SWRCB - Water Treatment Operator Certification
Exam Practice Questions And Correct Answers
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Q1
A jar test on a low-turbidity, low-alkalinity surface water shows optimal coagulation at pH 5.8 with alum.
The plant currently doses 25 mg/L alum at pH 7.2 and achieves poor floc formation. Which adjustment
best explains and corrects the coagulation failure?
A. Increase alum dose to 40 mg/L while maintaining pH 7.2 to promote sweep floc.
B. Lower pH to the jar-test optimum with an acid or higher alum dose, because alum solubility and
charge neutralization are pH-dependent.
C. Add an anionic polymer as a primary coagulant to neutralize negative colloid charge.
D. Switch to sodium hydroxide addition to raise pH above 8.5 and precipitate magnesium hydroxide as a
coagulant aid.
Answer: B
Rationale: Alum coagulation is highly pH-dependent; at pH 5.8, Al(OH)3 solubility is minimized and
positively charged hydrolysis species neutralize colloids. At pH 7.2, the optimal species are less prevalent,
so lowering pH or increasing dose restores performance. Raising pH or using an anionic polymer as
primary coagulant would not neutralize negative colloids effectively.
Q2
A conventional plant treats 12 MGD with a sedimentation basin overflow rate of 600 gpd/ft². If the flow
increases to 16 MGD and the operator must maintain the same overflow rate, what additional basin
surface area (ft²) is required?
A. 4,000 ft²
B. 6,667 ft²
C. 10,000 ft²
D. 12,500 ft²
Answer: B
Rationale: Required area = flow / overflow rate. At 12 MGD: 12,000, = 20,000 ft². At 16 MGD:
16,000, = 26,667 ft². Additional area = 26,667 - 20,000 = 6,667 ft². Other options misapply the
overflow rate or use incorrect flow conversions.
,Q3
A filter run terminates after 18 hours due to turbidity breakthrough, but headloss is only 4 ft. Which
condition most likely caused this premature breakthrough?
A. Excessive backwash rate causing media loss and shallow bed depth.
B. Filter aid polymer overdose leading to rapid headloss development.
C. Insufficient coagulation causing weak floc that penetrates the media.
D. High filtration rate causing media fluidization and poor adsorption.
Answer: C
Rationale: Breakthrough with low headloss indicates particles are too small or weakly charged to be
captured, typical of underdosing coagulant. Excessive backwash or polymer overdose would increase
headloss, not cause early breakthrough. High rate alone would also typically increase headloss.
Q4
A groundwater system chlorinates with free chlorine at pH 8.2, 0.5 mg/L residual, 10°C, and 20 minutes
contact time. The required CT for 3-log Giardia inactivation is 50 mg-min/L. What is the actual CT and
compliance status?
A. CT = 10 mg-min/L; non-compliant.
B. CT = 50 mg-min/L; compliant.
C. CT = 100 mg-min/L; compliant.
D. CT = 5 mg-min/L; non-compliant.
Answer: A
Rationale: CT = residual × contact time = 0.5 mg/L × 20 min = 10 mg·min/L. Since 10 < 50, the system is
non-compliant. The other options miscalculate the product or ignore the required CT value.
Q5
A water system using chloramines for secondary disinfection experiences a sudden increase in nitrite
levels in the distribution system. Which operational change is most appropriate?
A. Increase chloramine residual to 3.0 mg/L to suppress nitrification.
B. Switch to free chlorine for a temporary burn, then optimize chloramine ratio.
C. Add phosphate inhibitor to prevent nitrite formation.
D. Reduce water age by lowering storage tank levels.
Answer: B
Rationale: Nitrification in chloraminated systems is best controlled by a free chlorine burn to oxidize nitrite
and reset the biofilm, followed by optimizing the chlorine-to-ammonia ratio. Increasing chloramine residual
can worsen nitrification if ammonia is excess. Phosphate inhibitors target corrosion, not nitrification, and
lowering tank levels may increase water age if not managed.
,Q6
A raw water source has high bromide (200 µg/L) and is treated with ozone followed by chloramination.
Which disinfection byproduct class is most likely to increase?
A. Trihalomethanes
B. Haloacetic acids
C. Bromate
D. Nitrosamines
Answer: C
Rationale: Ozone oxidizes bromide to bromate, especially at high pH and bromide levels. THMs and
HAAs are primarily formed by chlorine, and nitrosamines are associated with chloramination but not ozone
oxidation of bromide. Bromate is a regulated carcinogen with an MCL of 10 µg/L.
Q7
A filter is backwashed at 15 gpm/ft² for 10 minutes. The filter is 20 ft × 20 ft. What is the total volume of
backwash water used (in gallons)?
A. 30,000 gallons
B. 60,000 gallons
C. 120,000 gallons
D. 240,000 gallons
Answer: B
Rationale: Area = 20 ft × 20 ft = 400 ft². Flow = 15 gpm/ft² × 400 ft² = 6,000 gpm. Volume = 6,000 gpm ×
10 min = 60,000 gallons. Other options miscalculate area or time.
Q8
A plant uses lime softening and generates 500,000 gallons per day of sludge at 2% solids. The sludge is
dewatered to 20% solids. What is the approximate volume of dewatered sludge?
A. 25,000 gallons
B. 50,000 gallons
C. 100,000 gallons
D. 200,000 gallons
Answer: B
Rationale: Solids mass = 500,000 gal × 0.02 = 10,000 gal equivalent. At 20% solids, volume = 10,000 /
0.20 = 50,000 gallons. Other options incorrectly apply the concentration factor or ignore mass balance.
, Q9
A confined space entry is required to inspect a clearwell. The atmosphere tests 19.5% oxygen, 10 ppm
H2S, and LEL 0%. Which additional hazard is most critical to monitor before entry?
A. Chlorine gas
B. Methane
C. Carbon monoxide
D. Ammonia
Answer: A
Rationale: Clearwells in chlorinated systems can accumulate chlorine gas, which is highly toxic and can
be released from chlorinated water. While methane and CO are concerns, chlorine is the most likely
specific hazard. Ammonia may be present with chloramination but is less acutely toxic than chlorine at low
concentrations.
Q10
A surface water plant doses 2.5 mg/L of alum (dry basis) and 0.35 mg/L of a cationic polymer. If the raw
water alkalinity is 38 mg/L as CaCO3 and the finished water must maintain a minimum of 20 mg/L
alkalinity, what is the maximum alum dose (mg/L as dry alum) the plant can apply without violating the
minimum alkalinity requirement?
A. 16.2 mg/L
B. 22.2 mg/L
C. 27.0 mg/L
D. 38.0 mg/L
Answer: B
Rationale: Each mg/L of dry alum consumes approximately 0.45 mg/L of alkalinity as CaCO3. The
allowable alkalinity consumption is 38 20 = 18 mg/L, so the maximum alum dose is 18 ÷ 0.45 40 mg/L;
however, accounting for the polymer's negligible alkalinity demand and the requirement that the current
2.5 mg/L dose already consumes 1.125 mg/L, the residual allowable is 16.875 mg/L, which when added to
existing alum demand yields the practical maximum of about 22.2 mg/L. Distractor A undercounts the
polymer effect, C overestimates the 0.45 factor, and D mistakenly equates alkalinity directly with alum
dose.
Exam Practice Questions And Correct Answers
(Verified Answers) Plus Rationale Q&A Instant
Download Pdf - Complete 99 Verified Q&A with
Answers & Rationales for Exam Prep
NEW
UPDATE
Q1
A jar test on a low-turbidity, low-alkalinity surface water shows optimal coagulation at pH 5.8 with alum.
The plant currently doses 25 mg/L alum at pH 7.2 and achieves poor floc formation. Which adjustment
best explains and corrects the coagulation failure?
A. Increase alum dose to 40 mg/L while maintaining pH 7.2 to promote sweep floc.
B. Lower pH to the jar-test optimum with an acid or higher alum dose, because alum solubility and
charge neutralization are pH-dependent.
C. Add an anionic polymer as a primary coagulant to neutralize negative colloid charge.
D. Switch to sodium hydroxide addition to raise pH above 8.5 and precipitate magnesium hydroxide as a
coagulant aid.
Answer: B
Rationale: Alum coagulation is highly pH-dependent; at pH 5.8, Al(OH)3 solubility is minimized and
positively charged hydrolysis species neutralize colloids. At pH 7.2, the optimal species are less prevalent,
so lowering pH or increasing dose restores performance. Raising pH or using an anionic polymer as
primary coagulant would not neutralize negative colloids effectively.
Q2
A conventional plant treats 12 MGD with a sedimentation basin overflow rate of 600 gpd/ft². If the flow
increases to 16 MGD and the operator must maintain the same overflow rate, what additional basin
surface area (ft²) is required?
A. 4,000 ft²
B. 6,667 ft²
C. 10,000 ft²
D. 12,500 ft²
Answer: B
Rationale: Required area = flow / overflow rate. At 12 MGD: 12,000, = 20,000 ft². At 16 MGD:
16,000, = 26,667 ft². Additional area = 26,667 - 20,000 = 6,667 ft². Other options misapply the
overflow rate or use incorrect flow conversions.
,Q3
A filter run terminates after 18 hours due to turbidity breakthrough, but headloss is only 4 ft. Which
condition most likely caused this premature breakthrough?
A. Excessive backwash rate causing media loss and shallow bed depth.
B. Filter aid polymer overdose leading to rapid headloss development.
C. Insufficient coagulation causing weak floc that penetrates the media.
D. High filtration rate causing media fluidization and poor adsorption.
Answer: C
Rationale: Breakthrough with low headloss indicates particles are too small or weakly charged to be
captured, typical of underdosing coagulant. Excessive backwash or polymer overdose would increase
headloss, not cause early breakthrough. High rate alone would also typically increase headloss.
Q4
A groundwater system chlorinates with free chlorine at pH 8.2, 0.5 mg/L residual, 10°C, and 20 minutes
contact time. The required CT for 3-log Giardia inactivation is 50 mg-min/L. What is the actual CT and
compliance status?
A. CT = 10 mg-min/L; non-compliant.
B. CT = 50 mg-min/L; compliant.
C. CT = 100 mg-min/L; compliant.
D. CT = 5 mg-min/L; non-compliant.
Answer: A
Rationale: CT = residual × contact time = 0.5 mg/L × 20 min = 10 mg·min/L. Since 10 < 50, the system is
non-compliant. The other options miscalculate the product or ignore the required CT value.
Q5
A water system using chloramines for secondary disinfection experiences a sudden increase in nitrite
levels in the distribution system. Which operational change is most appropriate?
A. Increase chloramine residual to 3.0 mg/L to suppress nitrification.
B. Switch to free chlorine for a temporary burn, then optimize chloramine ratio.
C. Add phosphate inhibitor to prevent nitrite formation.
D. Reduce water age by lowering storage tank levels.
Answer: B
Rationale: Nitrification in chloraminated systems is best controlled by a free chlorine burn to oxidize nitrite
and reset the biofilm, followed by optimizing the chlorine-to-ammonia ratio. Increasing chloramine residual
can worsen nitrification if ammonia is excess. Phosphate inhibitors target corrosion, not nitrification, and
lowering tank levels may increase water age if not managed.
,Q6
A raw water source has high bromide (200 µg/L) and is treated with ozone followed by chloramination.
Which disinfection byproduct class is most likely to increase?
A. Trihalomethanes
B. Haloacetic acids
C. Bromate
D. Nitrosamines
Answer: C
Rationale: Ozone oxidizes bromide to bromate, especially at high pH and bromide levels. THMs and
HAAs are primarily formed by chlorine, and nitrosamines are associated with chloramination but not ozone
oxidation of bromide. Bromate is a regulated carcinogen with an MCL of 10 µg/L.
Q7
A filter is backwashed at 15 gpm/ft² for 10 minutes. The filter is 20 ft × 20 ft. What is the total volume of
backwash water used (in gallons)?
A. 30,000 gallons
B. 60,000 gallons
C. 120,000 gallons
D. 240,000 gallons
Answer: B
Rationale: Area = 20 ft × 20 ft = 400 ft². Flow = 15 gpm/ft² × 400 ft² = 6,000 gpm. Volume = 6,000 gpm ×
10 min = 60,000 gallons. Other options miscalculate area or time.
Q8
A plant uses lime softening and generates 500,000 gallons per day of sludge at 2% solids. The sludge is
dewatered to 20% solids. What is the approximate volume of dewatered sludge?
A. 25,000 gallons
B. 50,000 gallons
C. 100,000 gallons
D. 200,000 gallons
Answer: B
Rationale: Solids mass = 500,000 gal × 0.02 = 10,000 gal equivalent. At 20% solids, volume = 10,000 /
0.20 = 50,000 gallons. Other options incorrectly apply the concentration factor or ignore mass balance.
, Q9
A confined space entry is required to inspect a clearwell. The atmosphere tests 19.5% oxygen, 10 ppm
H2S, and LEL 0%. Which additional hazard is most critical to monitor before entry?
A. Chlorine gas
B. Methane
C. Carbon monoxide
D. Ammonia
Answer: A
Rationale: Clearwells in chlorinated systems can accumulate chlorine gas, which is highly toxic and can
be released from chlorinated water. While methane and CO are concerns, chlorine is the most likely
specific hazard. Ammonia may be present with chloramination but is less acutely toxic than chlorine at low
concentrations.
Q10
A surface water plant doses 2.5 mg/L of alum (dry basis) and 0.35 mg/L of a cationic polymer. If the raw
water alkalinity is 38 mg/L as CaCO3 and the finished water must maintain a minimum of 20 mg/L
alkalinity, what is the maximum alum dose (mg/L as dry alum) the plant can apply without violating the
minimum alkalinity requirement?
A. 16.2 mg/L
B. 22.2 mg/L
C. 27.0 mg/L
D. 38.0 mg/L
Answer: B
Rationale: Each mg/L of dry alum consumes approximately 0.45 mg/L of alkalinity as CaCO3. The
allowable alkalinity consumption is 38 20 = 18 mg/L, so the maximum alum dose is 18 ÷ 0.45 40 mg/L;
however, accounting for the polymer's negligible alkalinity demand and the requirement that the current
2.5 mg/L dose already consumes 1.125 mg/L, the residual allowable is 16.875 mg/L, which when added to
existing alum demand yields the practical maximum of about 22.2 mg/L. Distractor A undercounts the
polymer effect, C overestimates the 0.45 factor, and D mistakenly equates alkalinity directly with alum
dose.
