20 Laboratory Analysis & Sampling Practice Questions & Answers

1. 1. An operator collects a routine distribution coliform sample in a sterile bottle supplied by the certified laboratory. Why must the bottle not be rinsed with sample water before it is filled?

   • A.Rinsing wets the cap threads and lets the bottle leak on the way to the laboratory.
   • B.Rinsing warms the bottle above the temperature the laboratory wants for transit.
   • C.Rinsing washes out the sodium thiosulfate that neutralizes the disinfectant residual.✓ Answer
   • D.Rinsing washes out the nutrient broth the laboratory added for coliform growth.

   The laboratory seals a small amount of sodium thiosulfate inside the bottle before sterilizing it. That dechlorinating agent destroys the chlorine or chloramine residual at the instant of collection, so the disinfectant cannot keep inactivating organisms during transit. Rinsing pours the thiosulfate down the drain, disinfection continues in the bottle, and any bacteria present are under-counted.

   Source: Standard Methods for the Examination of Water and Wastewater, collection of samples for microbiological examination; 40 CFR 141.21Report a problem with this question

2. 2. Before collecting a routine coliform sample from a customer's tap, which preparation gives a sample that represents the water in the distribution main?

   • A.Draw the water standing in the service line first so the sample reflects overnight conditions.
   • B.Run the hot side until the line is heated, then cool the sample before it is shipped.
   • C.Fill the bottle from the nearest fire hydrant while it is discharging at full flow.
   • D.Remove the aerator and any hose, then flush the cold tap at a moderate rate for several minutes.✓ Answer

   Aerators, screens, hoses and swivel spouts hold biofilm and give a positive that came from the fixture rather than the water, and the first water out of the tap is stagnant premise plumbing, not main water. Flushing a smooth-nosed cold tap at moderate flow clears the service line so the bottle is filled with water actually moving in the main; a hydrant stream and a heated hot-side line are both unrepresentative and easily contaminated.

   Source: AWWA Water System Operations, Water Treatment - bacteriological sample collection; Standard Methods, sampling for microbiological examinationReport a problem with this question

3. 3. Samples for trihalomethanes are collected in glass vials that are filled completely so no bubble remains, and the vial already contains a dechlorinating agent. Why must all headspace be excluded?

   • A.Any air space lets the dechlorinating agent settle out and stop quenching the disinfectant.
   • B.Any air space lets the glass leach organic compounds into the water and bias the result high.
   • C.The volatile byproducts move into any air space, so the reported concentration is biased low.✓ Answer
   • D.The laboratory doses the vial by volume, so a bubble changes the dilution it calculates.

   Trihalomethanes are volatile, so they partition out of the water into any gas space in the vial and are lost when the vial is opened, which reports less byproduct than the water actually contained. Glass with an inert liner is used because plastic sorbs organics and can leach its own, and the dechlorinating agent stops further byproduct formation between collection and analysis.

   Source: 40 CFR 136 Table II, container and preservation requirements for purgeable halocarbons (trihalomethanes)Report a problem with this question

4. 4. A cooled and properly preserved sample reaches the certified laboratory after its holding time has already expired. What does that mean for the result?

   • A.The result may be reported as an estimate, because the sample was cooled in transit.
   • B.The result cannot be used, because preservation only slows change and never stops it.✓ Answer
   • C.The result stands, because the preservative stopped all chemical and biological change.
   • D.The result stands, because the holding time restarts when the preservative is added.

   Preservation retards biological action, hydrolysis and volatilization; it does not freeze the sample chemically. The holding time is the interval over which the method was shown to still describe the water as it was at the moment of collection, so once it lapses the value no longer represents that water and cannot support a compliance decision. The correct action is to document the exceedance and collect a new sample.

   Source: 40 CFR 136 Table II, sample preservation and maximum holding timesReport a problem with this question

5. 5. What does a signed chain-of-custody form establish for a sample that leaves the operator's possession?

   • A.An unbroken record of everyone who held the sample from collection to analysis.✓ Answer
   • B.A guarantee that the sample stayed at the required temperature during transit.
   • C.A record that the analytical method used meets the state certification rules.
   • D.A record of the calibration standards run on the instrument for that batch.

   Chain of custody is a documentary chain, not a technical one: each person who takes possession signs and dates the transfer, so it can be shown that the container analyzed is the one collected and that no unaccounted person could have altered it. A gap in that chain makes the sample legally indefensible even when the chemistry is flawless, which is why the form travels with the cooler and a copy stays in the plant record.

   Source: Standard Methods for the Examination of Water and Wastewater, chain-of-custody proceduresReport a problem with this question

6. 6. In a jar test run at six coagulant doses, jar 3 produces the largest and fastest-settling floc, while jar 4 leaves the lowest supernatant turbidity after the settling period. Which jar identifies the optimum dose?

   • A.Jar 4, because the lowest settled turbidity is the result the plant is trying to reach.✓ Answer
   • B.Jar 6, because the highest dose gives the widest margin against raw water changes.
   • C.Jar 3, because the largest and fastest-settling floc is the sign of best coagulation.
   • D.Jar 1, because the lowest dose that forms visible floc is the cheapest to feed.

   The jar test is judged by the water it leaves behind, not by the floc it makes: the optimum dose is the one giving the lowest settled or filtered supernatant turbidity, drawn from just below the surface, together with the best organic removal. Impressive floc often appears at a dose past the optimum, where excess coagulant reverses particle charge and restabilizes the smallest particles, so settled turbidity rises again while sludge production and chemical cost climb.

   Source: CSUS Office of Water Programs, Water Treatment Plant Operation - jar test procedureReport a problem with this question

7. 7. A jar test is set up to find the best coagulant dose for today's raw water. How should the jars be run?

   • A.Each jar holds the same raw water and all are stirred together at one speed, only the dose differing.✓ Answer
   • B.Each jar is dosed and then adjusted to a different pH so both variables are covered at once.
   • C.Each jar is given a different mixing speed so the best pairing of energy and dose appears.
   • D.Each jar is dosed and stirred in turn so floc formation can be watched one jar at a time.

   A jar test is a controlled comparison, so everything except the variable under study must be identical: same raw water, same volume, same rapid-mix and slow-mix speeds and times, same settling period, all jars on one gang stirrer started together. Only then can a difference in settled turbidity be attributed to the dose. Changing dose and pH or mixing energy at the same time leaves the operator unable to say which change produced the result.

   Source: CSUS Office of Water Programs, Water Treatment Plant Operation - jar test procedureReport a problem with this question

8. 8. A jar test uses a 1 percent alum stock solution containing 10,000 mg/L. The operator adds 3.0 mL of stock to a jar holding 2.0 liters (2,000 mL) of raw water. What dose does that jar receive?

   • A.15 mg/L✓ Answer
   • B.60 mg/L
   • C.30 mg/L
   • D.7.5 mg/L

   Find the mass of coagulant added, then spread it over the jar volume: 3.0 mL is 0.0030 L, and 0.0030 L x 10,000 mg/L = 30 mg of alum. Dividing 30 mg by the 2.0 L in the jar gives 15 mg/L. The same arithmetic gives the handy bench rule that 1 mL of a 1 percent stock added to a 1-liter jar is a dose of 10 mg/L.

   Source: CSUS Office of Water Programs, Water Treatment Plant Operation - jar test stock solution dosingReport a problem with this question

9. 9. A bench turbidimeter reads well above the online analyzer on the same filtered water, and the reading drifts steadily downward while the operator watches it. What is the most likely cause?

   • A.True color in the water, which raises the reading until the sample is filtered clear.
   • B.Air bubbles entrained in the sample, scattering light until they rise out of the beam.✓ Answer
   • C.Fingerprints and scratches on the sample cell, which read low until the cell is wiped clean.
   • D.A stale calibration, which shifts the bench reading up and then holds it steady.

   A nephelometer reports the light scattered at right angles to the incident beam, and a bubble scatters light exactly as a particle does, so entrained air always biases the reading high. Because bubbles rise out of the light path, the reading falls as the sample stands, which is the signature of this interference. True color absorbs light and biases turbidity low, and a calibration error would offset the reading without any drift.

   Source: EPA Method 180.1, Determination of Turbidity by Nephelometry - interferencesReport a problem with this question

10. 10. A DPD test on finished water gives a total chlorine residual of 2.4 mg/L and a free chlorine residual of 0.4 mg/L. What does the 2.0 mg/L difference represent?

    • A.Chlorate and chlorite byproducts that the total test counts and the free test misses.
    • B.Combined chlorine, mostly chloramines formed when chlorine reacts with ammonia.✓ Answer
    • C.Chlorine demand exerted by the water between the feed point and the sample tap.
    • D.Hypochlorous acid in reserve that has not yet dissociated in the sample bottle.

    Total chlorine is free plus combined, so subtracting free from total leaves the combined fraction, which is chloramine formed when chlorine reacts with ammonia and organic nitrogen. Free chlorine develops DPD color instantly and must be read at once; the total test needs potassium iodide and about three minutes. Reading the free test late lets combined chlorine bleed into the color and falsely suggests a free residual the water does not hold.

    Source: Standard Methods, DPD colorimetric method for free and combined chlorineReport a problem with this question

11. 11. Plant samples normally fall near pH 7.6. Which practice correctly standardizes the bench pH meter before the day's analyses?

    • A.Calibrate with distilled water as the zero point, then set the slope with one buffer.
    • B.Calibrate with a single buffer nearest the expected value and repeat it once a week.
    • C.Calibrate with any two buffers on hand, since the electrode design fixes the slope.
    • D.Calibrate with two buffers that bracket the expected pH, at the sample temperature.✓ Answer

    A pH measurement is electrometric: the meter converts electrode millivolts to pH using an offset and a slope, and both are established by two buffers that bracket the sample so the reading is interpolated rather than extrapolated. Electrode response and buffer values both shift with temperature, so buffers and sample must be at the same temperature, the meter is standardized each day of use, and distilled water is never a calibration point because it has no buffering capacity.

    Source: Standard Methods, electrometric determination of pHReport a problem with this question

12. 12. A 100-mL sample is titrated to the pH 4.5 endpoint with 12.5 mL of 0.02 N sulfuric acid. Using alkalinity (mg/L as CaCO3) = (mL of titrant x normality x 50,000) / mL of sample, what is the total alkalinity?

    • A.250 mg/L as CaCO3
    • B.62.5 mg/L as CaCO3
    • C.25 mg/L as CaCO3
    • D.125 mg/L as CaCO3✓ Answer

    Substituting into the formula given: 12.5 mL x 0.02 N x 50,000 = 12,500, divided by the 100-mL sample gives 125 mg/L as CaCO3. Titrating to the pH 4.5 endpoint converts all of the bicarbonate, carbonate and hydroxide present, so the number expresses the water's total capacity to neutralize acid, which is the buffer that alum or ferric coagulation consumes.

    Source: Standard Methods, alkalinity by titration to the pH 4.5 endpointReport a problem with this question

13. 13. A raw water sample is visibly turbid. What must the analyst do before reporting true color rather than apparent color?

    • A.Shake the sample so that the particles are evenly suspended, then read the color.
    • B.Warm the sample to room temperature to release gases, then read the color.
    • C.Acidify the sample to dissolve the suspended particles, then read the color.
    • D.Filter or centrifuge the sample to remove the turbidity, then read the color.✓ Answer

    Apparent color is measured on the unfiltered sample and therefore includes light scattered and absorbed by suspended matter, while true color is the color contributed by dissolved substances alone, chiefly natural organic matter and dissolved iron or manganese. Removing turbidity by filtration or centrifugation before comparing the sample against platinum-cobalt standards is what separates the two; shaking or acidifying changes the sample instead of clarifying it.

    Source: Standard Methods, color - apparent and true colorReport a problem with this question

14. 14. Finished water leaving the clearwell has a measured pH of 7.2 and a calculated saturation pH (pHs) of 7.9. Using LSI = pH - pHs, what does the value indicate?

    • A.The water is oversaturated and scale-forming, tending to deposit carbonate scale.
    • B.The water is aggressive, but the negative sign means a protective film is forming.
    • C.The water is balanced, because any index within one unit of zero is called stable.
    • D.The water is undersaturated and aggressive, tending to dissolve carbonate scale.✓ Answer

    LSI is 7.2 minus 7.9, or -0.7. A negative index means the water is undersaturated with respect to calcium carbonate, so it tends to dissolve carbonate rather than deposit it and behaves aggressively toward pipe, solder and fixtures; a positive index indicates a scale-forming water and a value near zero a balanced one. That sign is why corrosion control adjusts pH and alkalinity or feeds orthophosphate to build a protective film.

    Source: AWWA Water System Operations, Water Treatment - Langelier Saturation IndexReport a problem with this question

15. 15. Why does drinking water microbiology test for coliform bacteria instead of testing directly for each waterborne pathogen?

    • A.Coliforms are detected quickly and signal a possible route for fecal contamination.✓ Answer
    • B.Coliforms are the only organisms a state-certified laboratory is allowed to count.
    • C.Coliforms outlast every pathogen through disinfection, so they set the safety margin.
    • D.Coliforms are the most dangerous organisms that survive in treated drinking water.

    Coliforms are indicator organisms, not pathogens. Testing for every waterborne pathogen would be slow, costly and insensitive, because pathogens appear intermittently and in small numbers, while coliforms are abundant wherever fecal material or a treatment and distribution failure creates a pathway, and they grow readily on simple media within a day. E. coli within that group points specifically at fecal contamination.

    Source: 40 CFR 141 Subpart Y, Revised Total Coliform Rule; Standard Methods, the coliform group as indicator organismsReport a problem with this question

16. 16. A routine distribution sample is total coliform positive and E. coli negative. What is the operator's next step?

    • A.Issue a boil water notice at once, because any coliform detection is an acute risk.
    • B.Collect repeat samples at the original site and at points upstream and downstream.✓ Answer
    • C.Wait for next month's routine sample to see whether the detection repeats itself.
    • D.Discard the result as a false positive, because E. coli was absent from the sample.

    A total coliform positive is a warning that a pathway may exist, and it triggers repeat sampling rather than an immediate health action. Repeats taken at the original site and at points upstream and downstream show whether the finding is confined to one tap or extends along the main, which is what tells the operator where to look; repeated positives lead to an assessment that hunts for sanitary defects to correct. E. coli is the acute finding that drives urgent public notice.

    Source: 40 CFR 141 Subpart Y, Revised Total Coliform Rule - repeat monitoring and assessmentsReport a problem with this question

17. 17. Why does a negative coliform result fail to show that Cryptosporidium is absent from finished water?

    • A.Coliform media do not culture protozoan oocysts, which resist chlorine and need microscopy.✓ Answer
    • B.Oocysts are destroyed by the same free chlorine residual that inactivates the coliforms.
    • C.Oocysts pass through the membrane filter pores and end up reported with the total plate count.
    • D.Coliform media grow oocysts only when the plates are held at the fecal coliform temperature.

    Coliform methods culture bacteria on selective media; protozoan oocysts are not bacteria, do not grow on those media, and are found instead by filtering a large volume, concentrating the particles and identifying the oocysts microscopically. Cryptosporidium oocysts also resist free chlorine at practical doses and contact times, so their control depends on the physical barrier of coagulation, sedimentation and filtration, judged by turbidity, and on ultraviolet light or ozone.

    Source: EPA Method 1623, Cryptosporidium and Giardia in water; CSUS Water Treatment Plant Operation, waterborne protozoaReport a problem with this question

18. 18. A laboratory splits a sample and analyzes the second portion alongside the first. What does that duplicate measure?

    • A.The accuracy of the analysis, seen as recovery of a known amount added in.
    • B.The detection limit of the method, the lowest level distinguishable from zero.
    • C.The contamination carried into the batch by reagents, glassware and sampling gear.
    • D.The precision of the analysis, seen as the difference between the two results.✓ Answer

    Each quality control sample answers a different question. A duplicate or split repeats the same water and therefore measures repeatability, reported as relative percent difference, which is precision. A matrix spike or a known check standard measures accuracy as percent recovery, and a method, equipment or field blank measures contamination. The distinction matters because an analysis can be highly precise and still be biased away from the true value.

    Source: Standard Methods, quality assurance and quality control for water analysesReport a problem with this question

19. 19. A laboratory plots each day's check standard result on a control chart. How are the limits on that chart normally set?

    • A.Warning limits at one standard deviation and control limits at two standard deviations.
    • B.Warning limits at two standard deviations and control limits at three deviations.✓ Answer
    • C.Warning limits at three standard deviations and control limits at five deviations.
    • D.Warning limits at the method detection limit and control limits at twice that level.

    The chart is built from the mean and standard deviation of repeated measurements of the same standard. Because roughly ninety-five percent of results fall within two standard deviations and almost all within three, warning limits are drawn at two and control or action limits at three. A point outside the control limits, or a run of points on one side of the mean, means the analysis is out of control: find and correct the cause and reanalyze the batch rather than reporting the data.

    Source: Standard Methods, quality control - control chartsReport a problem with this question

20. 20. Raw water entering the plant measures 12 NTU and the settled water leaving the sedimentation basin measures 1.8 NTU. What percent of the turbidity has been removed?

    • A.68%
    • B.85%✓ Answer
    • C.94%
    • D.15%

    Percent removal is the amount removed divided by the amount that came in: 12 NTU minus 1.8 NTU is 10.2 NTU removed, and 10.2 divided by 12 is 0.85, or 85 percent. The influent value is always the denominator, which is why dividing the effluent by the influent, a common slip, gives the fraction remaining rather than the fraction removed.

    Source: CSUS Office of Water Programs, Water Treatment Plant Operation - percent removal calculationReport 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 →