, OH()
2
K
w
γ
Mn OH()
3
c
K
Mn OH()
3

c
γ
H
H
+
{}

+
H
opt
+
2
γ
H
2
H
opt
+
[]
3
γ
Mn
K
w
2

γ
H
H
opt

+
[]
2
γ
Mn OH()c
K
MnOHc
K
w

+
K
w
γ
Mn OH()
3
c
K
Mn OH()
3
c
γ
H

=
TX249_frame_C13.fm Page 602 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
Example 13.2 Calculate the optimum pH for precipitating Mn(OH)
2
. Assume

the water contains 140 mg/L of dissolved solids.
Solution:
Therefore,
Solving by trial and error, let Y = 2.30(10
28
)[ ]
3
+ 2.51(10
17
)[ ]
2

Example 13.3 From the optimum pH’s of 11.97, calculate the concentrations
sp
Mn
in mg/L. Assume the water contains 140 mg/L of dissolved solids.
Solution:
[ ] 2.30(10
28
)[ ]
3
2.51(10
17
)[ ]
2
Y
10
−11
2.3(10
−5

) 2.51(10
−5
) 4.81(10
−5
)
10
−12
2.3(10
−8
) 2.51(10
−7
) 2.74(10
−7
)
2
γ
H
2
H
opt
+
[]
3
γ
Mn
K
w
2

γ

H
H
opt
+
[]
2
γ
Mn OH()c
K
MnOHc
K
w

+
K
w
γ
Mn OH()
3
c
K
Mn OH()
3
c
γ
H

=
µ
2.5 10

5–
()140()3.5 10
3–
()==
γ
H
γ
Mn OH()c
γ
Mn OH()
3
c
10

0.5 1()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]

–
0.94== = =
γ
Mn
10

0.5 2()

2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]

–
0.77 K
w
10
14–
()===K
MnOHc
10
3.4–
=
K
Mn OH()
3
c
10
7.8–
=
2 0.94()
2
H
opt
+

[]
3
0.77 10
14–
()
2

0.94 H
opt
+
[]
2
0.94()10
3.4–
()10
14–
()

+
10
14–
0.94 10
7.8–
()0.94()

=
2.30 10
28
()H
opt

+
[]
3
2.51 10
17
()H
opt
+
[]
2
+ 7.14 10
7–
()=
H
opt
+
H
opt
+
H
opt
++
++
H
opt
++
++
H
opt
++

++
10
11–
4.81 10
5–
()
y 10
11–
–
10
11–
10
12–
–

7.14 10
7–
()4.81 10
5–
()–
4.81 10
5–
()2.74 10
7–
()–

=
y 7.14 10
7–
() y 1.08 10

12–
() gmol/L H
opt
+
[]==
10
12–
2.74 10
7–
() Therefore, pH log
10
1.08 10
12–
()– 11.97 Ans==
sp
Mn
[]
K
sp Mn, OH()
2
γ
H
2
H
+
[]
2
γ
Mn
K

w
2

K
sp Mn, OH()
2
γ
H
H
+
[]
γ
Mn OH()c
K
MnOHc
K
w

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

++=

K
sp Mn, OH()

2
K
w
γ
Mn OH()
3
c
K
Mn OH()
3
c
γ
H
H
+
[]

+
TX249_frame_C13.fm Page 603 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
Therefore,
13.4.2 PRACTICAL OPTIMUM
p
H RANGE
FOR THE REMOVAL OF MANGANESE
From Example 13.2, at 25°C and at a solids concentration of 140 mg/L, the optimum
pH for the removal of manganese is 11.97. The corresponding concentration for
sp
Mn
is 0.0179 mg/L. As in the case for the removal of ion, there is a practical pH

range at which units used for the removal of manganese can be operated and effect
good results. Tables 13.6 and 13.7 show the respective concentrations of sp
Mn
at
other conditions of pH and total solids. The values for [sp
Mn
] were obtained using
Equation (13.21).
Again, as in the case of the removal of iron, the tables show that total solids (or
equivalently, the activities of the ions) do not have a significant effect on the optimum
pH value for the removal of manganese as Mn(OH)
2
. The optimum pH remains at
around 12.0. For the practical optimum pH, we adopt the following:
manganese removal as Mn(OH)
2
: 11.5 ≤ optimum pH ≤ 12.5,
which is the same as that for ferrous removed as Fe(OH)
2
.
13.5 OXIDATION OF IRON AND MANGANESE
TO REDUCE PRECIPITATION pH
To summarize, the optimum pH’s of precipitation are as follows: ferrous removed
as Fe(OH)
2
, 11.5 ≤ optimum pH ≤ 12.5; ferrous removed as Fe(OH)
3
, 5.5 ≤ optimum
pH ≤ 12.5; and Mn removed as Mn(OH)
2

, 11.5 ≤ optimum pH ≤ 12.5. The reason
for the high pH values for the removal of iron as Fe(OH)
2
and for the removal of
manganese as Mn(OH)
2
is the formation of the complex ions. These ions are FeOH
+
and Fe for the ferrous ion and Mn(OH)
+
, Mn(OH)
0
, and Mn for the
K
sp Mn, OH()
2
4.5 10
14–
()
γ
H
γ
Mn OH()c
γ
Mn OH()
3
c
0.94
γ
Mn

= 0.77=== =
K
MnOHc
10
3.4–
=
K
Mn OH()
2
c
10
6.8–
K
Mn OH()
3
c
10
7.8–
==
sp
Mn
[]
4.5()10
14–
()0.94()
2
10
11.97–
[]
2

0.77()10
14–
()
2

4.5()10
14–
()0.94()10
11.97–
[]
0.94()10
3.4–
()10
14–
()

+=
4.5()10
14–
()
10
6.8–

4.5()10
14–
()10
14–
()
0.94()10
7.8–

()0.94()10
11.97–
[]

++
4.57 10
38–
()
7.7 10
29–
()

4.53 10
26–
()
3.74 10
18–
()

2.84 10
7–
()
4.5 10
28–
()
1.5 10
20–
()

+++=

3.27 10
7–
() geq/L 0.0179 mg/L Ans==
(OH)
3
−
(OH)
3
−
TX249_frame_C13.fm Page 604 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
TABLE 13.6
Concentration of sp
Mn
as a
Function of pH at 25°°
°°
C
pH, Dissolved
Solids = 140 mg/L [sp
Mn
], mg/L
0 1.4(10
+19
)
1 1.4(10
+17
)
2 1.4(10
+15

)
3 1.4(10
+13
)
4 1.4(10
+11
)
5 1.4(10
+9
)
6 1.4(10
+7
)
7 1.4(10
+5
)
8 1.4(10
+3
)
9 1.5(10
+1
)
10 1.8(10
−1
)
11 1.8(10
−2
)
12 9.0(10
−3

)
13 1.7(10
−2
)
14 9.5(10
−1
)
15 9.0(10
−1
)
TABLE 13.7
Concentration of sp
Mn
as a
Function of pH at 25°°
°°
C
pH, Dissolved
Solids = 35,000 mg/L [sp
Mn
], mg/L
0 3.5(10
+19
)
1 3.5(10
+17
)
2 3.5(10
+15
)

3 3.5(10
+13
)
4 3.5(10
+11
)
5 3.5(10
+9
)
6 3.5(10
+7
)
7 3.5(10
+5
)
8 3.5(10
+3
)
9 3.5(10
+1
)
10 3.9(10
−1
)
11 1.5(10
−2
)
12 1.1(10
−2
)

13 3.1(10
−2
)
14 2.3(10
−1
)
15 2.2(10
+0
)
TX249_frame_C13.fm Page 605 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
manganous ion. The values of the equilibrium constants of these complex ions are
such that their concentrations diminish the hydroxide ions needed to precipitate the solid
hydroxides Fe(OH)
2
and Mn(OH)
2
. For these solids to precipitate out, more hydrox-
ide ions must be added, resulting in a high pH. If these complex ions were destroyed,
then the metal ions released would precipitate as the respective hydroxides, prevent-
ing the complex equilibria to occur.
To destroy the complex ions requires the use of oxidants. Chlorine, potassium
permanganate, and ozone are normally used for this purpose. In the case of iron,
the ferrous state is oxidized to the ferric state. This oxidation includes the oxidation
of the ferrous complexes. Because Fe(OH)
3
precipitates at a wider optimum pH
range, the precipitation pH can therefore be reduced to a much lower value, i.e., to
even a pH of 5.5.
Let us tackle the situation of precipitating manganese at a lower pH range. Once

the complex ions have been destroyed, the reactions left would be those for a pre-
cipitation product of a higher oxidation state than 2+. The reactions for the destruction
using chlorine, potassium permanganate, and ozone are, respectively, as follows:
(13.23)
(13.24)
(13.25)
The previous reactions show that manganese is oxidized from Mn(II) to Mn(IV).
This oxidation poses the possibility of Mn(IV) forming a complex with the hydrox-
ides. The review of the literature, however, did not uncover any evidence for this to
be so. It did reveal that Mn(IV) forms a complex with fluorine and potassium
(Holtzclaw and Robinson, 1988). This complex is K
2
[MnF
6
]. The review also did
not reveal any solubility product constant for MnO
2
. It could very well be that it
does not have any. If, in fact, MnO
2
does not have any solubility product constant,
then the previous reactions can be construed to be possible at any pH. Until a study
is done to show the complex formation of Mn(IV) and the accompanying solubility
constant and to be compatible with the removal of ferrous as Fe(OH)
3
, we will adopt
the practical pH for removal of manganese through its oxidation by the above
reactions as 5.5 ≤ optimum pH ≤ 12.5. This range is, however, arbitrary and applies
when the complexes have been destroyed. For more accurate results, a pilot plant
investigation for a given raw water should be undertaken.

13.6 UNIT OPERATIONS FOR IRON AND
MANGANESE REMOVAL
The unit operations involved can be divided into two general categories: removal at
the pH range of 11.5 to 12.5 and removal at the pH range of 5.5 to 12.5. The former
range is called the high pH range and the latter is called the low pH range. The
latter is arbitrarily categorized as low, because it is possible to adjust the operation
at a low pH value. In these ranges, both iron and manganese can be removed at the
Mn
2+
Cl
2
2H
2
O → MnO
2 s()
↓ 2Cl
−
4H
+
++++
3Mn
2+
2KMnO
4
2H
2
O → 5MnO
2 s()
↓ 2K
+

4H
+
++++
3Mn
2+
O
3
3H
2
O → 3MnO
2
s() ↓ 6H
+
+++
TX249_frame_C13.fm Page 606 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
same time and at the same unit. The differences and similarities of these categories
of operations will be explored.
13.6.1 HIGH pH RANGE
Under this category, iron is removed as Fe(OH)
2
and manganese is removed as
Mn(OH)
2
.

The unit operations will then involve a unit (tank) for containing the
addition of the hydroxide source (lime) to raise the pH and for mixing the chemicals
and allowing completion of the precipitation of Fe(OH)
2

and Mn(OH)
2
. The precip-
itates formed will still be in suspension; thus, it is necessary for them to flocculate
by adding a flocculation tank. After the flocculation tank will be a settling tank. The
settling tanks removes the flocculated solids. A filter then follows to filter out any
solids that were not removed by the settling tank.
The sequence of unit operations discussed above may not all be needed in a
given application. What will actually be used depends upon the concentrations of
iron and manganese in the raw water. If the concentrations are small such as less
than 10 mg/L, the resulting floc concentration will be small; the chemically treated
water, in this instance, may simply be introduced directly into the filter, without
using any flocculation or settling basin. In all cases, however, the mixing chamber
for the reactions to take place and the filter are always necessary. The unit operations
of mixing, flocculation, settling, and filtration were already discussed in the earlier
chapters of this book.
As can be deduced from the addition of lime, the unit operations above are the
same as those that would be applied in water softening. Thus, for reasons of economics,
iron and manganese removal are normally incorporated into water softening and all
the unit operations involved are identical. In fact, iron and manganese are hardness ions.
13.6.2 LOW pH RANGE
Under this category, iron is removed as Fe(OH)
3
and manganese is removed as
MnO
2
, which involves oxidation of the metal ions to their higher oxidation states.
The unit operations under the previous category are applicable under the present
category, only that under the present scheme, oxidation of the metals is involved.
The unit process involves one of Eqs. (13.23) through (13.25) for the case of

manganese and similar equations for iron. The equation designed into the unit
operations depends upon the unit process contemplated. The unit process part,
however, will only be at the mixing tank where the reactions should occur.
As in the previous category, the unit operations of flocculation and settling may
not be necessary. Again, in all cases, the mixing chamber for the reaction to take
place and filtration is always necessary. The reaction does not immediately produce
particles capable of being filtered, therefore, sufficient detention time should be
provided in the reaction chamber to allow the particles to grow into filtrable sizes.
This normally ranges from 20 to 30 minutes. In any case, a pilot plant may be
necessary to determine the exact detention time.
When oxidizing iron and manganese using dissolved oxygen, the process is
usually carried out under catalytic reaction on some contact surfaces. To accomplish
TX249_frame_C13.fm Page 607 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
the reaction, the water is generally made to trickle over small rock surfaces such as
limestone, coke, or pyrolusite (MnO
2
). Pyrolusite possesses high catalytic power,
springing from the iron and manganese oxides precipitated from the raw water that
coat over and around the rocks. These coatings act as catalysts between the reaction
of the ferrous and manganous ions in the water and the oxygen from the air as they
come into intimate contact over the contact surfaces of the rocks.
The contact medium sizes vary from 3.5 to 5 cm. The accumulated flocs produced
from the deposition of the hydroxides or oxides are periodically flushed out by rapid
drainage. This is done by filling the tank containing the bed of rocks to capacity
and quickly releasing the water.
13.7 CHEMICAL REQUIREMENTS
The chemical requirements are those for the hydroxide source, chlorine, permanga-
nate, ozone, and oxygen. The discussion will be subdivided into requirements in the
ferrous reactions and into requirements in the manganous reactions. The treatment

on chemical requirements, in effect, is reduced to the determination of the equivalent
masses of the pertinent chemicals.
13.7.1 REQUIREMENTS IN THE FERROUS REACTIONS
Under these requirements are the reactions at the high and low pH ranges. At the
high range, the hydroxide reaction is
(13.26)
The equivalent mass of Fe
2+
in the previous reaction is Fe/2 = 27.9 and that of the
hydroxide ion is 2OH
−
/2 = 17.
The possible hydroxide sources are caustic and lime. To make these sources jointly
equivalent to Equation (13.26), they must all be made equivalent to 2OH
−
. Thus,
(13.27)
(13.28)
From the previous equivalence, the equivalent of mass of NaOH is 2NaOH/2 = 40
and the equivalent mass of CaO is CaO/2 = 28.05.
Chlorine, permanganate, and ozone, will only be used at the low pH range. There
is no need for them to be used at the high pH range, because under these conditions,
the ferrous ion will be forced to precipitate as the ferrous hydroxide. It is true that
after precipitating, chlorine, permanganate, or ozone could then be used to oxidize
ferrous hydroxide to the ferric state, but no practical reason exists for designing the
unit process this way. Thus, the unit process design will be to oxidize the ferrous
Fe
2+
2OH
−

→ Fe OH()
2 s()
↓+
2NaOH 2OH
−
⇒
Ca OH()
2
2OH
−
⇒
TX249_frame_C13.fm Page 608 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
ion at the low pH range, converting it directly to the ferric hydroxide solid. Including
dissolved oxygen, the reactions are, respectively, as follows:
(13.29)
(13.30)
(13.31)
(13.32)
The previous reactions are oxidation-reduction reactions, so it is appropriate to
use as the reference species the number of electron moles involved, although the
choice of the reference species is completely arbitrary. One atom of Fe
2+
is being
oxidized to Fe
3+
, therefore, one electron mole is involved (thus, the number of
reference species = 1) and the equivalent masses are ferrous = Fe/1 = 55.8, chlorine =
Cl
2

/1 = 35.5, potassium permanganate = KMnO
4
/1 = 52.67, ozone = O
3
/1 = 8.0,
and oxygen O
2
/1 = 8.0.
Occasionally, the hypochlorites are used as the chlorinating–oxidizing agent. To
determine the equivalent masses of the hypochlorites, they must be made to interact
with electrons. On the basis of one electron mole,
(13.33)
From this reaction, the equivalent mass of sodium hypochlorite is NaOCl/1 = 37.25
and that of calcium hypochlorite, it is Ca(OCl)
2
/1 = 35.78.
13.7.2 REQUIREMENTS IN THE MANGANOUS REACTIONS
Under these requirements are also the reactions at the high and low pH ranges. At
the high range, the hydroxide reaction is
The equivalent mass of Mn
2+
in the previous reaction is Mn/2 = 27.45 and that of
the hydroxide ion is 2OH
−
/2 = 17. If the reactions for the caustic and lime are written,
the equivalent masses will be found to be the same as those in the case of ferrous,
which are 40 and 28.05, respectively.
For the same reasons as in the case of ferrous, chlorine, permanganate, and
ozone will only be used at the low pH range. The unit process design is to oxidize
Fe

2+
1
2

Cl
2
3H
2
OFeOH()
3 s()
↓→+ Cl
−
3H
+
+++
Fe
2+
1
3

KMnO
4
7
3

H
2
OFeOH()
3 s()
↓→+

1
3

MnO
2
↓
1
3

K
+
5
3

H
+
++++
Fe
2+
1
6

O
3
15
6

H
2
OFeOH()

3 s()
↓→+ 2H
+
++
Fe
2+
5
2

H
2
O
1
4

O
2
Fe OH()
3 s()
↓→+ 2H
+
++
1
2

1
3

1
6

1
4

1
2

OCl
−
e
−
H
+
1
2

Cl
−
→
1
2

H
2
O++ +
1
2

1
4

Mn
2+
2OH
−
Mn OH()
2 s()
↓→+
TX249_frame_C13.fm Page 609 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
Mn
2+
at the low pH range converting it directly to MnO
2
as shown in Eqs. (13.23)
through (13.25). Converting these equations in terms of one atom of Mn
2+
, along
with the reaction of oxygen, the following equations are produced:
(13.34)
(13.35)
(13.36)
(13.37)
In each of the previous reactions, the manganese atom is oxidized from 2+ to
4+, involving 2 electron moles. Thus, the equivalent masses are manganese Mn/2 =
27.45, chlorine Cl
2
/2 = 35.5, potassium permanganate = KMnO
4
/2 = 52.67, ozone

= O
3
/2 = 8, and oxygen = O
2
/2 = 8. The equivalent masses of sodium hypochlorite
and calcium hypochlorite are the same as those in the case of the ferrous, which are
37.25 and 35.78, respectively.
Tables 13.8 and 13.9 summarize the equivalent masses.
TABLE 13.8
Summary of Equivalent Masses, Ferrous Reactions
Fe
2+
OH
−
CaO NaOH Cl
2
NaClO Ca(ClO)
2
KMnO
4
O
3
O
2
High pH
range
27.9 17.0 28.05 40.0 — — — — — —
Low pH
range
55.8 — — — 35.5 37.25 35.78 52.67 8.0 8.0

TABLE 13.9
Summary of Equivalent Masses, Manganous Reactions
Mn
2+
OH
−
CaO NaOH Cl
2
NaClO Ca(ClO)
2
KMnO
4
O
3
O
2
High pH
range
27.45 17.0 28.05 40.0 — — — — — —
Low pH
range
27.45 — — — 35.5 37.25 35.78 52.67 8.0 8.0
Mn
2+
Cl
2
2H
2
O MnO
2 s()

↓→ 2Cl
−
4H
+
++ + +
Mn
2+
2
3

KMnO
4
2
3

H
2
O
5
3

MnO
2 s()
↓
2
3

K
+
4

3

H
+
++→++
Mn
2+
1
3

O
3
H
2
O MnO
2 s()
↓ 2H
+
+→++
Mn
2+
1
2

O
2
H
2
O MnO
2 s()

↓ 2H
+
+→++
2
3

1
3

1
2

TX249_frame_C13.fm Page 610 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
These tables show that, except for Fe
2+
and Mn
2+
, the equivalent masses are the
same in both the ferrous and manganese reactions. Now, let
M
CaOFe
= kilograms of lime used for the ferrous reaction
M
NaOHFe
= kilograms of caustic used for the ferrous reaction
= kilograms of chlorine used for the ferrous reaction
M
NaClOFe
= kilograms of sodium hypochlorite used in the ferrous reaction

= kilograms of calcium hypochlorite used in the ferrous reaction
= kilograms of potassium permanganate used in the ferrous reaction
= kilograms of ozone used in the ferrous reaction
= kilograms of dissolved oxygen used in the ferrous reaction
M
CaOMn
= kilograms of lime used for the manganous reaction
M
NaOHMn
= kilograms of caustic used for the manganous reaction
= kilograms of chlorine used for the manganous reaction
M
NaClOMn
= kilograms of sodium hypochlorite used in the manganous reaction
= kilograms of calcium hypochlorite used in the manganous reaction
= kilograms of potassium permanganate used in the manganous
reaction
= kilograms of ozone used in the manganous reaction
= kilograms of dissolved oxygen used in the manganous reaction
= mg/L of ferrous to be removed
= mg/L of manganous removed
= cubic meters of water treated
Therefore,
(13.38)
(13.39)
(13.40)
(13.41)
(13.42)
(13.43)
(13.44)

M
Cl
2
Fe
M
Ca ClO()
2
Fe
M
KMnO
4
Fe
M
O
3
Fe
M
O
2
Fe
M
Cl
2
Mn
M
Ca ClO()
2
Mn
M
KMnO

4
Mn
M
O
3
Mn
M
O
2
Mn
∆ Fe[]
mg
∆ Mn[]
mg
V
M
CaOFe
28.05∆ Fe[]
mg
V
1000 27.9()

0.0010∆ Fe[]
mg
V
==
M
NaOHFe
40∆ Fe[]
mg

V
1000 27.9()

0.0014∆ Fe[]
mg
V
==
M
Cl
2
Fe
35.5∆ Fe[]
mg
V
1000 55.8()

0.00064∆ Fe[]
mg
V
==
M
NaClOFe
37.25∆ Fe[]
mg
V
1000 55.8()

0.00067∆ Fe[]
mg
V

==
M
Ca ClO()
2
Fe
35.78∆ Fe[]
mg
V
1000 55.8()

0.00064∆ Fe[]
mg
V
==
M
KMnO
4
Fe
52.67∆ Fe[]
mg
V
1000 55.8()

0.00094∆ Fe[]
mg
V
==
M
O
3

Fe
8∆ Fe[]
mg
V
1000 55.8()

0.00014∆ Fe[]
mg
V
==
TX249_frame_C13.fm Page 611 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
(13.45)
(13.46)
(13.47)
(13.48)
(13.49)
(13.50)
(13.51)
(13.52)
(13.53)
Example 13.4 A raw water contains 2.5 mg/L Mn. Calculate the kilograms
of lime per cubic meter needed to meet the limit concentration of 0.05 mg/L, if
removal is to be done at the high pH range.
Solution:
Example 13.5 A raw water contains 40 mg/L iron. Calculate the kilograms of
lime per cubic meter needed to meet the limit concentration of 0.3 mg/L, if removal
is to be done at the high pH range.
Solution:
M

O
2
Fe
8∆ Fe[]
mg
V
1000 55.8()

0.00014∆ Fe[]
mg
V
==
M
CaOMn
28.05∆ Mn[]
mg
V
1000 27.45()

0.0010∆ Mn[]
mg
V
==
M
NaOHMn
40.0∆ Mn[]
mg
V
1000 27.45()

0.0015∆ Mn[]
mg
V
==
M
Cl
2
Mn
35.5∆ Mn[]
mg
V
1000 27.45()

0.0013∆ Mn[]
mg
V
==
M
NaClOMn
37.25∆ Mn[]
mg
V
1000 27.45()

0.0014∆ Mn[]
mg
V
==
M
Ca ClO()

2
Mn
35.78∆ Mn[]
mg
V
1000 27.45()

0.0013∆ Mn[]
mg
V
==
M
KMnO
4
Mn
52.67∆ Mn[]
mg
V
1000 27.45()

0.0019∆ Mn[]
mg
V
==
M
O
3
Mn
8.0∆ Mn[]
mg

V
1000 27.45()

0.00029∆ Mn[]
mg
V
==
M
O
2
Mn
8.0∆ Mn[]
mg
V
1000 27.45()

0.00029∆ Mn[]
mg
V
==
M
CaOMn
0.0010∆ Mn[]
mg
V
0.0010 2.5 0.05–()1() 2.45 10
3–
() kg/m
3
Ans== =

M
CaOFe
0.0010∆ Fe[]
mg
V
0.0010 40 0.3–()1() 0.040 kg/m
3
Ans== =
TX249_frame_C13.fm Page 612 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
13.8 ALKALINITY EXPRESSED IN OH
−−
−−
AND ACIDITY
EXPRESSED IN H
++
++
Knowledge of the equivalent mass is needed in the succeeding discussions on the
chemical requirements for pH adjustments, which entails adding acids or bases. As
acids are added to water and wastewater to lower the pH, they need to react with existing
alkalinities first before the pH changes. For the purpose of determining the equivalent
masses of the acids as a result of the reaction due to this addition, the constituent species
of the alkalinity need to be expressed in terms of a single unifying species. Once this
unifying species has been established, the process of determining the equivalent mass
becomes simpler. This particular species happens to be the hydroxide ion.
Similar statements also hold with respect to bases added to raise the pH. They
need to react first with existing acidities before the pH can increase, and their
equivalent masses also need to be determined by referring to a single unifying
species. This single unifying species is the hydrogen ion. We will discuss the
hydroxide ion first.

In the carbonate system, the total alkalinity species has been identified as com-
posed of the constituent species carbonate ion, bicarbonate ion, and hydroxide ion.
To express all these constituent species in terms of a single parameter, total alkalinity,
they need to be expressed in terms of the number of equivalents. Thus, if the
concentrations are in terms of equivalents per liter, they can all be added to sum up
to the total alkalinity, of course, also in terms of equivalents per liter.
All the concentrations are expressed in terms of equivalents, so any one species
may represent the total alkalinity. For example, the total alkalinity may be expressed
in terms of equivalents of the carbonate ion, or in terms of equivalents of the bicarbonate
ion, or in terms of the equivalents of the hydroxide ion. The number of equivalents of
the hydroxide ion is equal to its number of moles, so it is convenient to expresses total
alkalinity in terms of the equivalent hydroxide ion, and the reaction of total alkalinity
may then be simply represented by the reaction of the hydroxide ion OH
−
.
Without going into a detailed discussion, the total acidity may also be expressed
in terms of the equivalent hydrogen ion. Thus, the reaction of total acidity may be
represented by the H
+
ion.
13.9 CHEMICAL REQUIREMENTS FOR pH
ADJUSTMENTS
The formulas for the kilograms of carbon dioxide, , needed to lower the pH
and the kilograms of lime, , needed to raise the pH were derived in Chapter 12.
For convenience, they are reproduced below:
(13.54)
(13.55)
M
CO
2

pH
M
CaOp
H
M
CO
2
pH
22 A
cadd
[]
geq
A
cur
[]
geq
10
pH–
to
10
pH–
cur
–
φ
a

+=




V
=
M
CaOpH
28.05 A
add
[]
geq
V
28.05 A
ccur
[]
geq
10
pH–
cur
10
pH–
to
–
φ
b

+


=
V
=
TX249_frame_C13.fm Page 613 Friday, June 14, 2002 2:32 PM

© 2003 by A. P. Sincero and G. A. Sincero
where,
[A
cadd
] = gram equivalents of acidity to be added
[A
add
] = gram equivalents of base to be added
[A
cur
]
geq
= gram equivalents of current alkalinity
[A
ccur
]
geq
= gram equivalents of current acicity
pH
to
= the destination pH
pH
cur
= current pH
= cubic meters of water treated
= fractional dissociation of the hydrogen ion from the acid provided
= fractional dissociation of the hydroxide ion from the base provided
Let us derive the formulas for calculating the quantities of sulfuric acid, hydrochlo-
ric acid, and nitric acid and the formulas for calculating the quantities of caustic soda
and soda ash that may be needed to lower and to raise the pH, respectively. To find the

equivalent masses of the acids, they must be reacted with the hydroxyl ion. Reaction
with this ion is necessary, since total alkalinity may be represented by the hydroxyl
ion. Remember that the acids must first consume all the existing alkalinity represented
in the overall by the before they can lower the pH. Thus, proceed as follows:
(13.56)
(13.57)
(13.58)
From these equations, the equivalent masses are sulfuric acid = Η
2
SO
4
/2 = 49.05,
hydrochloric acid = ΗCl/1 = 36.5, and nitric acid = ΗΝΟ
3
/1 = 63.01.
To find the equivalent masses of the caustic soda and soda ash, they must be
reacted with an acid, which may be represented by the hydrogen ion. Again, remem-
ber that the bases specified previously must first consume all the existing acidity of
the water represented in the overall by H
+
before they can raise the pH. Thus,
(13.59)
(13.60)
From the equations, the equivalent masses are caustic soda = NaOH/1 = 40 and soda
ash = Na
2
CO
3
/2 = 2 × 23 + 12 + 48/2 = 53.
Let M

HClpH
,

and be the kilograms of sulfuric acid, hydro-
chloric acid, or nitric acid used to lower the pH from the current pH to pH
to
. Gleaning
from Equation (13.54) and the respective equivalent masses of H
2
SO
4
, HCl, and
HNO
3
and the cubic meters, , of water treated,
(13.61)
V
φ
a
φ
b
OH
−
H
2
SO
4
2OH
−
2Η

2
O→+ SO
4
2−
+
HCl OH
−
Η
2
OCl
−
+→+
HNO
3
OH
−
Η
2
ONO
3
−
+→+
NaOH H
+
→ H
2
ONa
+
++
Na

2
CO
3
2H
+
→ H
2
CO
3
2Na
+
++
M
H
2
SO
4
pH
, M
HNO
3
pH
V
M
H
2
SO
4
pH
49.05 A

cadd
[]
geq
A
cur
[]
geq
10
pH–
to
10
pH–
cur
–
φ
a

+=



V
=
TX249_frame_C13.fm Page 614 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
(13.62)
(13.63)
And, also, gleaning from Equation (13.55) and the respective equivalent masses of
NaOH and Na
2

CO
3
and the cubic meters of water treated, ,
(13.64)
(13.65)
where
Μ
NaOHpH
and are the kilograms of caustic soda or soda ash, respec-
tively, used to raise the pH from the current pH to pH
to
.
They are strong acids, therefore, the
φ
a
’s for H
2
SO
4
, HCl, and HNO
3
are unity.
The hydrogen ion resulting from the second ionization of sulfuric is very small so
it can be neglected. Also, because NaOH is a strong base, its
φ
b
is equal to unity.
The
φ
b

for Na
2
CO
3
is not as straightforward, and we need to calculate it. Sodium
carbonate ionizes completely, as follows:
(13.66)
The carbonate ion then proceeds to react with water to produce the hydroxide ion,
as follows (Holtzclaw and Robinson, 1988):
(13.67)
Now, proceed to calculate the fractional dissociation of the hydroxide ion from
sodium carbonate. Begin by assuming a concentration for the carbonate of 0.1 M.
This will then produce, also, 0.1 M of the carbonate ion. By its subsequent reaction
with water, however, its concentration at equilibrium will be smaller. Let x be the
concentrations of and at equilibrium. Then the concentration of
is 0.1 − x (at equilibrium). Thus,
M
HClpH
36.5 A
cadd
[]
geq
A
cur
[]
geq
10
pH–
to
10

pH–
cur
–
φ
a

+=



V
=
M
HNO
3
pH
63.01 A
cadd
[]
geq
A
cur
[]
geq
10
pH–
to
10
pH–
cur

–
φ
a

+=



V
=
V
M
NaOHpH
40.0 A
ccur
[]
geq
10
pH–
cur
10
pH–
to
–
φ
b

+




V
=
M
Na
2
CO
3
pH
53.0 A
ccur
[]
geq
10
pH–
cur
10
pH–
to
–
φ
b

+



V
=
M

Na
2
CO
3
pH
Na
2
CO
3
2Na
+
→ CO
3
2−
+
CO
3
2−
HOH  HCO
3
−
OH
−
K
b
++1.4 10
4–
()=
HCO
3

−
OH
−
CO
3
2−
K
b
1.4 10
4–
()
x
2
0.1 x–

==
x 0.00367=
TX249_frame_C13.fm Page 615 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
Therefore,
Note that, in the previous calculation, the activity coefficients have been ignored.
Lastly, find the
φ
b
for lime. The chemical reaction for lime after it has been
slaked is
(13.68)
Begin by assuming a concentration 0.1 M. Let x be the concentrations of Ca
2+
at

equilibrium; the corresponding concentration of OH
−
will then be 2x at equilibrium.
Thus,
Therefore,
Table 13.10 shows some other values of
φ
a
and
φ
b
.
TABLE 13.10
φφ
φφ
a
or
φφ
φφ
b
for Acids and Bases Used in
Water and Wastewater Treatment
Acid or
Base
φφ
φφ
a
or
φφ
φφ

b
0.1 M 0.5 M 1.0 M
H
2
SO
4
1.0 1.0 1.0
HCl 1.0 1.0 1.0
HNO
3
1.0 1.0 1.0
H
2
CO
3
0.00422 0.00094 0.000668
NaOH 1.0 1.0 1.0
Na
2
CO
3
0.0367 0.0166 0.0118
Ca(OH)
2
0.125 0.025 0.0125
φ
b Na
2
CO
3

,
0.00367
0.1

0.0367==
Ca OH()
2
 Ca
2+
2OH
−
+ K
sp,Ca OH()
2
7.9 10
6–
()=
K
sp,Ca OH()
2
7.9 10
6–
()x 2x()
2
4x
3
===
x 0.0125=
φ
b Ca OH()

2
,
0.0125
0.1

0.125==
x
1.4 10
4–
() 1.4 10
4–
()[]
2
4 1.0()1.4 10
4–
()[]++–
2

0.0083==
TX249_frame_C13.fm Page 616 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero
Example 13.6 A raw water containing 3 mg/L of manganese has a pH 4.0.
To remove the manganese, the pH needs to be raised to 6.0. The current acidity is
30 mg/L as CaCO
3
[Ca
2+
] = 0.7 mgmol/L. Assume the temperature is 25°C and
φ
b

of 0.0367. Calculate the amount of soda ash needed.
Solution:
Therefore,
13.10 SLUDGE PRODUCTION
For convenience, the chemical reactions responsible for the production of sludge
solids are summarized next. These equations have been derived at various points.
(13.69)
(13.70)
(13.71)
(13.72)
(13.73)
(13.74)
(13.75)
(13.76)
M
Na
2
CO
3
pH
53.0 A
ccur
[]
geq
10
pH–
cur
10
pH–
to

–
φ
b

+



V
=
10
pH
to
–
10
6–
gmols/L; 10
pH
cur
–
10
4–
gmols/L;
V
1.0 m
3
===
M
Na
2

CO
3
pH
53.0
30
1000 50()

10
4–
10
6–
–
0.0367

+



1() 0.175 kg/m
3
Ans==
Fe
2+
2OH
−
Fe OH()
2 s()
↓→+
Fe
2+

1
2

Cl
2
3H
2
OFe→ OH()
3 s()
↓ Cl
−
3H
+
++ ++
Fe
2+
1
3

KMnO
4
7
3

H
2
OFe→ OH()
3 s()
↓
1

3

MnO
2
↓
1
3

K
+
5
3

H
+
++ +++
Fe
2+
1
6

O
3
15
6

H
2
OFe→ OH()
3 s()

↓ 2H
+
++ +
Fe
2+
5
2

H
2
O
1
4

O
2
Fe→ OH()
3 s()
↓ 2H
+
++ +
Mn
2+
2OH
−
Mn OH()
2 s()
↓→+
Mn
2+

Cl
2
2H
2
OMn→ O
2 s()
↓ 2Cl
−
4H
+
+++ +
Mn
2+
2
3

KMnO
4
2
3

H
2
O
5
3

MnO
2 s()
↓

2
3

K
+
4
3

H
+
++→++
TX249_frame_C13.fm Page 617 Friday, June 14, 2002 2:32 PM
© 2003 by A. P. Sincero and G. A. Sincero