20 Coagulation, Sedimentation & Process Control Practice Questions & Answers

Every Coagulation, Sedimentation & Process Control practice question from the Water Treatment Operator Practice Test, with the correct answer and a short explanation.

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1. Spring runoff drives raw river turbidity from 15 NTU to over 900 NTU, and the plant has a presedimentation basin ahead of the rapid mix. What is the main reason to place that basin in service?
- A.It drops out heavy silt and sand so coagulation sees a steadier load✓ Answer
- B.It changes dissolved organic colour into settleable floc before the rapid mix
- C.It supplies the disinfection contact time the plant loses when flow rises
- D.It takes out dissolved iron and manganese so no coagulant is consumed
A presedimentation basin removes heavy, fast settling silt and sand by plain gravity before any chemical is added, so the coagulation and flocculation steps receive a lower and much steadier solids load and the coagulant dose does not have to chase every swing in raw turbidity. It is not a disinfection contact basin, and it acts on settleable solids rather than on dissolved colour or dissolved metals.
Source: Recommended Standards for Water Works (Ten States Standards), presedimentation basin provisionsReport a problem with this question
2. A groundwater has a strong rotten egg odour and a high carbon dioxide content. Why is aeration placed ahead of the rest of the treatment train?
- A.It strips hydrogen sulfide and carbon dioxide and adds dissolved oxygen✓ Answer
- B.It kills the bacteria that make sulfide so no disinfectant is needed later
- C.It lowers hardness by driving calcium out of solution as air is mixed in
- D.It adds alkalinity so that less lime is required in the softening step
Aeration is a gas transfer process that works in both directions: it strips dissolved gases and volatile taste and odour compounds such as hydrogen sulfide and carbon dioxide out of the water, and it dissolves oxygen into the water, which helps oxidise iron and manganese downstream. Driving off carbon dioxide raises pH but does not add alkalinity, and aeration neither disinfects nor softens.
Source: CSUS Office of Water Programs, Water Treatment Plant Operation — aeration and gas transferReport a problem with this question
3. A cyanobacteria bloom is building at the intake and that raw water is being drawn through the plant. Which pretreatment approach is correct?
- A.Feed a heavy free chlorine dose at the intake to kill the cells as early as possible
- B.Apply powdered activated carbon right at the pre-oxidant feed point so both work together
- C.Wait until oxidation has broken the cells open, then raise the coagulant dose
- D.Remove the intact cells by coagulation and settling before any strong oxidant✓ Answer
Oxidants lyse cyanobacteria cells and release the toxin held inside them into the water, so the cells must be removed physically while they are still intact. Coagulation, flocculation and settling take out the cells and the toxin they contain, but they do nothing for dissolved toxin already in the water, which needs carbon adsorption or a strong oxidant. Powdered carbon is fed downstream of, and well separated from, any pre-oxidant, because the oxidant otherwise occupies the carbon surface.
Source: Recommended Standards for Water Works, harmful algal bloom pretreatment provisions; US EPA cyanotoxin treatment guidanceReport a problem with this question
4. Clay turbidity in the raw water passes straight through the sedimentation basin when no coagulant is being fed. What keeps these colloidal particles in suspension?
- A.The particles carry like charges that repel and keep them dispersed✓ Answer
- B.The particles are fully dissolved, so no solid phase is present to settle
- C.The particles are lighter than water and rise slowly toward the surface
- D.The particles are held up by dissolved oxygen bubbles on their surface
Colloids carry like surface charges, usually negative, and the resulting electrostatic repulsion keeps them from colliding and joining together; they are also so small that gravity cannot settle them within any practical detention time. The coagulant neutralises that charge so the short range attractive forces can act and the particles can grow into settleable floc. They are suspended solids, not dissolved matter.
Source: CSUS Office of Water Programs, Water Treatment Plant Operation — coagulation and colloidal stabilityReport a problem with this question
5. An operator sharply raises the aluminium sulfate dose on a raw water that has low natural alkalinity. What happens to the alkalinity and the pH?
- A.Both fall, because the coagulant consumes alkalinity and frees carbon dioxide✓ Answer
- B.Both hold steady, because the coagulant settles out before it can react
- C.Both rise, because the coagulant contributes alkalinity as it hydrolyses
- D.Alkalinity falls and pH rises, because sulfate buffers the treated water
Metal salt coagulants hydrolyse in water, consuming alkalinity and producing carbon dioxide, so alkalinity and pH both drop. On a low alkalinity raw water a heavy dose can pull the pH below the range where the coagulant works, hydroxide floc stops forming properly and turbidity carries to the filters, so the operator must add a base such as lime, caustic soda or soda ash and keep a residual alkalinity for the reaction to finish.
Source: CSUS Office of Water Programs, Water Treatment Plant Operation — coagulant hydrolysis and alkalinityReport a problem with this question
6. As the coagulant dose is raised step by step, settled turbidity falls, reaches a minimum, and then begins climbing again at the highest doses. What explains the climb?
- A.The extra coagulant strips all alkalinity and stops any floc from forming
- B.Excess coagulant raises the pH and redissolves the settled sludge blanket
- C.The extra coagulant makes the water denser so the floc can no longer sink
- D.Excess coagulant reverses the particle charge and restabilises the suspension✓ Answer
Past the optimum dose the excess positively charged hydrolysis products do more than neutralise the particle surface, they reverse its charge; the particles repel one another again, stay dispersed, and turbidity rises. That is why the dose response curve has a minimum and why more coagulant is not automatically better: overdosing wastes chemical, consumes alkalinity, and makes more sludge.
Source: AWWA Water System Operations, Water Treatment — coagulant overdosing and charge reversalReport a problem with this question
7. In a six jar test at 20, 25, 30, 35, 40 and 45 mg/L, every jar settles to a turbidity at or below the plant target, and the 45 mg/L jar forms the largest floc. Which dose should be selected?
- A.40 mg/L, because a dose near the top of the range adds a safety margin
- B.20 mg/L, because it is the lowest dose that meets the settled turbidity target✓ Answer
- C.30 mg/L, because a mid range dose leaves room to move in either direction
- D.45 mg/L, because the largest floc gives the best settling and filter run
The optimum dose from a jar test is the lowest dose that meets the settled water turbidity goal, not the jar with the biggest or fastest forming floc. Paying for chemical beyond that point consumes alkalinity, depresses pH, produces extra sludge, and can push the water into charge reversal where turbidity gets worse instead of better.
Source: CSUS Office of Water Programs, Water Treatment Plant Operation — jar test procedure and optimum doseReport a problem with this question
8. Floc leaving the flocculation basin has gone small and pinpoint and settled turbidity is rising, while raw water quality and coagulant dose are unchanged. What is the most likely cause?
- A.The coagulant is injected too far upstream of the rapid mix injection point
- B.The flocculator drives are turning too fast and are shearing the floc apart✓ Answer
- C.The sludge blanket in the basin has been allowed to build up too deeply
- D.The raw water has warmed, which slows floc growth and weakens the floc
Flocculation uses gentle, tapered energy: strong enough to keep particles colliding, gentle enough to leave the growing floc intact. Once floc has formed, too much paddle speed shears it back into pinpoint particles that settle poorly and carry over to the filters, which is exactly the pattern described. Warm water speeds flocculation; it is cold water that slows the reaction and gives weak floc.
Source: CSUS Office of Water Programs, Water Treatment Plant Operation — flocculation and tapered mixing energyReport a problem with this question
9. Winter water near 2 degrees C is producing weak floc that breaks up in the launders even at the optimum coagulant dose. Which adjustment addresses the problem most directly?
- A.Raise the flocculator speed so more collisions build a denser floc
- B.Feed a low dose of polymer as a coagulant aid to toughen the floc✓ Answer
- C.Double the coagulant dose so the heavier metal salt weights the floc
- D.Lower the basin water level so the floc has less distance to settle
Cold water reacts more slowly and is more viscous, so the floc it forms is light and fragile. A small dose of polymer used as a coagulant aid bridges particles together and produces a tougher, denser floc that survives the trip to the settling zone. Doubling the primary coagulant instead risks charge reversal and adds sludge, and raising paddle speed shears the fragile floc further.
Source: AWWA Water System Operations, Water Treatment — coagulant aids and cold water flocculationReport a problem with this question
10. A rectangular sedimentation basin is 80 ft long by 25 ft wide with a side water depth of 12 ft and receives 1.5 MGD. Using surface overflow rate equal to flow divided by basin surface area, what is the rate in gpd/ft2?
- A.1,500 gpd/ft2
- B.7,143 gpd/ft2
- C.750 gpd/ft2✓ Answer
- D.63 gpd/ft2
Surface area is 80 ft x 25 ft = 2,000 ft2, and 1.5 MGD is 1,500,000 gpd, so 1,500,000 / 2,000 = 750 gpd/ft2. Depth does not enter this calculation; it affects detention time instead. The surface overflow rate is numerically the settling velocity of the slowest particle the basin removes completely, so pushing the rate up by raising flow lowers removal efficiency.
Source: Standard sedimentation calculation, surface overflow rate = flow / surface area (AWWA Water System Operations, Water Treatment)Report a problem with this question
11. Tube settlers are installed in an existing rectangular basin and the plant is then able to treat more flow in the same basin. Why does that work?
- A.They weight the floc particles, so heavier solids reach the floor sooner
- B.They raise the horizontal velocity, so more solids reach the sludge zone
- C.They add effective settling area, so each particle falls a short distance✓ Answer
- D.They deepen the settling zone, so each particle is given longer to settle
Inclined tubes or plates divide the depth of the basin into many shallow settling cells, so the effective settling area is many times the plan area of the same footprint and a particle only has to fall the short distance to the nearest surface. Since surface overflow rate is flow divided by settling area, more area allows more flow at the same removal efficiency.
Source: AWWA Water System Operations, Water Treatment — tube and inclined plate settlersReport a problem with this question
12. How does an upflow solids contact unit differ in operation from a conventional rectangular basin train?
- A.It carries out mixing, flocculation and settling in one vessel with a sludge blanket✓ Answer
- B.It settles solids without coagulant because upward flow strains them from water
- C.It separates mixing, flocculation and settling into three basins set in series
- D.It uses chlorine instead of a coagulant to bring the fine solids together
A solids contact or upflow clarifier combines rapid mix, flocculation and sedimentation in a single vessel and recirculates previously formed solids, so the raw water is contacted with existing floc and rises through a slurry blanket that captures fine particles. It still needs a coagulant, and the blanket level and solids concentration become the main operating controls.
Source: Recommended Standards for Water Works, solids contact unit provisionsReport a problem with this question
13. Settled turbidity climbs each sunny afternoon while raw turbidity and coagulant dose hold steady, and a dye test shows water reaching the outlet weirs far ahead of the calculated detention time. What is the most likely cause?
- A.A coagulant feed pump losing its prime for part of each afternoon shift
- B.A sludge collector drive that has stopped and left solids on the floor
- C.A density current from warm surface water short circuiting to the weirs✓ Answer
- D.A raw water pump that is running below its normal speed all afternoon
Solar heating warms the top layer of the basin; that warmer, less dense water floats over the colder water below and runs straight to the outlet, so part of the flow never uses the settling zone. The dye result is the giveaway, because it shows a hydraulic short circuit rather than a chemical feed fault, and the daily on sunny days pattern matches solar heating.
Source: CSUS Office of Water Programs, Water Treatment Plant Operation — short circuiting and density currentsReport a problem with this question
14. In a solids contact softening unit, large clumps of floc are rising to the surface and drifting over the weirs. What is the most likely cause and the correct response?
- A.The unit is overdosed with lime; stop the lime feed and drain the basin
- B.The blanket has grown too thin; cut sludge withdrawal and add coagulant
- C.The influent is too cold; raise the recirculation rate to warm the unit
- D.The blanket has grown too deep; increase sludge withdrawal to lower it✓ Answer
When the slurry blanket is allowed to build above its normal operating level, solids reach the collection zone and wash over the weirs; increasing sludge withdrawal, or blowdown, brings the blanket back down and stops the carryover. Rising clumps can also come from gas released by solids left too long in the unit, which is another reason to keep the blanket turning over.
Source: CSUS Office of Water Programs, Water Treatment Plant Operation — solids contact unit sludge blanket controlReport a problem with this question
15. A groundwater carries both carbonate and noncarbonate hardness and the plant uses lime soda softening. Which statement matches each chemical to the hardness it removes?
- A.Soda ash removes carbonate hardness and lime removes noncarbonate hardness
- B.Soda ash removes both forms and lime is added to settle the residual floc
- C.Lime removes carbonate hardness and soda ash removes noncarbonate hardness✓ Answer
- D.Lime removes both forms and soda ash is added only to raise the finished pH
Carbonate, or temporary, hardness has alkalinity associated with it, so raising pH with lime alone precipitates it as calcium carbonate. Noncarbonate, or permanent, hardness has no such alkalinity, so soda ash must supply the carbonate ion to precipitate it. Softened water is then recarbonated with carbon dioxide to lower the pH and stop calcium carbonate from scaling filters and mains.
Source: CSUS Office of Water Programs, Water Treatment Plant Operation — lime and lime soda ash softeningReport a problem with this question
16. An earthy taste and odour episode occurs for about three weeks each summer. Why does a plant usually meet it with powdered activated carbon rather than a granular carbon bed?
- A.It is dosed during the episode and then discarded with the settled sludge✓ Answer
- B.It holds far more adsorption capacity per pound than granular carbon does
- C.It is regenerated in place after the episode and then returned to service
- D.It removes the odour compounds by oxidising rather than by adsorbing them
Powdered carbon is fed into the process stream for as long as the seasonal episode lasts and leaves the plant with the sludge, so a short annual problem needs only a feeder, not a permanent structure. Granular carbon is a fixed bed that must be built, watched for breakthrough and thermally reactivated, which suits a continuous problem; per unit weight it is the granular carbon that has the greater capacity.
Source: CSUS Office of Water Programs, Water Treatment Plant Operation — taste and odour control with activated carbonReport a problem with this question
17. A well water contains both iron and manganese. Oxidation and filtration is clearing the iron, but manganese is still reaching the distribution system. What is the most likely reason?
- A.Manganese is passing because the water is being oxidised at too high a pH
- B.Manganese forms inside the distribution mains, so plant treatment cannot help
- C.Manganese cannot be oxidised, so only filtration will take it out of water
- D.Manganese needs a higher pH and more oxidant and contact time than iron✓ Answer
Both metals are soluble in their reduced form in anoxic groundwater and must be oxidised to an insoluble form before a filter can catch them, but manganese oxidises much more slowly than iron and needs a higher pH and a longer reaction time. A dose and pH that finish the iron reaction can leave manganese still dissolved, so it passes the filter and precipitates later in the mains as black water and staining.
Source: AWWA Water System Operations, Water Treatment — iron and manganese oxidation and removalReport a problem with this question
18. A small groundwater system switches from oxidation and filtration to polyphosphate sequestration for its iron. What does that change about the water delivered?
- A.The iron is taken out just as completely, with less chemical handling
- B.The iron is changed to a harmless gas that leaves the water at the tap
- C.The iron stays in the water in a held form instead of being taken out✓ Answer
- D.The iron is destroyed by phosphate so the metallic taste disappears
Sequestration binds the iron in a soluble complex so it will not precipitate and stain fixtures, but the metal is still delivered to the customer and the metallic taste remains. The complex can be broken by heating or by a strong oxidant, releasing the iron again, so sequestration is a holding measure for low concentrations and the polyphosphate must be fed ahead of any oxidant, never onto a removal process.
Source: US EPA guidance on iron and manganese in drinking water; Recommended Standards for Water Works, sequestration provisionsReport a problem with this question
19. System demand peaks sharply for about two hours each evening, well above what the plant can produce at its rated capacity. How is that peak normally met?
- A.The plant is pushed above its rated capacity for the length of the peak
- B.Filters are run at twice their normal rate until the evening peak passes
- C.Finished water storage covers the peak while the plant runs at a steady rate✓ Answer
- D.Raw water is blended around the filters to make up the extra volume
Source and treatment facilities are sized for maximum day demand, while peak hour demand and fire flow are supplied from finished water storage. Storage is what decouples the plant from instantaneous demand, so filters, chemical feeds and disinfectant contact can be held at a uniform rate; chasing the peak with the plant is what produces turbidity spikes and short contact time.
Source: Recommended Standards for Water Works — treatment sized for maximum day demand, peak demand met from storageReport a problem with this question
20. Plant flow is raised from 2.0 MGD to 3.0 MGD and the coagulant dose must stay at 12 mg/L. Using feed rate in lb/day equal to dose in mg/L times flow in MGD times 8.34, what is the new feed rate?
- A.300 lb/day✓ Answer
- B.200 lb/day
- C.450 lb/day
- D.100 lb/day
12 mg/L x 3.0 MGD x 8.34 = 300 lb/day. Chemical feed is proportional to flow, so a flow change forces every feeder to be reset: left at the old 200 lb/day setting the dose would drift down to about 8 mg/L. The same flow change simultaneously shortens detention times, raises surface overflow and filter loading rates, and cuts disinfectant contact time, which is why plant flow should be ramped gradually.
Source: Standard chemical feed calculation, lb/day = dose (mg/L) x flow (MGD) x 8.34 (AWWA Water System Operations, Water Treatment)Report a problem with this question

Practice questions written against the standardized Water Treatment Operator Need-to-Know Criteria published by Water Professionals International (formerly the Association of Boards of Certification) and standard references from the CSUS Office of Water Programs and AWWA. This site is not affiliated with or endorsed by WPI/ABC, AWWA, or the US EPA. Operator certification is issued by your state's certifying authority, which sets plant classification tiers, operator grades, eligibility, and the passing standard — confirm those with your state before testing. Contaminant limits and monitoring requirements are set federally and are revised over time, so no answer here should be relied on as a current regulatory value; consult the regulations in force for your system. This bank covers the drinking-water treatment exam only — wastewater treatment, wastewater collection, and water distribution are separate certifications. About the Need-to-Know Criteria →

1. Spring runoff drives raw river turbidity from 15 NTU to over 900 NTU, and the plant has a presedimentation basin ahead of the rapid mix. What is the main reason to place that basin in service?

2. A groundwater has a strong rotten egg odour and a high carbon dioxide content. Why is aeration placed ahead of the rest of the treatment train?

3. A cyanobacteria bloom is building at the intake and that raw water is being drawn through the plant. Which pretreatment approach is correct?

4. Clay turbidity in the raw water passes straight through the sedimentation basin when no coagulant is being fed. What keeps these colloidal particles in suspension?

5. An operator sharply raises the aluminium sulfate dose on a raw water that has low natural alkalinity. What happens to the alkalinity and the pH?

6. As the coagulant dose is raised step by step, settled turbidity falls, reaches a minimum, and then begins climbing again at the highest doses. What explains the climb?

7. In a six jar test at 20, 25, 30, 35, 40 and 45 mg/L, every jar settles to a turbidity at or below the plant target, and the 45 mg/L jar forms the largest floc. Which dose should be selected?

8. Floc leaving the flocculation basin has gone small and pinpoint and settled turbidity is rising, while raw water quality and coagulant dose are unchanged. What is the most likely cause?

9. Winter water near 2 degrees C is producing weak floc that breaks up in the launders even at the optimum coagulant dose. Which adjustment addresses the problem most directly?

10. A rectangular sedimentation basin is 80 ft long by 25 ft wide with a side water depth of 12 ft and receives 1.5 MGD. Using surface overflow rate equal to flow divided by basin surface area, what is the rate in gpd/ft2?

11. Tube settlers are installed in an existing rectangular basin and the plant is then able to treat more flow in the same basin. Why does that work?

12. How does an upflow solids contact unit differ in operation from a conventional rectangular basin train?

13. Settled turbidity climbs each sunny afternoon while raw turbidity and coagulant dose hold steady, and a dye test shows water reaching the outlet weirs far ahead of the calculated detention time. What is the most likely cause?

14. In a solids contact softening unit, large clumps of floc are rising to the surface and drifting over the weirs. What is the most likely cause and the correct response?

15. A groundwater carries both carbonate and noncarbonate hardness and the plant uses lime soda softening. Which statement matches each chemical to the hardness it removes?

16. An earthy taste and odour episode occurs for about three weeks each summer. Why does a plant usually meet it with powdered activated carbon rather than a granular carbon bed?

17. A well water contains both iron and manganese. Oxidation and filtration is clearing the iron, but manganese is still reaching the distribution system. What is the most likely reason?

18. A small groundwater system switches from oxidation and filtration to polyphosphate sequestration for its iron. What does that change about the water delivered?

19. System demand peaks sharply for about two hours each evening, well above what the plant can produce at its rated capacity. How is that peak normally met?

20. Plant flow is raised from 2.0 MGD to 3.0 MGD and the coagulant dose must stay at 12 mg/L. Using feed rate in lb/day equal to dose in mg/L times flow in MGD times 8.34, what is the new feed rate?