SWRCB - Water Treatment Operator Certification
Examination Grade II 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
Q1
A surface water plant uses alum coagulation at pH 6.5. Jar tests show optimal turbidity removal at 15
mg/L, but plant effluent turbidity spikes after a rain event. Which adjustment is MOST appropriate?
A. Increase alum dose to 25 mg/L without pH adjustment
B. Decrease alum dose and raise pH to 7.5
C. Increase coagulant dose and lower pH to 6.0-6.3
D. Switch to polymer alone and maintain pH
Answer: C
Rationale: Higher turbidity and NOM from stormwater consume coagulant and depress pH; increasing
dose while slightly lowering pH optimizes charge neutralization and sweep floc. Simply raising dose
without pH control can lead to restabilization or inadequate floc formation. Decreasing dose or switching to
polymer alone would likely worsen removal.
Q2
A filter has a loading rate of 6 gpm/ft² and a bed depth of 30 inches. If the media effective size is 0.5 mm
and uniformity coefficient is 1.7, what is the primary concern for turbidity breakthrough?
A. Excessive headloss development
B. Insufficient media depth for adequate L/D ratio
C. High uniformity coefficient causing stratification
D. Low filtration rate leading to anaerobic conditions
Answer: B
Rationale: The L/D ratio (bed depth to effective size) is 30 in / 0.5 mm "H 60, which is below the
recommended minimum of 1000 for this media size, increasing risk of turbidity breakthrough. Headloss
and stratification are secondary to insufficient depth for the given media. Low filtration rate is not the issue
here.
,Q3
Which disinfection byproduct class is most likely to increase when switching from free chlorine to
chloramines for secondary disinfection?
A. Trihalomethanes (THMs)
B. Haloacetic acids (HAAs)
C. N-nitrosodimethylamine (NDMA)
D. Bromate
Answer: C
Rationale: Chloramines can form NDMA, especially in the presence of organic nitrogen precursors,
whereas THM and HAA formation typically decreases. Bromate is associated with ozone, not chloramines.
Thus NDMA is the key DBP of concern when switching to chloramines.
Q4
A plant must achieve 3-log removal of Cryptosporidium. Which treatment barrier combination is MOST
effective and commonly approved?
A. Conventional filtration plus free chlorine
B. Ozone plus chloramines
C. UV disinfection plus conventional filtration
D. Chlorine dioxide plus granular activated carbon
Answer: C
Rationale: UV disinfection provides effective Cryptosporidium inactivation, and conventional filtration
physically removes oocysts, together achieving required log removal. Free chlorine and chloramines are
ineffective against Cryptosporidium. Ozone is effective but may form bromate; the combination with
chloramines is not the standard approved approach for 3-log credit.
Q5
A solids contact clarifier is showing a rising sludge blanket and high effluent turbidity. Which operational
change is LEAST likely to correct the problem?
A. Increasing sludge withdrawal rate
B. Adjusting coagulant dose based on jar tests
C. Increasing influent flow rate to dilute solids
D. Checking and adjusting pH for optimal coagulation
Answer: C
Rationale: Increasing flow rate would worsen the problem by reducing detention time and increasing
hydraulic loading, likely causing more solids carryover. The other actions address the root causes:
excessive sludge blanket, poor coagulation, or pH imbalance. Thus increasing flow is least likely to help.
,Q6
A water system must maintain a minimum chlorine residual of 0.2 mg/L in the distribution system. If the
plant doses 2.0 mg/L and the system has a 4-hour travel time with a decay constant k = 0.15/hr, what is
the estimated residual at the farthest point?
A. 1.10 mg/L
B. 0.90 mg/L
C. 0.55 mg/L
D. 0.20 mg/L
Answer: A
Rationale: Using first-order decay: C = C0 * e^(-kt) = 2.0 * e^(-0.15*4) = 2.0 * e^(-0.6) "H 2.0 * 0.5488 "H
1.10 mg/L. The other options reflect incorrect application of decay or ignoring time. This residual exceeds
the minimum, so compliance is met.
Q7
Which statement BEST describes the role of the Langelier Saturation Index (LSI) in corrosion control?
A. Positive LSI indicates water is corrosive and should be adjusted
B. Negative LSI indicates water is scale-forming and should be adjusted
C. Near-zero LSI indicates water is balanced and minimally corrosive
D. LSI is only applicable to wastewater treatment
Answer: C
Rationale: LSI near zero suggests water is at equilibrium and neither highly scaling nor corrosive.
Positive LSI indicates scaling tendency; negative LSI indicates corrosivity. LSI is used in drinking water
treatment, not solely wastewater.
Q8
A plant uses potassium permanganate to control taste and odors. Which condition would MOST likely
reduce its effectiveness?
A. Presence of reduced iron and manganese
B. High pH and alkalinity
C. Low temperature and high organic load
D. Excess free chlorine residual
Answer: C
Rationale: Low temperature slows reaction kinetics, and high organic load consumes permanganate,
reducing its effectiveness for taste and odor control. Reduced iron and manganese are actually targets for
oxidation. High pH and alkalinity may affect speciation but are less directly inhibitory than cold, organically
loaded water.
, Q9
A membrane filtration plant experiences a sudden increase in transmembrane pressure (TMP) and
decreased permeability. Which is the MOST likely cause?
A. Increased feedwater temperature
B. Membrane fouling or scaling
C. Decreased feedwater turbidity
D. Higher backwash frequency
Answer: B
Rationale: Rising TMP and declining permeability are classic indicators of membrane fouling or scaling,
which increase resistance to flow. Higher temperature would decrease viscosity and TMP, while lower
turbidity and more frequent backwashing would typically improve permeability. Thus fouling/scaling is the
most likely cause.
Q10
A plant must reduce lead levels at the tap. Which corrosion control strategy is MOST effective and
commonly required under the Lead and Copper Rule?
A. Increasing chlorine residual to 2.0 mg/L
B. Adjusting pH and alkalinity to make water less corrosive
C. Adding phosphate-based corrosion inhibitor
D. Switching to chloramines for disinfection
Answer: C
Rationale: Phosphate-based inhibitors form a protective film on pipe surfaces, effectively reducing lead
solubility and release. pH/alkalinity adjustment can help but is often insufficient alone; chlorine and
chloramines can exacerbate lead release. Thus phosphate addition is the most targeted and commonly
required strategy.
Q11
A conventional surface-water plant treats 12 MGD with an alum dose of 35 mg/L. The raw water alkalinity
is 22 mg/L as CaCO3. Which operational consequence is most likely if no alkalinity supplement is added?
A. pH depression below the optimal coagulation range, impairing floc formation and risking lead/copper
corrosion in distribution
B. Elevated finished-water pH requiring additional acid feed for corrosion control
C. Increased floc density and faster settling due to lower pH
D. No effect because alum consumes only turbidity, not alkalinity
Answer: A
Rationale: Alum consumes approximately 0.45–0.5 mg/L of alkalinity (as CaCO3) per mg/L of alum, so a
35 mg/L dose requires roughly 16-18 mg/L of alkalinity; with only 22 mg/L present, residual alkalinity is
marginal and pH can drop below the coagulant's effective range. Low pH also destabilizes the carbonate
film in distribution piping, increasing plumbosolvency. The other options misstate alum chemistry or ignore
the stoichiometric alkalinity demand.
Examination Grade II 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
Q1
A surface water plant uses alum coagulation at pH 6.5. Jar tests show optimal turbidity removal at 15
mg/L, but plant effluent turbidity spikes after a rain event. Which adjustment is MOST appropriate?
A. Increase alum dose to 25 mg/L without pH adjustment
B. Decrease alum dose and raise pH to 7.5
C. Increase coagulant dose and lower pH to 6.0-6.3
D. Switch to polymer alone and maintain pH
Answer: C
Rationale: Higher turbidity and NOM from stormwater consume coagulant and depress pH; increasing
dose while slightly lowering pH optimizes charge neutralization and sweep floc. Simply raising dose
without pH control can lead to restabilization or inadequate floc formation. Decreasing dose or switching to
polymer alone would likely worsen removal.
Q2
A filter has a loading rate of 6 gpm/ft² and a bed depth of 30 inches. If the media effective size is 0.5 mm
and uniformity coefficient is 1.7, what is the primary concern for turbidity breakthrough?
A. Excessive headloss development
B. Insufficient media depth for adequate L/D ratio
C. High uniformity coefficient causing stratification
D. Low filtration rate leading to anaerobic conditions
Answer: B
Rationale: The L/D ratio (bed depth to effective size) is 30 in / 0.5 mm "H 60, which is below the
recommended minimum of 1000 for this media size, increasing risk of turbidity breakthrough. Headloss
and stratification are secondary to insufficient depth for the given media. Low filtration rate is not the issue
here.
,Q3
Which disinfection byproduct class is most likely to increase when switching from free chlorine to
chloramines for secondary disinfection?
A. Trihalomethanes (THMs)
B. Haloacetic acids (HAAs)
C. N-nitrosodimethylamine (NDMA)
D. Bromate
Answer: C
Rationale: Chloramines can form NDMA, especially in the presence of organic nitrogen precursors,
whereas THM and HAA formation typically decreases. Bromate is associated with ozone, not chloramines.
Thus NDMA is the key DBP of concern when switching to chloramines.
Q4
A plant must achieve 3-log removal of Cryptosporidium. Which treatment barrier combination is MOST
effective and commonly approved?
A. Conventional filtration plus free chlorine
B. Ozone plus chloramines
C. UV disinfection plus conventional filtration
D. Chlorine dioxide plus granular activated carbon
Answer: C
Rationale: UV disinfection provides effective Cryptosporidium inactivation, and conventional filtration
physically removes oocysts, together achieving required log removal. Free chlorine and chloramines are
ineffective against Cryptosporidium. Ozone is effective but may form bromate; the combination with
chloramines is not the standard approved approach for 3-log credit.
Q5
A solids contact clarifier is showing a rising sludge blanket and high effluent turbidity. Which operational
change is LEAST likely to correct the problem?
A. Increasing sludge withdrawal rate
B. Adjusting coagulant dose based on jar tests
C. Increasing influent flow rate to dilute solids
D. Checking and adjusting pH for optimal coagulation
Answer: C
Rationale: Increasing flow rate would worsen the problem by reducing detention time and increasing
hydraulic loading, likely causing more solids carryover. The other actions address the root causes:
excessive sludge blanket, poor coagulation, or pH imbalance. Thus increasing flow is least likely to help.
,Q6
A water system must maintain a minimum chlorine residual of 0.2 mg/L in the distribution system. If the
plant doses 2.0 mg/L and the system has a 4-hour travel time with a decay constant k = 0.15/hr, what is
the estimated residual at the farthest point?
A. 1.10 mg/L
B. 0.90 mg/L
C. 0.55 mg/L
D. 0.20 mg/L
Answer: A
Rationale: Using first-order decay: C = C0 * e^(-kt) = 2.0 * e^(-0.15*4) = 2.0 * e^(-0.6) "H 2.0 * 0.5488 "H
1.10 mg/L. The other options reflect incorrect application of decay or ignoring time. This residual exceeds
the minimum, so compliance is met.
Q7
Which statement BEST describes the role of the Langelier Saturation Index (LSI) in corrosion control?
A. Positive LSI indicates water is corrosive and should be adjusted
B. Negative LSI indicates water is scale-forming and should be adjusted
C. Near-zero LSI indicates water is balanced and minimally corrosive
D. LSI is only applicable to wastewater treatment
Answer: C
Rationale: LSI near zero suggests water is at equilibrium and neither highly scaling nor corrosive.
Positive LSI indicates scaling tendency; negative LSI indicates corrosivity. LSI is used in drinking water
treatment, not solely wastewater.
Q8
A plant uses potassium permanganate to control taste and odors. Which condition would MOST likely
reduce its effectiveness?
A. Presence of reduced iron and manganese
B. High pH and alkalinity
C. Low temperature and high organic load
D. Excess free chlorine residual
Answer: C
Rationale: Low temperature slows reaction kinetics, and high organic load consumes permanganate,
reducing its effectiveness for taste and odor control. Reduced iron and manganese are actually targets for
oxidation. High pH and alkalinity may affect speciation but are less directly inhibitory than cold, organically
loaded water.
, Q9
A membrane filtration plant experiences a sudden increase in transmembrane pressure (TMP) and
decreased permeability. Which is the MOST likely cause?
A. Increased feedwater temperature
B. Membrane fouling or scaling
C. Decreased feedwater turbidity
D. Higher backwash frequency
Answer: B
Rationale: Rising TMP and declining permeability are classic indicators of membrane fouling or scaling,
which increase resistance to flow. Higher temperature would decrease viscosity and TMP, while lower
turbidity and more frequent backwashing would typically improve permeability. Thus fouling/scaling is the
most likely cause.
Q10
A plant must reduce lead levels at the tap. Which corrosion control strategy is MOST effective and
commonly required under the Lead and Copper Rule?
A. Increasing chlorine residual to 2.0 mg/L
B. Adjusting pH and alkalinity to make water less corrosive
C. Adding phosphate-based corrosion inhibitor
D. Switching to chloramines for disinfection
Answer: C
Rationale: Phosphate-based inhibitors form a protective film on pipe surfaces, effectively reducing lead
solubility and release. pH/alkalinity adjustment can help but is often insufficient alone; chlorine and
chloramines can exacerbate lead release. Thus phosphate addition is the most targeted and commonly
required strategy.
Q11
A conventional surface-water plant treats 12 MGD with an alum dose of 35 mg/L. The raw water alkalinity
is 22 mg/L as CaCO3. Which operational consequence is most likely if no alkalinity supplement is added?
A. pH depression below the optimal coagulation range, impairing floc formation and risking lead/copper
corrosion in distribution
B. Elevated finished-water pH requiring additional acid feed for corrosion control
C. Increased floc density and faster settling due to lower pH
D. No effect because alum consumes only turbidity, not alkalinity
Answer: A
Rationale: Alum consumes approximately 0.45–0.5 mg/L of alkalinity (as CaCO3) per mg/L of alum, so a
35 mg/L dose requires roughly 16-18 mg/L of alkalinity; with only 22 mg/L present, residual alkalinity is
marginal and pH can drop below the coagulant's effective range. Low pH also destabilizes the carbonate
film in distribution piping, increasing plumbosolvency. The other options misstate alum chemistry or ignore
the stoichiometric alkalinity demand.
