PHYSICAL - CHEMICAL TREATMENT OF WATER AND WASTEWATER - CHAPTER 14 pps

Removal of Phosphorus
by Chemical Precipitation

Phosphorus is a very important element that has attracted much attention because of
its ability to cause eutrophication in bodies of water. For example, tributaries from as
far away as the farmlands of New York feed the Chesapeake Bay in Maryland and
Virginia. Because of the use of phosphorus in fertilizers for these farms, the bay receives
an extraordinarily large amount of phosphorus input that has triggered excessive growths
of algae in the water body. Presently, large portions of the bay are eutrophied. Without
any doubt, all coves and little estuaries that are tributaries to this bay are also
eutrophied. Thus, it is important that discharges of phosphorus be controlled in order
to avert an environmental catastrophe. In fact, the eutrophication of the Chesapeake
Bay and the clogging of the Potomac River by blue greens are two of the reasons
for the passage of the Federal Water Pollution Control Act Amendments of 1972.
This chapter discusses the removal of phosphorus by chemical precipitation. It
first discusses the natural occurrence of phosphorus, followed by a discussion on the
modes of removal of the element. The chemical reactions of removal, unit operations
of removal, chemical requirements, optimum pH range of operation, and sludge pro-
duction are all discussed. The chemical precipitation method employed uses alum,
lime, and the ferric salts, FeCl

3

and Fe

2

(SO

4

)

3

.

14.1 NATURAL OCCURRENCE OF PHOSPHORUS

The element phosphorus is a nonmetal. It belongs to

Group VA

in the Periodic
Table in the third period. Its electronic configuration is [Ne]3

s

2

3

p

3

. [Ne] means
that the neon configuration is filled. The valence configuration represented by the
3, the

M

shell, shows five electrons in the orbitals: 2 electrons in the

s

orbitals and
3 electrons in the

p

orbitals. This means that phosphorus can have a maximum
oxidation state of

+

5. The commonly observed oxidation states are 3

−

, 3

+

, and 5

+

.
Phosphorus is too active a nonmetal to be found free in nature. Our interest

in its occurrence is the form that makes it as fertilizer to plants. As a fertilizer, it
must be in the form of orthophosphate. Phosphorus occurs in three phosphate
forms: orthophosphate, condensed phosphates (or polyphosphates), and organic
phosphates. Phosphoric acid, being triprotic, forms three series of salts: dihydrogen
phosphates containing the ion, hydrogen phosphate containing the
ion, and the phosphates containing the ion. These three ions collectively are
called orthophosphates. As orthophosphates, the phosphorus atom exists in its
highest possible oxidation state of 5

+

. As mentioned, phosphorus can cause eutroph-
ication in receiving streams. Thus, concentrations of orthophosphates should be
controlled through removal before discharging the wastewater into receiving bodies
14
H
2
PO
4
−
HPO
4
2−
PO
4
3−

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© 2003 by A. P. Sincero and G. A. Sincero

628

of water. The orthophosphates of concern in wastewater are sodium phosphate
(Na

3

PO

4

), sodium hydrogen phosphate (Na

2

HPO

4

), sodium dihydrogen phosphate
(NaH

2

PO

4

), and ammonium hydrogen phosphate [(NH

4

)

2

HPO

4

]. They cause the
problems associated with algal blooms.
When phosphoric acid is heated, it decomposes, losing molecules of water forming
the P–O–P bonds.The process of losing water is called

condensation

, thus the term
condensed phosphates and, since they have more than one phosphate group in the
molecule, they are also called

polyphosphates

. Among the acids formed from the
condensation of phosphoric acid are

dipolyphosphoric

acid

or

pyrophosphoric acid

(H

4

P

2

O

7

, oxidation state

=

5

+

),

tripolyphosphoric acid

(H

5

P

3

O

10

, oxidation state

=

5

+

),
and

metaphosphoric acid

[(HPO

3

)

n

, oxidation state

=

5

+

]. Condensed phosphates
undergo hydrolysis in aqueous solutions and transform into the orthophosphates. Thus,
they must also be controlled. Condensed phosphates of concern in wastewater are
sodium hexametaphosphate (NaPO

3

)

6

, sodium dipolyphosphate (Na

4

P

2

O

7

), and sodium
tripolyphosphate (Na

5

P

3

O

10

).
When organic compounds containing phosphorus are attacked by microorganisms,
they also undergo hydrolysis into the orthophosphate forms. Thus, as with all the other
phosphorus species, they have to be controlled before wastewaters are discharged.
Figure 14.1 shows the structural formulas of the various forms of phosphates. Note
that the oxygen not bonded to hydrogen in orthophosphoric acid, trimetaphosphoric

FIGURE 14.1

Structural formulas of various forms of phosphates.
Orthophosphoric acid
Trimetaphosphoric acid
Tripolyphosphoric acid

Organic backbone
Organic phosphate
Dipolyphosphoric acid or
pyrophosphoric acid

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629

acid, dipolyphosphoric acid, and tripolyphosphoric acid has a single bond with the
central phosphorus atom. Oxygen has six electrons in its valence shell, therefore,
this indicates that phosphorus has shared two of its own valence electrons to oxygen
without oxygen sharing any of its electrons to phosphorus. The acceptance by oxygen
of these phosphorus electrons, completes its valence orbitals to the required eight
electrons for stability.
All of the hydrogen atoms are ionizable. This means that the largest negative
charge for the complete ionization of orthophosphoric acid is 3

−

; that of trimet-
aphosphoric acid is also 3

−

. Dipolyphosphoric acid will have 4

−

as the largest
negative charge and tripolyphosphoric acid will have 5

−

as the largest negative
charge. The charges in organic phosphates depend upon the organic backbone the
phosphates are attached to and how many of the phosphate radicals are being
attached.
The phosphorus concentration in domestic wastewaters varies from 3 to 15 mg/L
and that in lake surface waters, from 0.01 to 0.04 mg/L all measured as P. These
values include all the forms of phosphorus.

14.2 MODES OF PHOSPHORUS REMOVAL

Again, as in previous chapters, the best place to investigate for determining the
modes of removal is the table of solubility products constants as shown in Table 14.1.
A precipitation product that has the lowest

K

sp

means that the substance is the most
insoluble. As shown in the table, the phosphate ion can be precipitated using a
calcium precipitant producing either Ca

5

(PO

4

)

3

(OH)

(

s

)

or Ca

3

(PO

4

)

2

. Of these two
precipitates, Ca

5

(PO

4

)

3

(OH)

(

s

)

has the smaller

K

sp

of 10

−

55.9

; thus, it will be used as
the criterion for the precipitation of phosphates. Ca

5

(PO

3

)

3

(OH)

(

s

)

is also called

calcium hydroxy apatite

.
As shown in the table, the other mode of precipitation possible is through
precipitating the phosphate ion as AlPO

4(

s

)

and FePO

4

. The precipitant normally used
in this instances are alum and the ferric salts (ferric chloride and ferric sulfate),
respectively.

TABLE 14.1
Solubility Product Constants for Phosphate
Precipitation

Precipitation Product Solubility Product,

K

sp

at 25

°

C

Ca

5

(PO

3

)

3

(OH)

(

s

)

(10

−

55.9

)
Ca

3

(PO

4

)

2

(10

−

25

)
AlPO

4(

s

)

(10

−

21

)

FePO

4

(10

−

21.9

)

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630

14.3 CHEMICAL REACTION OF THE PHOSPHATE ION
WITH ALUM

To precipitate the phosphate ion as aluminum phosphate, alum is normally used.
The chemical reaction is shown next:
(14.1)
As shown in these reactions, the phosphorus must be in the phosphate form.
The reaction occurs in water, so the phosphate ion originates a series of equilibrium
orthophosphate reactions with the hydrogen ion. This series is shown as follows
(Snoeyink and Jenkins):
(14.2)
(14.3)
(14.4)

Let represent the species in solution containing the PO

4

species of the
orthophosphates, using alum as the precipitant. Therefore,
(14.5)
Express Equation (14.5) in terms of using Eqs. (14.2) through (14.4). This
will enable to be expressed in terms of [Al

3

+

] using Equation (14.1) and
. Proceed as follows:
(14.6)
(14.7)
Al
3+
PO
4
3−
 AlPO
4 s()
↓+
AlPO
4 s()
↓  Al
3

PO
4
3−
K
sp,AlPO
4
+ 10
−21
=
PO
4
3−
H
+
 HPO
4
2−
+
HPO
4
2−
 PO
4
3−
H
+
Κ
HPO
4
+⇒ 10

−12.3
=
HPO
4
2−
H
+
 H
2
PO
4
−
+
H
2
PO
4
−
 HPO
4
2−
H
+
Κ
H
2
PO
4
+⇒ 10
−7.2

=
H
2
PO
4
−
H
+
 H
3
PO
4
+
H
3
PO
4
 H
2
PO
4
−
H
+
Κ
H
3
PO
4
+⇒ 10

−2.1
=
sp
PO
4
Al
sp
PO
4
Al
[]PO
4
3−
[]HPO
4
2−
[]H
2
PO
4
−
[]H
3
PO
4
[]+++=
PO
4
3−
[]

sp
PO
4
Al
[]
K
sp,AlPO
4
HPO
4
2−
[]
HPO
4
2−
{}
γ
HPO
4

PO
4
3−
{}H
+
{}
γ
HPO
4
K

HPO
4

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

== =
H
2
PO
4
−
[]
H
2

PO
4
−
{}
γ
H
2
PO
4

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

γ
PO
4

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

== =
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© 2003 by A. P. Sincero and G. A. Sincero
(14.8)
Equations (14.6) through (14.8) may now be substituted into Equation (14.5) to
produce

(14.9)
But, from Equation (14.1), = / = / = /
. Substituting,
(14.10)
,
γ
Al
,
γ
H
, , and are, respectively, the activity coefficients of the phos-
phate, aluminum, hydrogen, hydrogen phosphate, and dihydrogen phosphate ions.
, , and are, respectively, the equilibrium constants of the hydro-
gen phosphate and dihydrogen phosphate ions and phosphoric acid.
[Al
3+
] needs to be eliminated for the equation to be expressed solely in terms of
[H
+
]. When alum is added to water, it will unavoidably react with the existing natural
alkalinity. For this reason, the aluminum ion will not only react with the phosphate ion
to precipitate AlPO
4(s)
, but it will also react with the OH
−
to precipitate Al(OH)
3(s)
. Also,
Al
3+

will form complexes Al(OH)
2+
, , , , and
in addition to the Al(OH)
3(s)
. All these interactions complicate our objective
of eliminating [Al
3+
].
Consider, however, the equilibrium of Al(OH)
3(s)
, which is as follows:
(14.11)
= 10
−33
may be compared with = 10
−21
. From = 10
−33
,
the concentration of Al
3+
needed to precipitate Al(OH)
3
may be calculated to be
2.0(10
−9
) gmol/L. Doing similar calculation from = 10
−21
, the concentration

of Al
3+
needed to precipitate AlPO
4
is 3.0(10
−11
) gmol/L. These two concentrations
H
3
PO
4
[]H
3
PO
4
{}
H
2
PO
4
−
{}H
+
{}
K
H
3
PO
4

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

== =
sp
PO

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

γ
PO
4
γ
H
2

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

++=
γ
PO
4
γ
H
3
PO
4

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

+
PO
4
3−
[]PO
4
3−
{}
γ
PO
4

K
sp,AlPO
4
γ
PO
4
Al
3+
{}K
sp,AlPO
4
γ
PO
4
γ
Al
Al
3+
[]
sp
PO
4
Al
[]
K
sp,AlPO
4
γ
PO
4

γ
Al
Al
3+
[]

γ
H
K
sp,AlPO
4
H
+
[]
γ
HPO
4
K
HPO
4
γ
Al
Al
3+
[]

γ
H
2
K

sp,AlPO
4
H
+
[]
2
γ
H
2
PO
4
K
H
2
PO
4
K
HPO
4
γ
Al
Al
3+
[]

++=
γ
H
3
K

sp,AlPO
4
H
+
[]
3
K
H
3
PO
4
K
H
2
PO
4
K
HPO
4
γ
Al
Al
3+
[]

+
γ
PO
4
γ

HPO
4
γ
H
2
PO
4
K
HPO
4
K
H
2
PO
4
K
H
3
PO
4
Al
7
(OH)
17
4+
Al
13
(OH)
34
5+

Al(OH)
4
−
Al
2
(OH)
2
4+
Al(OH)
3 s()
 Al
3+
3OH
−
K
sp,Al(OH)
3
+ 10
−33
=
K
sp,Al(OH)
3
K
sp,AlPO
4
K
sp,Al(OH)
3
K

sp,AlPO
4
TX249_frame_C14.fm Page 631 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
are so close to each other that it may be concluded that Al(OH)
3
and AlPO
4
are
coprecipitating. This finding allows us to eliminate [Al
3+
]. Thus,
Substituting into Equation (14.10),
(14.12)
Equation (14.12) portrays the equilibrium relationship between and [H
+
].
Example 14.1 Calculate when the pH is 10. Assume the water con-
tains 140 mg/L of dissolved solids.
Solution:
Al
3+
[]
Al
3+
{}
γ
Al

K

sp,Al OH()
3
γ
Al
OH
−
{}
3

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

.== =
sp
PO
4
Al

[]
K
sp,AlPO
4
K
w
3
γ
PO
4
K
sp,Al(OH)
3
γ
H
3
H
+
[]
3

K
sp,AlPO
4
K
w
3
γ
HPO
4

K
HPO
4
K
sp,Al(OH)
3
γ
H
2
H
+
[]
2

+=
K
sp,AlPO
4
K
w
3
γ
H
2
PO
4
K
H
2
PO

4
K
HPO
4
K
sp,Al(OH)
3
γ
H
H
+
[]

K
sp,AlPO
4
K
w
3
K
H
3
PO
4
K
H
2
PO
4
K

HPO
4
K
sp,Al(OH)
3

++
sp
PO
4
[]
sp
PO
4
Al
[]
sp
PO
4
Al
[]
K
sp,AlPO
4
K
w
3
γ
PO
4

K
sp,Al(OH)
3
γ
H
3
H
+
[]
3

K
sp,AlPO
4
K
w
3
γ
HPO
4
K
HPO
4
K
sp,Al(OH)
3
γ
H
2
H

+
[]
2

+=
K
sp,AlPO
4
K
w
3
γ
H
2
PO
4
K
H
2
PO
4
K
HPO
4
K
sp,Al(OH)
3
γ
H
H

+
[]

K
sp,AlPO
4
K
w
3
K
H
3
PO
4
K
H
2
PO
4
K
HPO
4
K
sp,Al(OH)
3

++
K
sp,AlPO
4

10
−21
=
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
PO
4
10
0.5 3()

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

–
0.56== = =
K
sp,Al(OH)
3
10
33–
=
γ
H
γ
H
2
PO
4
10
−
0.5 1()
2
3.5 10
3–
()[]

1+1.14 3.5 10
3–
()[]

0.94== =
γ
HPO
4
= 10
−
0.5 2()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]

0.77 K
HPO
4
= 10
12.3–
K
H
2
PO
4
= 10

7.2–
K
H
3
PO
4
= 10
2.1–
=
TX249_frame_C14.fm Page 632 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
Therefore,
The answer of 9.17(10
+6
) mg/L emphasizes a very important fact: phosphorus
cannot be removed at alkaline conditions. It will be shown in subsequent discussions
that the solution conditions must be acidic for effective removal of phosphorus using
alum.
14.3.1 DETERMINATION OF THE OPTIMUM
p
H RANGE
Equation (14.12) shows that is a function of the hydrogen ion concentration.
This means that the concentration of the species containing the PO
4
species of the
orthophosphates is a function of pH. If the equation is differentiated and the result
equated to zero, however, an optimum value cannot be guaranteed to be found. A range
of pH values can, however, be assigned and the corresponding values of calcu-
lated. By inspection of the result, the optimum range can be determined. Tables 14.2
and 14.3 show the results of assigning this range of pH and values of calculated

using Equation (14.12). These tables show that optimum removal of phosphorus using
alum results when the unit is operated at pH values less than 5.0.
Note: The dissolved solids content has only a negligible effect on the resulting
concentrations.
14.4 CHEMICAL REACTION OF THE PHOSPHATE
ION WITH LIME
Calcium hydroxy apatite contains the phosphate and hydroxyl groups. Using calcium
hydroxide as the precipitant, the chemical reaction is shown below:
(14.13)
sp
PO
4
Al
[]
10
21–
()10
14–
()
3
0.56()10
33–
()0.94()
3
10
10–
[]
3

10

21–
()10
14–
()
3
0.77()10
12.3–
()10
33–
()0.94()
2
10
10–
[]
2

+=
10
21–
()10
14–
()
3
0.94()10
7.2–
()10
12.3–
()10
33–
()0.94()10

10–
[]

+
10
21–
()10
14–
()
3
10
2.1–
()10
7.2–
()10
12.3–
()10
−33
()

+
1.0 10
63–
()
4.65 10
64–
()

1.0 10
63–

()
3.41 10
66–
()

1.0 10
63–
()
2.79 10
63–
()

1.0 10
63–
()
2.51 10
55–
()

+++=
295.76 gmols/L 9.17 10
6
() mg/L as P==
sp
PO
4
Al
[]
sp
PO

4
Al
sp
PO
4
Al
5Ca
2+
3PO
4
3−
OH
−
 Ca
5
PO
4
()
3
OH
s()
↓++
Ca
5
PO
4
()
3
OH
s()

↓  5Ca
2+
3PO
4
3−
OH
−
++K
sp,apatite
10
55.9–
=
TX249_frame_C14.fm Page 633 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
TABLE 14.2
Concentration of as a Function
of pH at 25°C (mg/L as P)
pH, Dissolved
Solids = 140 mg/L (mg/L)
0 6.6(10
−5
)
1 7.1(10
−5
)
2 1.2(10
−4
)
3 6.5(10
−4

)
4 5.9(10
−3
)
5 5.9(10
−2
)
6 6.3(10
−1
)
7 1.1(10
+1
)
8 5.4(10
+2
)
9 4.8(10
+4
)
10 4.8(10
+6
)
11 5.8(10
+8
)
12 8.3(10
+10
)
13 4.0(10
+13

)
14 2.3(10
+16
)
15 3.5(10
+19
)
TABLE 14.3
Concentration of as a Function
of pH at 25°C (mg/L as P)
pH, Dissolved
Solids ==
==
35,000 mg/L (mg/L)
0 6.6(10
−5
)
1 8.0(10
−5
)
2 2.1(10
−4
)
3 1.5(10
−3
)
4 1.5(10
−2
)
5 1.5(10

−1
)
6 2.0(10
−0
)
7 8.9(10
+1
)
8 7.6(10
+3
)
9 7.5(10
+5
)
10 8.2(10
+7
)
11 1.6(10
+10
)
12 9.2(10
+12
)
13 8.5(10
+15
)
14 8.5(10
+18
)
15 8.5(10

+21
)
sp
PO
4
Al
[sp
PO
4
Al
]
sp
PO
4
Al
[sp
PO
4
Al
]
TX249_frame_C14.fm Page 634 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
These equations also show that the phosphorus must be in the phosphate form.
As in the case of alum, the phosphate ion produces the set of reactions given by
Eqs. (14.2) through (14.4). Let be the species in solution containing the PO
4
species, using the calcium in lime as the precipitant. will be the same as
given by Equation (14.5) which, along with Eqs. (14.2) through (14.4), can be
manipulated to produce Equation (14.9). From Equation (14.13),
Substituting in Equation (14.9) and simplifying,

(14.14)
γ
Ca
and K
w
are, respectively, the activity coefficient of the calcium ion and ion product
of water.
The [Ca
2+
] in the denominator is analogous to [Al
3+
] in the case of alum; however,
the present case is different in that [Ca
2+
] cannot be eliminated in a straightforward
manner. [Al
3+
] was easily eliminated because Al(OH)
3(s)
was precipitating along with
AlPO
4
. Ca(OH)
2(s)
, which could be precipitating, has a K
sp
value of 7.9(10
−6
). This
is much, much greater than K

sp,apatite
= 10
−55.9
and calcium hydroxide will not be
precipitating along with Ca
5
(PO
4
)
3
OH
(s)
. The other possible precipitate is CaCO
3
that is produced when lime reacts with the natural alkalinity of the water; however,
calcium carbonate has a K
sp
value of 4.8(10
−9
). Again, this value is much, much
greater than K
sp,apatite
and calcium carbonate will not be precipitating along with
Ca
5
(PO
4
)
3
OH

(s)
, either. We will, therefore, let the equation stand and express
as a function of [Ca
2+
], along with [H
+
] and the constants.
Example 14.2 Calculate expressed as mg/L of P when the pH is 8.
Assume the water contains 140 mg/L of dissolved solids and that [Ca
2+
] = 130 mg/L
as CaCO
3
.
sp
PO
4
Ca
sp
PO
4
Ca
[]
PO
4
3−
[]
PO
4
3−

{}
γ
PO
4

K
sp,apatite
Ca
2+
{}
5
OH
−
{}



1/3
γ
PO
4

K
sp,apatite
1/3
γ
H
1/3
H
+

[]
1/3
γ
Ca
1/3
Ca
2+
[]
5/3
K
w
1/3
γ
PO
4

== =
sp
PO
4
Ca
[]
K
sp,apatite
1/3
γ
H
1/3
H
+

[]
1/3
γ
Ca
1/3
Ca
2+
[]
5/3
K
w
1/3
γ
PO
4

K
sp,apatite
1/3
γ
H
4/3
H
+
[]
4/3
γ
HPO
4
K

HPO
4
γ
Ca
1/3
Ca
2+
[]
5/3
K
w
1/3

+=
K
sp,apatite
1/3
γ
H
7/3
H
+
[]
7/3
γ
H
2
PO
4
K

H
2
PO
4
K
HPO
4
γ
Ca
1/3
Ca
2+
[]
5/3
K
w
1/3

+
K
sp,apatite
1/3
γ
H
10/3
H
+
[]
10/3
K

H
3
PO
4
K
H
2
PO
4
K
HPO
4
γ
Ca
1/3
Ca
2+
[]
5/3
K
w
1/3

+
sp
PO
4
Ca
[]
sp

PO
4
Ca
[]
TX249_frame_C14.fm Page 635 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
Solution:
Therefore,
sp
PO
4
Ca
[]
K
sp,apatite
1/3
γ
H
1/3
H
+
[]
1/3
γ
Ca
1/3
Ca
2+
[]
5/3

K
w
1/3
γ
PO
4

K
sp,apatite
1/3
γ
H
4/3
H
+
[]
4/3
γ
HPO
4
K
HPO
4
γ
Ca
1/3
Ca
2+
[]
5/3

K
w
1/3

+=
K
sp,apatite
1/3
γ
H
7/3
H
+
[]
7/3
γ
H
2
PO
4
K
H
2
PO
4
K
HPO
4
γ
Ca

1/3
Ca
2+
[]
5/3
K
w
1/3

+
K
sp,apatite
1/3
γ
H
10/3
H
+
[]
10/3
K
H
3
PO
4
K
H
2
PO
4

K
HPO
4
γ
Ca
1/3
Ca
2+
[]
5/3
K
w
1/3

+
K
sp,apatite
10
55.9–
γ
H
γ
H
2
PO
4
10
0.5 1()
2
3.5 10

3–
()[]
1+1.14 3.5 10
3–
()[]

–
0.94=== =
γ
Ca
γ
HPO
4
10
0.5 2()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]
–
0.77== =
CA
2+
[]
130
1000 40.1()

3.24 10
3–
()gmol/L==
γ
PO
4
10
0.5 3()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]
–
0.56==
K
HPO
4
10
12.3–
K
H
2
PO
4
10
7.2–
K

H
3
PO
4
10
2.1–
===
sp
PO
4
Ca
[]
10
55.9–
()
1/3
0.94()
1/3
10
8–
[]
1/3
0.77()
1/3
3.24 10
3–
()[]
5/3
10
14–

()
1/3
0.56()

=
10
55.9–
()
1/3
0.94()
4/3
10
8–
[]
4/3
0.77()10
12.3–
()0.77()
1/3
3.24 10
3–
()[]
5/3
10
14–
()
1/3

+
10

55.9–
()
1/3
0.94()
7/3
10
8–
[]
7/3
0.94()10
7.2–
()10
12.3–
()0.77()
1/3
3.24 10
3–
()[]
5/3
10
14–
()
1/3

+
10
55.9–
()
1/3
0.94()

10/3
10
8–
[]
10/3
10
2.1–
()10
7.2–
()10
12.3–
()0.77()
1/3
3.24 10
3–
()[]
5/3
10
14–
()
1/3

+
4.91 10
22–
()
7.85 10
10–
()

4.62 10
30–
()
5.41 10
22–
()

4.34 10
38–
()
4.16 10
29–
()

4.08 10
46–
()
3.52 10
31–
()

+++=
9.58 10
9–
()gmol/L=
6.25 10
13–
()8.54 10
9–
()1.04 10

9–
()1.16 10
15–
()+++=
2.97 10
4–
()mg/L Ans=
TX249_frame_C14.fm Page 636 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
14.4.1 DETERMINATION OF THE OPTIMUM pH AND THE OPTIMUM
pH R
ANGE
Analyzing Equation (14.14) shows that [H
+
]’s are all written in the numerator. This
means that if the equation is differentiated, the result will give positive terms on
one side of the equation which will then be equated to zero to get the minimum.
This kind of equation cannot guarantee a minimum. Thus, the optimum and the
optimum pH range will be determined by preparing a table as was done in the case
of alum. Tables 14.4 through 14.6 show the results. This table was prepared using
Equation (14.14).
Tables 14.4 and 14.5 reveal very important information. The efficiency of
removal of phosphorus increases as the concentration of calcium increases from 0 to
130 mg/L. Remember that we disallowed the use of the K
sp
of Ca(OH)
2
because it
was too large, and we concluded that the hydroxide would not be precipitating
alongside with the apatite. It is not, however, preventable to add more lime in order

to increase the concentration of the calcium ion to the point of saturation and, thus,
be able to use in calculations. In theory, before the hydroxide precipitation
can happen, all the apatite particles would have already precipitated, resulting,
indeed, in a very high efficiency of removal of phosphorus. Would this really happen?
The answer would be yes, but this could be a good topic for applied research.
A concentration of zero for the calcium ion is, of course, nonexistent. Thus,
Table 14.4 may be considered as purely imagined. Also, from the tables there is no
single optimum pH; the range, however, may be determined by inspection as ranging
TABLE 14.4
Concentration of as a Function of
pH at 25°C (mg/L as P)
pH, Dissolved Solids ==
==

140 mg/L [Ca
2+
] ==
==
0 mg/L (mg/L)
0 19.0(10
+23
)
1 3.4(10
+21
)
2 1.4(10
+20
)
3 6.6(10
+18

)
4 3.1(10
+17
)
5 1.4(10
+16
)
6 6.6(10
+14
)
7 3.1(10
+13
)
8 1.4(10
+12
)
9 6.6(10
+10
)
10 3.1(10
+9
)
11 1.4(10
+8
)
12 6.6(10
+6
)
13 3.1(10
+5

)
14 1.4(10
+4
)
15 6.6(10
+2
)
K
sp,Ca(OH)
2
sp
PO
4
Ca
[sp
PO
4
Ca
]
TX249_frame_C14.fm Page 637 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
TABLE 14.5
Concentration of as a Function of pH
at 25°C (mg/L as P)
pH, Dissolved Solids ==
==
140 mg/L
[Ca
2+
] ==

==
130 mg/L as CaCO
3
(mg/L)
0 7.7(10
+16
)
1 3.8(10
+13
)
2 3.1(10
+10
)
3 7.6(10
+7
)
4 3.2(10
+5
)
5 1.3(10
+3
)
6 7.4(10
+0
)
7 5.8(10
−2
)
8 2.97(10
−4

)
9 5.6(10
−5
)
10 2.6(10
−6
)
11 1.3(10
−7
)
12 9.7(10
−9
)
13 2.2(10
−9
)
14 9.0(10
−10
)
15 4.1(10
−10
)
TABLE 14.6
Concentration of as a Function of pH
at 25°C (mg/L as P)
pH, Dissolved Solids ==
==
35,000 mg/L
[Ca
2++

++
] ==
==
130 mg/L as CaCO
3
(mg/L)
0 3.0(10
+16
)
1 1.7(10
+13
)
2 2.1(10
+10
)
3 7.1(10
+7
)
4 3.2(10
+5
)
5 1.6(10
+3
)
6 1.0(10
+1
)
7 1.9(10
−1
)

8 7.6(10
−3
)
9 3.5(10
−4
)
10 1.8(10
−5
)
11 1.6(10
−6
)
12 4.3(10
−7
)
13 1.9(10
−7
)
14 8.6(10
−8
)
15 4.0(10
−8
)
sp
PO
4
Ca
[sp
PO

4
Ca
]
sp
PO
4
Ca
[sp
PO
4
Ca
]
TX249_frame_C14.fm Page 638 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
from pH 7.0 and above. The effect of dissolved solids has reduced this range to
8 and above; however, a concentration of 35,000 mg/L of total solids is already
very high and would not be encountered in the normal treatment of water and
wastewater. This concentration is representative of the dissolved solids concentra-
tion of sea water.
14.5 CHEMICAL REACTION OF THE PHOSPHATE ION
WITH THE FERRIC SALTS
The chemical reaction to precipitate the phosphate ion as ferric phosphate is shown
next:
(14.15)
As in the case of precipitation using alum and precipitation using lime, the
phosphate ion produces the set of reactions given by Eqs. (14.2) through (14.4). Let
be the species in solution containing the PO
4
species of the orthophosphates
using the ferric salts FeCl

3
and Fe
2
(SO
4
)
3
as the precipitants. Again, will
be the same as given by Equation (14.5) which, along with Eqs. (14.2) through
(14.4), can be manipulated to produce Equation (14.9). From Equation (14.15),
.
Substituting in Equation (14.9) and simplifying,
(14.16)
γ
FeIII
is the activity coefficient of the ferric ion.
Investigate the possibility of eliminating [Fe
3+
] from the denominator of the
above equation. This is, indeed, possible through the use of ferric hydroxide. The
dissociation reaction is
(14.17)
Fe
3+
PO
4
3−
 FePO
4
↓+

FePO
4
↓  Fe
3+
PO
4
3−
K
sp,FePO
4
+ 10
21.9–
=
sp
PO
4
FeIII
sp
PO
4
FeIII
[]
PO
4
3−
[]
PO
4
3−
{}

γ
PO
4

K
sp,FePO
4
γ
PO
4
Fe
3+
{}

K
sp,FePO
4
γ
PO
4
γ
FeIII
Fe
3+
[]

== =
sp
PO
4

FeIII
[]
K
sp,FePO
4
γ
PO
4
γ
FeIII
Fe
3+
[]

γ
H
K
sp,FePO
4
H
+
[]
γ
HPO
4
K
HPO
4
γ
FeIII

Fe
3+
[]

+=
γ
H
2
K
sp,FePO
4
H
+
[]
2
γ
H
2
PO
4
K
H
2
PO
4
K
HPO
4
γ
FeIII

Fe
3+
[]

γ
H
3
K
sp,FePO
4
H
+
[]
3
K
H
3
PO
4
K
H
2
PO
4
K
HPO
4
γ
FeIII
Fe

3+
[]

++
Fe OH()
3 s()
 Fe
3+
3OH
−
() K
sp,Fe OH()
3
+ 1.1 10
36–
()=
TX249_frame_C14.fm Page 639 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
From = 1.1(10
−36
), the concentration of Fe
3+
needed to precipitate Fe(OH)
3
can be calculated to be equal to 4.5(10
−10
) gmol/L. And, from = 10
−21.9
,
the concentration of Fe

3+
needed to precipitate FePO
4
can also be calculated to be
equal to 1.1(10
−11
) gmol/L. Because these two concentrations are practically equal,
Fe(OH)
3(s)
will definitely precipitate along with FePO
4
. Therefore,
.
Substituting in Equation (14.16),
(14.18)
Example 14.3 Calculate expressed as mg/L of P when the pH is 8.
Assume the water contains 140 mg/L of dissolved solids.
Solution:
K
sp,Fe OH()
3
K
sp,FePO
4
Fe
3+
[]
Fe
3+
{}

γ
FeIII

K
sp,Fe OH()
3
γ
FeIII
OH
−
{}
3

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

== =

sp
PO
4
FeIII
[]
K
sp,FePO
4
K
w
3
γ
PO
4
K
sp,Fe OH()
3
γ
H
3
H
+
[]
3

K
sp,FePO
4
K
w

3
γ
HPO
4
K
HPO
4
K
sp,Fe OH()
3
γ
H
2
H
+
[]
2

+=
K
sp,FePO
4
K
w
3
γ
H
2
PO
4

K
H
2
PO
4
K
HPO
4
K
sp,Fe OH()
3
γ
H
H
+
[]

K
sp,FePO
4
K
w
3
K
H
3
PO
4
K
H

2
PO
4
K
HPO
4
K
sp,Fe OH()
3

++
sp
PO
4
FeIII
[]
sp
PO
4
FeIII
[]
K
sp,FePO
4
K
w
3
γ
PO
4

K
sp,Fe OH()
3
γ
H
3
H
+
[]
3

K
sp,FePO
4
K
w
3
γ
HPO
4
K
HPO
4
K
sp,Fe OH()
3
γ
H
2
H

+
[]
2

+=
K
sp,FePO
4
K
w
3
γ
H
2
PO
4
K
H
2
PO
4
K
HPO
4
K
sp,Fe OH()
3
γ
H
H

+
[]

K
sp,FePO
4
K
w
3
K
H
3
PO
4
K
H
2
PO
4
K
HPO
4
K
sp,Fe OH()
3

++
K
sp,FePO
4

10
21.9–
=
γ
PO
4
10
0.5 3()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]
–
0.56 K
sp,Fe OH()
3
1.1 10
36–
()===
γ
H
γ
H
2
PO
4
10

0.5 1()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]
–
0.94== =
γ
HPO
4
10
0.5 2()
2
3.5 10
3–
()[]
1+1.14 3.5 10
3–
()[]
–
0.77==
K
HPO
4
10
12.3–
K

H
2
PO
4
10
7.2–
K
HPO
4
10
12.3–
===K
H
3
PO
4
10
2.1–
=
sp
PO
4
FeIII
[]
10
21.9–
()10
14–
()
3

0.56()1.1()10
36–
()0.94()
3
10
8–
[]
3

=
10
21.9–
()10
14–
()
3
0.77()10
12.3–
()1.1()10
36–
()0.94()
2
10
8–
[]
2

+
TX249_frame_C14.fm Page 640 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero

14.5.1 DETERMINATION OF THE OPTIMUM pH AND THE OPTIMUM
pH R
ANGE
As in the case of Eqs. (14.12) and (14.14), Equation (14.18) cannot guarantee an
optimum if it is differentiated and equated to 0. Hence, the optimum and the optimum
pH range will be determined by preparing a table as was done in the case of alum
and lime. Tables 14.7 and 14.8 show the results. This table was prepared using
Equation (14.18). As shown, the optimum range is equal to or less than 3.
TABLE 14.7
Concentration of as a
Function of pH at 25°C (mg/L as P)
pH, Dissolved
Solids ==
==
140 mg/L (mg/L)
0 1.1(10
−2
)
1 1.2(10
−2
)
2 2.1(10
−2
)
3 1.1(10
−1
)
4 1.0(10
−0
)

5 1.0(10
+1
)
6 1.1(10
+2
)
7 1.8(10
+3
)
8 1.17(10
+5
)
9 8.4(10
+6
)
10 8.3(10
+8
)
11 8.9(10
+10
)
12 1.4(10
+13
)
13 6.9(10
+15
)
14 6.1(10
+18
)

15 6.0(10
+21
)
10
21.9–
()10
14–
()
3
0.94()10
7.2–
()10
12.3–
()1.1()10
36–
()0.94()10
8–
[]

+
10
21.9–
()10
14–
()
3
10
2.1–
()10
7.2–

()10
12.3–
()1.1()10
36–
()

+
1.26 10
64–
()
5.12 10
61–
()

=
1.26 10
64–
()
3.75 10
65–
()

1.26 10
64–
()
3.07 10
64–
()

1.26 10

64–
()
2.76 10
58–
()

+++
2.46 10
4–
()3.36 0.41 4.57 10
7–
()+++ 3.77 gmols/L==
1.17= 10
5
()mg/L Ans
sp
PO
4
FeIII
[sp
PO
4
FeIII
]
TX249_frame_C14.fm Page 641 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
14.6 COMMENTS ON THE OPTIMUM pH RANGES
When the ferric salts are used to precipitate FePO
4
, the ion must compete with

the OH
−
ion. When the pH is high, there will be a large concentration of the OH
−
and the phosphate ion can lose the competition and fail to precipitate; Fe(OH)
3
may
precipitate instead of FePO
4
. Thus, to precipitate the phosphate ion, the concentration
of OH
−
must be suppressed. This is done by adding more H
+
, which reacts with the
OH
−
. If the OH
−
is “busy” satisfying the hydrogen ion, the Fe
3+
ion is now available
for reaction with phosphate to precipitate FePO
4
instead of Fe(OH)
3
. The addition
of the hydrogen depresses the pH and the optimum pH is found to be equal to or
less than 3.
For the case of the AlPO

4
precipitation, alum also produces the hydroxide
Al(OH)
3
analogous to the ferric salts producing Fe(OH)
3
. Thus, for the phosphate
ion to outcompete the hydroxide ion, the pH must also be lowered in order to produce
the desired precipitate of AlPO
4
, instead of Al(OH)
3
. Again, the pH is reduced by
adding the hydrogen ion. The optimum pH range for AlPO
4
precipitation, we found
to be equal to or less than 5.
In the case of the use of lime, the OH
−
ion is a reactant for the production of the
apatite, Ca
5
(PO
4
)
3
OH
(s)
. This is the reason why the pH must be raised, which is found
to be equal to or greater than 7. In addition, no other precipitate competes with the

precipitation of the apatite. Thus, overall, of the three precipitants, lime is the best.
TABLE 14.8
Concentration of as a Function
of pH at 25°C (mg/L as P)
pH, Dissolved
Solids ==
==
35,000 mg/L (mg/L)
0 1.2(10
−2
)
1 1.4(10
−2
)
2 3.7(10
−2
)
3 2.6(10
−1
)
4 2.6(10
−0
)
5 2.7(10
+1
)
6 3.8(10
+2
)
7 1.5(10

+3
)
8 1.3(10
+6
)
9 1.3(10
+8
)
10 1.4(10
+10
)
11 2.8(10
+12
)
12 1.6(10
+15
)
13 1.5(10
+18
)
14 1.5(10
+21
)
15 1.5(10
+24
)
sp
PO
4
FeIII

[sp
PO
4
FeIII
]
PO
4
3−
TX249_frame_C14.fm Page 642 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
14.7 EFFECT OF THE K
sp
’s ON THE PRECIPITATION
OF PHOSPHORUS
The worst of the precipitants are the ferric salts. Compare = 1.1(10
−36
) and
= 10
−21.9
. These two K
sp
’s produce comparable concentrations of the ferric
ion to precipitate either Fe(OH)
3
or FePO
4
. Thus, at high pH conditions, the phos-
phate ion would have a big competitor in the form of the hydroxide ion. An Fe
3+
available in solution is grabbed by the OH

−
ion to form the ferric hydroxide, leaving
less amount of Fe
3+
to precipitate ferric phosphate. The ferric salts are, therefore, a
poor performer for removing phosphorus. Also, to be effective requires adjusting
the pH to the pH of almost mineral acidity of less than 3.
Now, compare = 10
−33
and = 10
−21
for the case of using alum.
This situation is similar to the case of the ferric salts. The concentrations of the
aluminum ion to precipitate either the aluminum hydroxide or the aluminum phos-
phate are comparable. This means that the hydroxyl ion is also a competitor in the
phosphate precipitation. The process operates poorly at high pH, although it has a
better pH range of equal to or less than 5 compared to that of the ferric salts.
In the case of using lime to precipitate phosphorus, we know that calcium
hydroxide has a much, much larger K
sp
than that of the apatite, the required precip-
itate. The former has a K
sp
of = 7.9(10
−6
). Compare this to that of apatite
which is K
sp,apatite
= 10
−55.9

. Thus, adding lime to the water means that a large amount
of the calcium ion is available to precipitate the apatite, Ca
5
(PO
4
)
3
OH
(s)
. Also, no
other precipitate competes with this precipitation. Therefore, using lime is the best
method for removing phosphorus. Adding a quantity of lime to the point of satura-
tion, Ca(OH)
2
has the potential of removing phosphorus “100%.” This process must,
however, be investigated. In the overall, the process using lime entails adjusting the
pH to greater than 7, which is much, much better than the range for the use of either
alum or ferric salts.
14.8 UNIT OPERATIONS FOR PHOSPHORUS REMOVAL
The removal of phosphorus involves provision for the addition of chemicals both
for adjusting the pH and for precipitating the phosphate, mixing them with the raw
water, settling the resulting precipitates, and filtering those precipitates that escape
removal by settling. The precipitates formed are still in suspension; thus, it is also
necessary for them to flocculate by adding a flocculation tank.
14.9 CHEMICAL REQUIREMENTS
The chemicals required are alum, lime, and the ferric salts FeCl
3
and Fe
2
(SO

4
)
3
.
These are in addition to the acid or bases needed to adjust the pH. The formulas
used to calculate the amounts of these chemicals for pH adjustment were already
derived in Chapters 11, 12, and 13. The chemical requirements to be discussed here
will only be for alum, lime, and the ferric salts.
K
sp,Fe OH()
3
K
sp,FePO
4
K
sp,Al OH()
3
K
sp,AlPO
4
K
sp,Ca OH()
2
TX249_frame_C14.fm Page 643 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
The amount of alum needed to precipitate the phosphate is composed of the
alum required to satisfy the natural alkalinity of the water and the amount needed
to precipitate the phosphate. Satisfaction of the natural alkalinity will bring the
equilibrium of aluminum hydroxide. Remember, however, that even if these quan-
tities of alum were provided, the concentration of phosphorus that will be discharged

from the effluent of the unit still has to conform to the equilibrium reaction that
depends upon the pH level at which the process was conducted. The optimum pH,
we have found, is equal to or less 5.
Consider the full alum formula in the chemical reaction,
(14.19)
Note: Only one arrow is used and not the forward and backward arrows. The
reaction is not in equilibrium.
As mentioned before, the previous reaction for phosphate removal takes place
in conjunction with the Al(OH)
3
equilibrium. Bicarbonate alkalinity is always present
in natural waters, so this equilibrium is facilitated by alum reacting with calcium
bicarbonate so that, in addition to the amount of alum required for the precipitation
of phosphorus, more is needed to neutralize the bicarbonate. The reaction is repre-
sented by
(14.20)
From the previous reactions, the equivalent masses are alum = Al
2
(SO
4
)
3
⋅ xH
2
O/
6 = 57.05 + 3x and calcium bicarbonate = 3Ca(HCO
3
)
2
/6 = 81.05. The number of

moles of phosphorus in is one; thus, from the balanced chemical reaction, the
equivalent mass of phosphorus is 2P/6 = 10.33.
Although the alkalinity found in natural waters is normally calcium bicarbonate,
it is conceivable that natural alkalinities could also be associated with other cations.
Thus, to be specific, as shown by Equation (14.20), the alkalinity discussed here is
calcium bicarbonate, with the understanding that “other forms” of natural alkalinities
may be equated to calcium bicarbonate alkalinity when expressed in equivalent
concentrations. Let A
Ca
be this calcium bicarbonate alkalinity.
The quantity of alum needed to react with the calcium bicarbonate alkalinity is
exactly the quantity of alum needed to establish the aluminum hydroxide equilib-
rium. Let [A
Ca
]
geq
be the gram equivalents per liter of this bicarbonate alkalinity in
the raw water of volume cubic meters and let M
AlAlk
be the kilograms of alum at
a fractional purity of P
Al
required to react with it. Then,
(14.21)
Remember that alkalinity is also expressed analytically in terms of calcium
carbonate in which the equivalent mass is considered 50. Let us digress from our
main discussion and explore this matter further in order to determine how the various
Al
2
SO

4
()
3
⋅ xH
2
O2PO
4
3−
2 AlPO
4 s()
↓ 3SO
4
2−
xH
2
O++→+
Al
2
SO
4
()
3
⋅ xH
2
O 3 Ca HCO
3
()
2
2Al OH()
3

↓
6CO
2
3 CaSO
4
xH
2
O+++→+
PO
4
3−
V
M
AlAlk
A
Ca
[]
geq
57.05 3x+()
V
P
Al
=
TX249_frame_C14.fm Page 644 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
methods of expressing alkalinities are related. How will the analytical result be
converted to an equivalent concentration to be used in Equation (14.21)?
Consider the following reactions:
CaCO
3

+ 2H
+
→ H
2
CO
3
+ CA
2+
(14.22)
2Ca(HCO
3
)
2
+ 2H
+
→ H
2
CO
3
+ 2Ca
2+
(14.23)
From these reactions, the equivalent mass of calcium bicarbonate in relation to the
equivalent mass of calcium carbonate of 50 is 2Ca(HCO
3
)
2
/2 = 162.1 (much different
in comparison with 81.05). Now, assume gram equivalents per liter of
calcium bicarbonate alkalinity being computed on the basis of Equation (14.23).

This computation, in turn, is obtained from an analytical result based on Equation
(14.22). The number of grams of the bicarbonate alkalinity is then equal to 162.1
. The quantity of mass remains the same in any system. Thus, referring
to Equation (14.20), there will also be the same 162.1 grams of the
bicarbonate there. To find the number of equivalents, however, it is always equal to
the mass divided by the equivalent mass. Therefore, based on Equation (14.20), the
corresponding number of equivalents of the 162.1 grams of calcium
bicarbonate alkalinity is 162.1 /81.05 = 2 . And,
(14.24)
Again, is the calcium bicarbonate alkalinity referred to the reaction of alum
with calcium bicarbonate and is the same bicarbonate alkalinity referred
to calcium carbonate with equivalent mass of 50. This equation converts analytical
results expressed as calcium carbonate to a form that Equation (14.21) can use.
Now, let [Phos]
mg
be the milligram per liter of phosphorus as P in phosphate
and let M
AlPhos
be the kilograms of alum at fractional purity of P
Al
required to
precipitate this phosphate phosphorus. Letting f
P
represent the fractional removal of
phosphorus, the following equation is then produced:
(14.25)
Letting M
Al
be the total kilograms of alum needed for the removal of phosphorus,
(14.26)

Now, let us tackle the problem of calculating the chemical requirements for
removing calcium apatite. Although calcium carbonate has a very high K
sp
that it
would not be precipitating with the apatite, the natural alkalinity would have to be
A
Ca
[]
geqCaCO
3
A
Ca
[]
geqCaCO
3
A
Ca
[]
geqCaCO
3
A
Ca
[]
geqCaCO
3
A
Ca
[]
geqCaCO
3

A
Ca
[]
geqCaCO
3
A
Ca
[]
geq
2 A
Ca
[]
geqCaCO
3
=
A
Ca
[]
geq
A
Ca
[]
geqCaCO
3
M
AlPhos
f
P
Phos[]
mg

1000 10.33()P
Al

57.05 3x+()
V
9.7 10
5–
()f
P
Phos[]
mg
57.05 3x+()
V
P
Al

==
M
Al
M
AlAlk
M
AlPhos
+
57.05 3x+()
P
Al

A
Ca

[]
geq
9.7 10
5–
()f
P
Phos[]
mg
+()
V
==
TX249_frame_C14.fm Page 645 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
neutralized nonetheless, producing the carbonate. The formation of the calcium
carbonate solid will only occur, however, after the apatite formation has completed.
Thus, the amount of lime needed to precipitate the phosphate is composed of the
lime needed to precipitate the phosphate in apatite and the lime needed to neutralize
the natural alkalinity of the water.
Note: Even if these quantities were provided, the concentration of phosphorus
that would be discharged from the effluent of the unit still has to conform
to the equilibrium reaction that depends upon the pH level at which the
process was conducted.
The optimum pH, we have found, is equal to or greater than 7.
The applicable chemical reactions are:
Ca(OH)
2
+ Ca(HCO
3
)
2

→ H
2
O + CaCO
3(s)
↓ (14.27)
(14.28)
To make the previous equations compatible, use 2 as the number of reference species,
based on Ca(OH)
2
. Therefore the equivalent masses are lime = CaO/2 = 37.05;
calcium bicarbonate = Ca(HCO
3
)
2
/2 = 81.05; and phosphorus = 3/5P/2 = 9.3. In this
particular case, because the equivalent mass of calcium bicarbonate is also equal to
81.05, .
Let M
CaOAlk
be the kilograms of lime at fractional purity of P
CaO
required to react
with the calcium carbonate alkalinity. Then,
(14.29)
Now, letting M
CaOPhos
be the kilograms of lime required to react with the phosphorus,
(14.30)
Let M
CaO

be the total kilograms of lime needed for the removal of phosphorus.
Thus,
(14.31)
Last, let us tackle the calculation of the amount of ferric salts needed to precip-
itate the phosphate. This amount is composed of the amount needed to neutralize
the natural alkalinity of the water and the amount needed to precipitate the phosphate.
Again, remember, that even if these quantities were provided, the concentration of
phosphorus that will be discharged from the effluent of the unit still has to conform
to the equilibrium reaction that depends upon the pH level at which the process was
conducted. The optimum pH, we have found, is equal to or less than 3.
Ca OH()
2
3
5

PO
4
3−

1
5

Ca
5
PO
4
()
3
OH
s()

↓
9
5

OH
−
++
A
Ca
[]
geq
2 A
Ca
[]
geqCaCO
3
=
M
CaOAlk
37.05 A
Ca
[]
geq
V
P
CaO
=
M
CaOPhos
37.05 f

P
Phos[]
mg
1000 9.3()P
CaO

V
4.0 10
3–
()f
P
Phos[]
mg
V
P
CaO

==
M
CaO
M
CaOAlk
M
CaOPhos
+ 37.05 A
Ca
[]
geq
4.0 10
3–

()f
P
Phos[]
mg
+()
V
P
CaO

==
TX249_frame_C14.fm Page 646 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
The applicable respective chemical reactions for the satisfaction of the bicar-
bonate alkalinity are
2FeCl
3
+ 3Ca(HCO
3
)
2
→ 2Fe(OH)
3
↓ + 6CO
2
+ 3CaCl
2
(14.32)
Fe
2
(SO

4
)
3
+ 3Ca(HCO
3
)
2
→ 2Fe(OH)
3
↓ + 6CO
2
+ 3CaSO
4
(14.33)
Upon satisfaction of the previous reactions, the normal precipitation of the phosphate
occurs. The respective reactions are
(14.34)
(14.35)
Note that the coefficients of Equation (14.34) have been adjusted to make the
equation compatible with Equation (14.32) and all the rest of the above equations. From
these equations, the equivalent masses are ferric chloride = 2FeCl
3
/6 = 54.1; calcium
bicarbonate = 3Ca(HCO
3
)
2
/6 = 81.05; ferric sulfate = Fe
2
(SO

4
)
3
/6 = 66.65; and phos-
phorus = 2P/6 = 15.5. Since the equivalent mass is still 81.05, .
Let M
FeIIIClAlk
and be the kilograms of ferric chloride and ferric sulfate,
respectively, required to react with the calcium bicarbonate alkalinity. Also, let the
respective purity be P
FeIIICl
and . Then,
(14.36)
(14.37)
Now, letting and be the kilograms of ferric chloride and
ferric sulfate, respectively, required to react with the phosphorus,
(14.38)
(14.39)
Let and be the total kilograms of ferric chloride and ferric
sulfate, respectively, needed for the removal of phosphorus. Thus,
(14.40)
(14.41)
2FeCl
3
2PO
4
3−
→ 2FePO
4
↓ 6Cl

−
++
Fe
2
SO
4
()
3
2PO
4
3−
→ 2FePO
4
↓ 3SO
4
2−
++
A
Ca
[]
geq
2 A
Ca
[]
geqCaO
3
=
M
FeIIISO
4

Alk
P
FeIIISO
4
M
FeIIIClAlk
54.1 A
Ca
[]
geq
V
P
FeIIICl

=
M
FeIIISO
4
Alk
66.65 A
Ca
[]
geq
V
P
FeIIISO
4

=
M

FeIIIClPhos
M
FeIIISO
4
Phos
M
FeIIIClPhos
54.1 f
P
Phos[]
mg
1000 15.5()P
FeIIICl

V
3.5 10
3–
()f
P
Phos[]
mg
V
P
FeIIICl

==
M
FeIIISO
4
Phos

66.5 f
P
Phos[]
mg
1000 15.5()P
FeIIISO
4

V
4.3 10
3–
()f
P
Phos[]
mg
V
P
FeIIISO
4

==
M
FeIIICl
M
FeIIISO
4
M
FeIIICl
M
FeIIIClAlk

M
FeIIIClPhos
+ 54.1 A
Ca
[]
geq
3.5 10
3–
()f
P
Phos[]
mg
+()
V
P
FeIIICl

==
M
FeIIISO
4
M
FeIIISO
4
Alk
M
FeIIISO
4
Phos
+=

66.5 A
Ca
[]
geq
4.3 10
3–
()f
P
Phos[]
mg
+()
V
P
FeIIISO
4

=
TX249_frame_C14.fm Page 647 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
14.10 SLUDGE PRODUCTION
Phosphorus is normally removed from wastewaters and, as such, the solids produced
would include the original suspended solids or turbidity. Therefore, two sources
contribute to the production of solids in phosphorus removal: the original suspended
solids and the solids produced from chemical reaction. If is the cubic meters of
water treated and sp
ss
are the solids removed at a fractional efficiency of f
ss
, the
kilograms of solids produced from the suspended solids is (f

ss
[sp
ss
]
mg
/1000) , where
[sp
ss
]
mg
is in mg/L.
For convenience, the chemical reactions responsible for production of chemical
sludge are reproduced below. They were derived previously.
Al
2
(SO
4
)
3
⋅ xH
2
O + 2 → 2AlPO
4(s)
↓ + 3 + xH
2
O (14.42)
Al
2
(SO
4

)
3
⋅ xH
2
O + 3Ca(HCO
3
)
2
→ 2Al(OH)
3
↓ + 6CO
2
+ 3CaSO
4
+ xH
2
O
(14.43)
Ca(OH)
2
+ Ca(HCO
3
)
2
→ H
2
O + CaCO
3(s)
↓ (14.44)
Ca(OH)

2
+  Ca
5
(PO
4
)
3
OH
(s)
↓ + OH
−
(14.45)
2FeCl
3
+ 3Ca(HCO
3
)
2
→ 2Fe(OH)
3
↓ + 6CO
2
+ 3CaCl
2
(14.46)
Fe
2
(SO
4
)

3
+ 3Ca(HCO
3
)
2
→ 2Fe(OH)
3
↓ + 6CO
2
+ 3CaSO
4
(14.47)
2FeCl
3
+ 2 → 2FePO
4
↓ + 6Cl
−
(14.48)
Fe
2
(SO
4
)
3
+ 2 → 2FePO
4
↓ + 3 (14.49)
Referring to the previous reactions, the solids produced from the removal of
phosphorus are AlPO

4(s)
, Al(OH)
3
, CaCO
3
, Ca
5
(PO
4
)
3
OH
(s)
, Fe(OH)
3
, and FePO
4
. The
equivalent masses are aluminum phosphate = 2AlPO
4
/6 = 40.67; aluminum hydrox-
ide is 2Al(OH)
3
/6 = 26.0; calcium carbonate = CaCO
3
/2 = 50; calcium hydroxyap-
atite = 1/5Ca
5
(PO
4

)
3
OH/2 = 50.25; ferric hydroxide = 2Fe(OH)
3
/6 = 35.6; and ferric
phosphate = 2FePO
4
/6 = 50.27. The equivalent mass of phosphorus varies. For Eqs.
(14.42), (14.48), and (14.49), the equivalent mass of phosphorus = 2P/6 = 10.33;
from Equation (14.45), equivalent mass = 3/5P/2 = 9.3. For calcium carbonate, the
equivalent mass in all the reactions is 3Ca(HCO
3
)
2
/6 or Ca(HCO
3
)
2
/2 = 81.05. Thus,
(14.50)
From treating m
3
of water with [Phos]
mg
mg/L of phosphate phosphorus being
removed, let = the kilograms of aluminum phosphate solids produced;
V
V
PO
4

3−
SO
4
2−
3
5

PO
4
3−
1
5

9
5

PO
4
3−
PO
4
3−
SO
4
2−
A
Ca
[]
geq
2 A

Ca
[]
geqCaCO
3
=
V
M
AlPO
4
TX249_frame_C14.fm Page 648 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
= kilograms of apatite solids produced; and = kilograms of ferric
phosphate solids produced. Also, from reacting with [A
cur
]
geq
gram moles per liter
of natural alkalinity in cubic meters of water, let = kilograms of alum solids
produced; = kilograms calcium carbonate solids produced; and =
kilograms of ferric hydroxide solids produced. Finally, let M
ss
= kilograms of solids
produced from the original suspended solids. Considering the respective equivalent
masses for the above equations, the following equations are produced:
(14.51)
(14.52)
(14.53)
(14.54)
(14.55)
(14.56)

(14.57)
Now, let = total solids produced in removing phosphorus using alum;
= total solids produced in removing phosphorus using lime; and
= total solids produced in removing phosphorus using the ferric salts.
= + + M
ss
; = + + M
ss
;
and = + + M
ss
. Therefore,
(14.58)
(14.59)
(14.60)
Example 14.4 A wastewater contains 10 mg/L of phosphorus expressed as P.
The pH is 8.0. Assume the wastewater contains 100 mg/L of alkalinity expressed
M
Ca
5
PO
4
()
3
OH
M
FePO
4
V
M

Al OH()
3
M
CaCO
3
M
Fe OH()
3
M
AlPO
4
40.67 f
P
Phos[]
mg
V
1000 10.33()

3.94 10
3–
()f
P
Phos[]
mg
V
==
M
Ca
5
PO

4
()
3
OH
50.25 f
P
Phos[]
mg
V
1000 9.3()

5.40 10
3–
()f
P
Phos[]
mg
V
==
M
FePO
4
50.27 f
P
Phos[]
mg
V
1000 10.33()

4.87 10

3–
()f
P
Phos[]
mg
V
==
M
Al OH()
3
26.0 A
Ca
[]
geq
V
=
M
CaCO
3
50 A
Ca
[]
geq
V
=
M
Fe OH()
3
35.6 A
Ca

[]
geq
V
=
M
ss
f
ss
sp
ss
[]
mg
1000

V
=
M
AlPO
4
Solids
M
CaOPO
4
Solids
M
FeIIIPO
4
Solids
M
AlPO

4
Solids
M
AlPO
4
M
Al OH()
3
M
CaOPO
4
Solids
M
Ca
5
PO
4
()
3
OH
M
CaCO
3
M
FeIIIPO
4
Solids
M
FePO
4

M
Fe OH()
3
M
AlPO
4
Solids
3.94 10
3–
()f
P
Phos[]
mg
26.0 A
Ca
[]
geq
f
ss
sp
ss
[]
mg
1000

++


V
=

M
CaOPO
4
Solids
5.40 10
3–
()f
P
Phos[]
mg
50 A
Ca
[]
geq
f
ss
sp
ss
[]
mg
1000

++


V
=
M
FeIIIPO
4

Solids
4.87 10
3–
()f
P
Phos[]
mg
35.6 A
Ca
[]
geq
f
ss
sp
ss
[]
mg
1000

++


V
=
TX249_frame_C14.fm Page 649 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
as CaCO
3
. The phosphorus is to be removed using lime. If the wastewater flow is
0.75 m

3
/sec, calculate the kilograms of solids produced per day. Assume the sus-
pended solids content is 200 mg/L.
Solution:
Assume all of the phosphorus and suspended solids removed. Therefore, f
P
= 1; f
ss
= 1
Therefore,
GLOSSARY
Calcium hydroxy apatite—Ca
5
(PO
3
)
3
(OH)
(s)
.
Condensation—The formation of water molecules when substances react.
Condensed phosphates—Substances containing phosphates characterized by
the presence of the P–O–P bonds formed when phosphoric acid combines
in a process called condensation.
Metaphosphoric acid—A condensation product of phosphoric acid forming a
chain of (HPO
3
)
n
.

Optimum pH—The pH at which maximum precipitation occurs or at which the
concentration of the species to be removed is at the lowest.
Optimum pH range—A range of pH at which the quantity of precipitates pro-
duced or the concentration of the species to be removed remaining in
solution is acceptable.
Orbitals—A volume of atomic space in which a maximum of two electrons
occupy.
Orthophosphates—The group of species containing the of , and
groups.
Oxidation states—A condition of being an atom in relation to the extent of the
atom losing or gaining electrons in its valence shell.
Polyphosphates—A compound containing two or more phosphate groups in the
molecule.
Shell—A volume of atomic space holding electronic orbitals.
Solubility product constant—The product of the concentrations of ions, raised
to appropriate powers, in equilibrium with their solid.
M
CaOPO
4
Solids
5.40 10
3–
()f
P
Phos[]
mg
50 A
Ca
[]
geq

f
ss
sp
ss
[]
mg
1000

++


=
V
A
Ca
[]
geq
2 A
Ca
[]
geqCaCO
3
2 100()
1000()50()

0.004 geq/L===
M
CaOPO
4
Solids

5.40 10
3–
()1()10()50 0.004()
1()200()
1000

++


0.75()60()60()24()=
29,419.2 kg/d Ans=
H
2
PO
4
−
, HPO
4
2−
PO
4
3−
TX249_frame_C14.fm Page 650 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero
Valence configuration—The arrangement of electrons in an atomic shell that
accounts for the chemical reactivity of the atom.
Valence shell—The shell of an atom responsible for chemical reactions.
SYMBOLS
A
ca

Calcium bicarbonate alkalinity
[A
Ca
]
geq
Gram equivalents per liter of calcium bicarbonate alkalinity
Gram equivalents per liter of alkalinity being computed on the
basis of CaCO
3
having an equivalent mass of 50
Al(OH)
2+
Hydroxo Al(III) ion
17-hydroxo, 7-Al(III) complex ion
34-hydroxo, 13-Al(III) complex ion
Tetrahydroxo Al(III) complex ion
Dihydroxo 2-Al(III) complex ion
Fractional removal of original suspended of raw water
Equilibrium constant of the hydrogen phosphate ion
Equilibrium constant of the dihydrogen phosphate ion
Equilibrium constant of phosphoric acid
Solubility product constant of aluminum hydroxide
Solubility product constant of aluminum phosphate
Solubility product constant of calcium hydroxy apatite phosphate
Solubility product constant of calcium hydroxide
Solubility product constant of ferric hydroxide
Solubility product constant of ferric phosphate
K
w
Ion product of water

M
Al
Total kilograms of alum needed for the removal of phosphorus
M
AlAlk
Kilograms of alum that react with the calcium bicarbonate alkalinity
Kilograms of alum solids produced in the removal of phosphorus
Kilograms of alum that react with phosphate phosphorus in raw water
Kilograms of aluminum phosphate solids produced in the removal
of phosphorus
Total solids produced in removing phosphorus using alum as the
precipitant
Total solids produced in removing phosphorus using lime as the
precipitant (including suspended solids in raw water)
Kilograms of apatite solids produced in the removal of phosphorus
Kilograms calcium carbonate solids produced in the removal of
phosphorus
M
CaO
Kilograms of lime needed for the removal of phosphorus
M
CaOAlk
Kilograms of lime that react with the calcium bicarbonate alkalin-
ity of the raw water
M
CaOPhos
Kilograms of lime that react with phosphate phosphorus in raw water
M
FeIIICl
Total kilograms of ferric chloride used for phosphate removal

M
FeIIIClAlk
Kilograms of ferric chloride that react with the calcium bicarbonate
of the raw water
A
Ca
[]
geqCaCO
3
Al
7
OH()
17
4+
Al
13
OH()
34
5+
Al OH()
4
−
Al
2
OH()
2
4+
f
ss
K

HPO
4
K
H
2
PO
4
K
H
3
PO
4
K
sp,Al OH()
3
K
sp,AlPO
4
K
sp,apatite
K
sp,Ca OH()
2
K
sp,Fe OH()
3
K
sp,FePO
4
M

Al OH()
3
M
AlPhos
M
AlPO
4
M
AlPO
4
Solids
M
CaOPO
4
Solids
M
Ca
5
PO
4
()
3
OH
M
CaCO
3
TX249_frame_C14.fm Page 651 Friday, June 14, 2002 2:34 PM
© 2003 by A. P. Sincero and G. A. Sincero