Page 244 - Water and wastewater engineering

COAGULATION AND FLOCCULATION 6-21

Example 6-4 illustrates the impact of the lack of alkalinity on the solution pH and the method
for estimating the amount of base to bring the solution to a pH range that is satisfactory for
coagulation.

Example 6-4. Estimate the pH that results from the addition of 100 mg/L of alum to a water
with no alkalinity, and estimate the amount of sodium hydroxide (NaOH) in mg/L required to
bring the pH to 7.0.

Solution:
4
a. From Example 6-2 , the number of moles of alum added is 1.68 10 moles/L.
b. From Equation 6-9, note that 3 moles of sulfuric acid are produced for each mole of
alum added. Therefore, the moles/L of sulfuric acid is

.
.
3 168 10 4 moles/L ) 5 04 10 4 moles/L
(

c. Sulfuric acid dissociates to form two moles of H for each mole of acid

HSO 4 2 H SO 4 2
2

so the moles/L of H formed is
(
.

.

2 5 04 10 4 moles/L ) 1 01 10 3 moles/L
d. The estimated pH is

pH log H ] log [101 10 3 moles/L 3 00
[
.
.
]
e. From Figure 6-9 a, it is evident that this is out of the range of coagulation with alum.
f. Using NaOH to neutralize the sulfuric acid, the reaction is
HSO 4 2 NaOH Na SO 4 2 H O
2
2
2
Therefore, 2 moles of sodium hydroxide are required to neutralize each mole of sulfuric acid

(
.
.
2 1 01 10 3 moles/L ) 2 02 10 3 moles/L
g. Converting to mg/L
3
3

(202 10 moles/L )(40 g/mole )(10 mg/g ) 80..64 or 81 mg/L
.
Comments:
1. To determine if base needs to be added when alkalinity is present, estimate the amount of
alkalinity present and calculate the amount of alkalinity “destroyed,” as in Example 6-2 .
If the amount destroyed exceeds the amount present, estimate the excess alum and use
this amount to estimate the amount of base to add.