WATER TREATMENT OIT TEST WITH 300 QUESTIONS AND
CORRECT ANSWERS/ WATER TREATMENT OPERATOR IN
TRAINING PRACTICE TEST LATEST 2025/2026 (NEWEST!)
COVERS CORE SUBJECTS INCLUDING WATER TREATMENT PROCESSES,
DISINFECTION, WATER QUALITY, SAFETY, LAB PROCEDURES, MONITORING,
REGULATIONS, AND PLANT OPERATIONS. DESIGNED TO REFLECT THE MOST
CURRENT EXAM FORMAT WITH DETAILED ANSWER EXPLANATIONS AND
PRACTICAL TEST-TAKING STRATEGIES. PERFECT FOR ENTRY-LEVEL OPERATORS
AND THOSE PREPARING FOR THE OIT LICENSE IN 2025 OR 2026.
What is the colour of water in a healthy system (conventional or lagoon)?
A green or brown
B red or purple
C grey
D black –
-Healthy wastewater systems (both conventional and lagoon) have green or brown water because
green means the algae and bacteria are thriving while brown suggests healthy bacterial activity
breaking down organics.
;What is the most common wastewater treatment process in the province of Ontario?
A activated sludge treatment
B lagoon treatment
C RBCs
D trickling filters –
-Activated sludge is the most common wastewater treatment process in Ontario because It
efficiently removes BOD and TSS using aeration and microbial activity. It can also be adapted to
different plant sizes and effluent requirements.
Why do algae raise the pH of a lagoon?
A they produce calcium carbonate which buffers the pH
B they consume Daphnia, which secrete acid
,C they consume the CO2 and carbonates which are precursors to acid formation
D they inhibit the acid-forming bacteria –
-Algae remove CO₂ from the water during photosynthesis. CO₂ helps form carbonic acid (H₂CO₃),
which lowers pH. When algae consume CO₂, the water becomes more alkaline, raising the pH.
Over aeration:
A can lower aerobic bacterial activity
B significantly improves BOD reduction
C significantly improves TSS removal
D wastes money –
-Over-aeration means excessive energy is used to provide oxygen, beyond what is needed. This does
not improve treatment efficiency significantly, but it increases energy costs.
1. Primary treatment is capable of removing up to how much of the TSS?
A 5%
B 30-40%
C 40-60%
D 95% -
-Primary treatment mainly removes suspended solids (TSS) through sedimentation in primary
clarifiers. TSS removal efficiency is typically between 40-60%. The remaining TSS consists of fine
particles and dissolved solids, which require secondary treatment.
Answer: C;Primary treatment is capable of removing up to how much of the BOD?
A 5%
B 30-40%
C 40-60%
D 95% -
-Primary treatment removes settleable organic matter, but most dissolved BOD remains in the
water. 30-40% of BOD is removed, with the rest requiring secondary treatment (biological processes
like activated sludge or lagoons).
;Sludge stabilization in the digesters:
A produces biologically active material that smells really bad
,B produces relatively inert material with a low biological activity, suitable for disposal in a landfill or
by land application
C produces material that can only be incinerated
D produces almost no solids at all –
Sludge stabilization (via anaerobic or aerobic digestion) reduces biological activity, odors, and
pathogens, making sludge safe for disposal or reuse. Digested sludge is relatively inert and can be
Land-applied as fertilizer (biosolids application) or Disposed of in landfills
Answer: B;How does an operator know that it's time to waste sludge to the digesters?
A the secondary clarifier starts to smell
B the secondary clarifier is full
C he/she measures the depth of the sludge blanket in the secondary clarifier
D the secondary clarifier has gone septic - CORRECT ANSWER-Sludge blanket depth monitoring is the
primary method used to determine when sludge should be wasted (removed) to the digesters.If the
sludge blanket gets too deep, solids carry over into the effluent, reducing treatment efficiency.
Raw influent wastewater is entering a 1000 m3 primary clarifier at a rate of 10 L/s. What is the
detention time in the clarifier, in days? (Note: Detention time = volume/flow rate)
A 1.16 d
B 11.6 d
C 27.8 d
D 69.4 d - CORRECT ANSWER-The Detention Time Formula is:
Detention Time = Volume / Flow Rate
Step 1: What do we know?
Volume: 1000m3
Flow Rate: 10 L/s
Detention Time: x d
Step 2: What do we need?
We need to determine our detention time in terms of days. That means converting our units from
seconds to days. In addition, we need our volume and flow rate to use the same units so that they
can calculated properly.
, Step 3: Need 1 - convert Liters to m3
10 L/s x 1m3/1000 L = 0.01 m3/s
[What did I do here?]
I need to change Liters to m3. I know that 1m3 = 1000 Liters, so I create a fraction 1m3/1000 Liters. I
then multiple 10 L/s by that fraction to change Liters into m3. This has no effect on the time unit.
[How did I know to put 1m3 on top and 1000L on bottom?] Because in terms of volume, when you
go from a smaller unit to a bigger unit you divide. When you go from a bigger unit to a smaller unit,
you multiply. In other words, because I am going from L to m3, I write the fraction as 1m3/1000 L so
that I divide by 1000L. If I had been going from m3 to L, the fraction would instead be 1000 L/1m3,
prompting me to multiply by 1000 L.
Step 4: Apply the Formula
Detention Time = 1000m.01 m3/s
Detention Time = 100,000 s
[What did I do here?]
Because I converted L to m3, the Volume and Flow Rate units now match. This means I can
successfully calculate them together, allowing me to cancel out m3 and just leave me with the unit
of time - seconds.
Step 5: Need 2 - Convert seconds to days
[How many seconds are in a day?]
To answer this, start with the standard units of time:
1 min = 60 s
1 hr = 60 min
1 d = 24 hr
60 x 60 x 24 = 86400 s/d
100,000 s / 86400 s/d = 1.1574 days; rounded: 1.16 d
What is the problem with too many filamentous organisms in the activated sludge process?
A not enough dissolved oxygen (DO) in the biological reactor
B a need to increase the RAS rate
C poor settling, and solids carryover in the effluent
D an overly thick sludge that is hard to pump –
CORRECT ANSWERS/ WATER TREATMENT OPERATOR IN
TRAINING PRACTICE TEST LATEST 2025/2026 (NEWEST!)
COVERS CORE SUBJECTS INCLUDING WATER TREATMENT PROCESSES,
DISINFECTION, WATER QUALITY, SAFETY, LAB PROCEDURES, MONITORING,
REGULATIONS, AND PLANT OPERATIONS. DESIGNED TO REFLECT THE MOST
CURRENT EXAM FORMAT WITH DETAILED ANSWER EXPLANATIONS AND
PRACTICAL TEST-TAKING STRATEGIES. PERFECT FOR ENTRY-LEVEL OPERATORS
AND THOSE PREPARING FOR THE OIT LICENSE IN 2025 OR 2026.
What is the colour of water in a healthy system (conventional or lagoon)?
A green or brown
B red or purple
C grey
D black –
-Healthy wastewater systems (both conventional and lagoon) have green or brown water because
green means the algae and bacteria are thriving while brown suggests healthy bacterial activity
breaking down organics.
;What is the most common wastewater treatment process in the province of Ontario?
A activated sludge treatment
B lagoon treatment
C RBCs
D trickling filters –
-Activated sludge is the most common wastewater treatment process in Ontario because It
efficiently removes BOD and TSS using aeration and microbial activity. It can also be adapted to
different plant sizes and effluent requirements.
Why do algae raise the pH of a lagoon?
A they produce calcium carbonate which buffers the pH
B they consume Daphnia, which secrete acid
,C they consume the CO2 and carbonates which are precursors to acid formation
D they inhibit the acid-forming bacteria –
-Algae remove CO₂ from the water during photosynthesis. CO₂ helps form carbonic acid (H₂CO₃),
which lowers pH. When algae consume CO₂, the water becomes more alkaline, raising the pH.
Over aeration:
A can lower aerobic bacterial activity
B significantly improves BOD reduction
C significantly improves TSS removal
D wastes money –
-Over-aeration means excessive energy is used to provide oxygen, beyond what is needed. This does
not improve treatment efficiency significantly, but it increases energy costs.
1. Primary treatment is capable of removing up to how much of the TSS?
A 5%
B 30-40%
C 40-60%
D 95% -
-Primary treatment mainly removes suspended solids (TSS) through sedimentation in primary
clarifiers. TSS removal efficiency is typically between 40-60%. The remaining TSS consists of fine
particles and dissolved solids, which require secondary treatment.
Answer: C;Primary treatment is capable of removing up to how much of the BOD?
A 5%
B 30-40%
C 40-60%
D 95% -
-Primary treatment removes settleable organic matter, but most dissolved BOD remains in the
water. 30-40% of BOD is removed, with the rest requiring secondary treatment (biological processes
like activated sludge or lagoons).
;Sludge stabilization in the digesters:
A produces biologically active material that smells really bad
,B produces relatively inert material with a low biological activity, suitable for disposal in a landfill or
by land application
C produces material that can only be incinerated
D produces almost no solids at all –
Sludge stabilization (via anaerobic or aerobic digestion) reduces biological activity, odors, and
pathogens, making sludge safe for disposal or reuse. Digested sludge is relatively inert and can be
Land-applied as fertilizer (biosolids application) or Disposed of in landfills
Answer: B;How does an operator know that it's time to waste sludge to the digesters?
A the secondary clarifier starts to smell
B the secondary clarifier is full
C he/she measures the depth of the sludge blanket in the secondary clarifier
D the secondary clarifier has gone septic - CORRECT ANSWER-Sludge blanket depth monitoring is the
primary method used to determine when sludge should be wasted (removed) to the digesters.If the
sludge blanket gets too deep, solids carry over into the effluent, reducing treatment efficiency.
Raw influent wastewater is entering a 1000 m3 primary clarifier at a rate of 10 L/s. What is the
detention time in the clarifier, in days? (Note: Detention time = volume/flow rate)
A 1.16 d
B 11.6 d
C 27.8 d
D 69.4 d - CORRECT ANSWER-The Detention Time Formula is:
Detention Time = Volume / Flow Rate
Step 1: What do we know?
Volume: 1000m3
Flow Rate: 10 L/s
Detention Time: x d
Step 2: What do we need?
We need to determine our detention time in terms of days. That means converting our units from
seconds to days. In addition, we need our volume and flow rate to use the same units so that they
can calculated properly.
, Step 3: Need 1 - convert Liters to m3
10 L/s x 1m3/1000 L = 0.01 m3/s
[What did I do here?]
I need to change Liters to m3. I know that 1m3 = 1000 Liters, so I create a fraction 1m3/1000 Liters. I
then multiple 10 L/s by that fraction to change Liters into m3. This has no effect on the time unit.
[How did I know to put 1m3 on top and 1000L on bottom?] Because in terms of volume, when you
go from a smaller unit to a bigger unit you divide. When you go from a bigger unit to a smaller unit,
you multiply. In other words, because I am going from L to m3, I write the fraction as 1m3/1000 L so
that I divide by 1000L. If I had been going from m3 to L, the fraction would instead be 1000 L/1m3,
prompting me to multiply by 1000 L.
Step 4: Apply the Formula
Detention Time = 1000m.01 m3/s
Detention Time = 100,000 s
[What did I do here?]
Because I converted L to m3, the Volume and Flow Rate units now match. This means I can
successfully calculate them together, allowing me to cancel out m3 and just leave me with the unit
of time - seconds.
Step 5: Need 2 - Convert seconds to days
[How many seconds are in a day?]
To answer this, start with the standard units of time:
1 min = 60 s
1 hr = 60 min
1 d = 24 hr
60 x 60 x 24 = 86400 s/d
100,000 s / 86400 s/d = 1.1574 days; rounded: 1.16 d
What is the problem with too many filamentous organisms in the activated sludge process?
A not enough dissolved oxygen (DO) in the biological reactor
B a need to increase the RAS rate
C poor settling, and solids carryover in the effluent
D an overly thick sludge that is hard to pump –
