Page 252 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

Coagulation 207

10 –2
Al(OH) 2+ Al(OH) 4– 1000
10 –3
Optimum sweep 100
Restabilization Optimum sweep
Al concentration (mol Al/L) 10 –5 neutralization Charge neutralization to 10 Al concentration (mg Al 2 (SO 4 ) 3 . 14H 2 O/L)
zones
–4
10
Combination
Charge
(sweep/adsorption)
1
ζ=0 with Al(OH) (s)
3
–6
10
0.1
10 –7
Al (OH) 20 4+ Al 3+ 0.01
8
10 –8
0 1 2 3 4 5 6 7 8 9 10 11121314
pH

FIGURE 9.11 Zones of coagulation as affected by alum dosage and pH. [Multiply left y scale in mol Al=L by (594,000=2) to get
concentration in mg Al 2 (SO 4 ) 3 14H 2 O=L.] (From Amirtharajah, A. and Mills, K.M., J. Am. Water Works Assoc., 74(4), 210, April 1982.)

demonstrates how to convert between concentration in mg=L Multiplying each side by ‘‘ ,’’
and mol=L, which is provided for convenience (often chem-
istry fundamentals are not at the fingertips of those who work log [Al(OH) 4 ] log [H ] ¼ log K 4 ¼þ12:35
þ

in the area infrequently). (Ex9:2:3)
Tables CD9.6 and CD9.7 demonstrate further, that is, in
specific detail, the construction of equilibrium diagrams for In p-form,
aluminum and ferric additions to water, respectively. The
p[Al(OH) 4 ] þ pH ¼ pK 4 ¼þ12:35 (Ex9:2:4)

solution steps are enumerated along with the linked plot
showing the associated equilibrium lines.
For plotting, the form is
p[Al(OH) 4 ] ¼ pH þ pK 4 , pK 4 ¼þ12:35 (Ex9:2:5)

Example 9.2 Illustrate Construction of an
Equilibrium Line for Coagulation Diagram or
Given p[Al(OH) 4 ] ¼ pH þ 12:35 (Ex9:2:6)

Let the reaction equation be the equilibrium between
aluminum hydroxide precipitate, Al(OH) 3 (am), and the From Equation Ex9.2.3, the equilibrium line that gives the
aluminate ion, Al(OH) 4 . Al(OH) 3 solid precipitate and [Al(OH) 4 ] relationship ver-

sus pH may be plotted, for example,
Solution
The reaction equation is, from Table 9.5, Equation Al.4,

. pH ¼ 1.0, p[Al(OH) 4 ] ¼ 11.35, or log[Al(OH) 4 ] ¼
11.35, or [Al(OH) 4 ] ¼ 10 11.35 mol=L ¼ 4.47

Al(OH) 3 (s) þ H 2 O ! Al(OH) 4 þ H þ (Al:4) 10 12 mol=L)

. pH ¼ 5.0, p[Al(OH) 4 ] ¼ 7.35, or log[Al(OH) 4 ] ¼
The equilibrium statement is 7.35, or [Al(OH) 4 ] ¼ 10 7.35 mol=L ¼ 4.47

10 8 mol=L)

K 4 ¼ [Al(OH) 4 ][H ] ¼ 10 12:35 (Ex9:2:1) 1..35 2
. pH ¼ 11, p[Al(OH) 4 ] ¼ 1.35, or log[Al(OH) 4 ] ¼

þ
1.35, or [Al(OH) 4 ] ¼ 10 mol=L ¼ 4.47 10

mol=L)
The logarithmic form is
pH ¼ 10, p[Al(OH) 4 ] ¼ 2.35, or log[Al(OH) 4 ] ¼ 2.35,

log [Al(OH) 4 ] þ log [H ] ¼ log K 4 ¼ 12:35 (Ex9:2:2) or [Al(OH) 4 ] ¼ 10 2.35 mol=L ¼ 4.47 10 3 mol=L).

þ