Louisiana Class III Water Treatment
Operator Exam Practice Questions And
Correct Answers (Verified Answers) Plus
1. A surface water treatment plant uses alum coagulation at pH 6.5. Jar tests show
that increasing pH to 7.2 improves turbidity removal but increases filtered water
aluminum to 0.15 mg/L. What is the most likely explanation for this observation?
A. At higher pH, alum solubility decreases, forming more Al(OH)3(s) that is poorly
removed by filtration.
B. At higher pH, the dominant aluminum species shifts to soluble Al(OH)4-, which
passes through filters.
C. At higher pH, natural organic matter becomes more hydrophobic, reducing alum
demand and leaving excess alum.
D. At higher pH, carbonate alkalinity precipitates aluminum as Al2(CO3)3, which is
filterable but causes turbidity.
Answer: B. At higher pH, the dominant aluminum species shifts to soluble Al(OH)4-,
which passes through filters.
Rationale: As pH increases above the minimum solubility range (pH 5.5-6.5 for alum),
aluminum solubility increases due to formation of soluble aluminate ion (Al(OH)4-), which
is not removed by filtration. Option A is incorrect because solubility increases, not
decreases. Option C is incorrect because NOM becomes more soluble at higher pH. Option
D is incorrect because aluminum carbonate is not a stable precipitate under these
conditions.
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,2. A conventional treatment plant treats 5 MGD with a filter loading rate of 2.5
gpm/ft². If the plant must maintain a minimum filter run time of 24 hours before
headloss reaches terminal, what is the minimum filter area required?
A. 1,389 ft²
B. 1,852 ft²
C. 2,778 ft²
D. 3,472 ft²
Answer: A. 1,389 ft²
Rationale: Flow in gpm = 5 MGD × 694.4 gpm/MGD = 3,472 gpm. Filter area = Flow /
Loading rate = 3,.5 = 1,389 ft². Options B, C, and D result from incorrect unit
conversions or misapplication of loading rate. This calculation ensures adequate filtration
capacity to meet run time goals.
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,3. A groundwater system chlorinates with free chlorine at pH 7.0 and 20°C. The
required CT for 4-log virus inactivation is 6 mg-min/L. If the residual free chlorine is
1.2 mg/L, what is the minimum contact time required?
A. 3.0 minutes
B. 5.0 minutes
C. 7.2 minutes
D. 10.0 minutes
Answer: B. 5.0 minutes
Rationale: CT = residual concentration × contact time. Contact time = CT / residual = 6 /
1.2 = 5.0 minutes. Options A, C, and D result from arithmetic errors or misapplying the CT
formula. This calculation is critical for demonstrating virus inactivation under the Surface
Water Treatment Rule.
4. Which of the following best explains why chloramine disinfection produces fewer
regulated disinfection byproducts (DBPs) than free chlorine?
A. Chloramines are stronger oxidants and react more rapidly with NOM, preventing DBP
formation.
B. Chloramines are less reactive with NOM, reducing the formation of trihalomethanes
and haloacetic acids.
C. Chloramines do not react with bromide, eliminating brominated DBP formation.
D. Chloramines are applied at lower doses, which inherently reduces all DBP formation.
Answer: B. Chloramines are less reactive with NOM, reducing the formation of
trihalomethanes and haloacetic acids.
Rationale: Monochloramine is a weaker oxidant than free chlorine and reacts more slowly
with natural organic matter, resulting in lower THM and HAA formation. Option A is
incorrect because chloramines are weaker oxidants. Option C is incorrect because
chloramines can still form brominated DBPs. Option D is incorrect because dose alone
does not explain the difference; reactivity is key.
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, 5. A lime-soda ash softening plant treats water with a total hardness of 250 mg/L as
CaCO3. If the plant achieves 85% hardness removal, what is the expected effluent
hardness?
A. 25 mg/L as CaCO3
B. 37.5 mg/L as CaCO3
C. 62.5 mg/L as CaCO3
D. 75 mg/L as CaCO3
Answer: B. 37.5 mg/L as CaCO3
Rationale: Hardness removed = 250 × 0.85 = 212.5 mg/L. Effluent hardness = 250 - 212.5
= 37.5 mg/L as CaCO3. Options A, C, and D result from miscalculating the removal
percentage or subtracting incorrectly. This is a fundamental calculation for softening
process control.
6. Under the Stage 2 Disinfectants and Disinfection Byproducts Rule (DBPR), which
compliance metric is used for TTHM and HAA5 at each monitoring location?
A. Annual average of all quarterly samples
B. Locational running annual average (LRAA)
C. System-wide running annual average (SRAA)
D. Maximum contaminant level (MCL) based on a single sample
Answer: B. Locational running annual average (LRAA)
Rationale: The Stage 2 DBPR requires compliance based on the locational running annual
average (LRAA) at each monitoring site, ensuring consistent control across the distribution
system. Option A is incorrect because it averages across all sites. Option C is the Stage 1
approach. Option D is incorrect because MCLs are not based on single samples.
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Operator Exam Practice Questions And
Correct Answers (Verified Answers) Plus
1. A surface water treatment plant uses alum coagulation at pH 6.5. Jar tests show
that increasing pH to 7.2 improves turbidity removal but increases filtered water
aluminum to 0.15 mg/L. What is the most likely explanation for this observation?
A. At higher pH, alum solubility decreases, forming more Al(OH)3(s) that is poorly
removed by filtration.
B. At higher pH, the dominant aluminum species shifts to soluble Al(OH)4-, which
passes through filters.
C. At higher pH, natural organic matter becomes more hydrophobic, reducing alum
demand and leaving excess alum.
D. At higher pH, carbonate alkalinity precipitates aluminum as Al2(CO3)3, which is
filterable but causes turbidity.
Answer: B. At higher pH, the dominant aluminum species shifts to soluble Al(OH)4-,
which passes through filters.
Rationale: As pH increases above the minimum solubility range (pH 5.5-6.5 for alum),
aluminum solubility increases due to formation of soluble aluminate ion (Al(OH)4-), which
is not removed by filtration. Option A is incorrect because solubility increases, not
decreases. Option C is incorrect because NOM becomes more soluble at higher pH. Option
D is incorrect because aluminum carbonate is not a stable precipitate under these
conditions.
Page 1 of 46
,2. A conventional treatment plant treats 5 MGD with a filter loading rate of 2.5
gpm/ft². If the plant must maintain a minimum filter run time of 24 hours before
headloss reaches terminal, what is the minimum filter area required?
A. 1,389 ft²
B. 1,852 ft²
C. 2,778 ft²
D. 3,472 ft²
Answer: A. 1,389 ft²
Rationale: Flow in gpm = 5 MGD × 694.4 gpm/MGD = 3,472 gpm. Filter area = Flow /
Loading rate = 3,.5 = 1,389 ft². Options B, C, and D result from incorrect unit
conversions or misapplication of loading rate. This calculation ensures adequate filtration
capacity to meet run time goals.
Page 2 of 46
,3. A groundwater system chlorinates with free chlorine at pH 7.0 and 20°C. The
required CT for 4-log virus inactivation is 6 mg-min/L. If the residual free chlorine is
1.2 mg/L, what is the minimum contact time required?
A. 3.0 minutes
B. 5.0 minutes
C. 7.2 minutes
D. 10.0 minutes
Answer: B. 5.0 minutes
Rationale: CT = residual concentration × contact time. Contact time = CT / residual = 6 /
1.2 = 5.0 minutes. Options A, C, and D result from arithmetic errors or misapplying the CT
formula. This calculation is critical for demonstrating virus inactivation under the Surface
Water Treatment Rule.
4. Which of the following best explains why chloramine disinfection produces fewer
regulated disinfection byproducts (DBPs) than free chlorine?
A. Chloramines are stronger oxidants and react more rapidly with NOM, preventing DBP
formation.
B. Chloramines are less reactive with NOM, reducing the formation of trihalomethanes
and haloacetic acids.
C. Chloramines do not react with bromide, eliminating brominated DBP formation.
D. Chloramines are applied at lower doses, which inherently reduces all DBP formation.
Answer: B. Chloramines are less reactive with NOM, reducing the formation of
trihalomethanes and haloacetic acids.
Rationale: Monochloramine is a weaker oxidant than free chlorine and reacts more slowly
with natural organic matter, resulting in lower THM and HAA formation. Option A is
incorrect because chloramines are weaker oxidants. Option C is incorrect because
chloramines can still form brominated DBPs. Option D is incorrect because dose alone
does not explain the difference; reactivity is key.
Page 3 of 46
, 5. A lime-soda ash softening plant treats water with a total hardness of 250 mg/L as
CaCO3. If the plant achieves 85% hardness removal, what is the expected effluent
hardness?
A. 25 mg/L as CaCO3
B. 37.5 mg/L as CaCO3
C. 62.5 mg/L as CaCO3
D. 75 mg/L as CaCO3
Answer: B. 37.5 mg/L as CaCO3
Rationale: Hardness removed = 250 × 0.85 = 212.5 mg/L. Effluent hardness = 250 - 212.5
= 37.5 mg/L as CaCO3. Options A, C, and D result from miscalculating the removal
percentage or subtracting incorrectly. This is a fundamental calculation for softening
process control.
6. Under the Stage 2 Disinfectants and Disinfection Byproducts Rule (DBPR), which
compliance metric is used for TTHM and HAA5 at each monitoring location?
A. Annual average of all quarterly samples
B. Locational running annual average (LRAA)
C. System-wide running annual average (SRAA)
D. Maximum contaminant level (MCL) based on a single sample
Answer: B. Locational running annual average (LRAA)
Rationale: The Stage 2 DBPR requires compliance based on the locational running annual
average (LRAA) at each monitoring site, ensuring consistent control across the distribution
system. Option A is incorrect because it averages across all sites. Option C is the Stage 1
approach. Option D is incorrect because MCLs are not based on single samples.
Page 4 of 46
