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Worksheet Grade 11 Careers & CTE

Corbenwood Corrosion Control: Orthophosphate Feed and Lead Results

A grade 11 careers worksheet in which a water quality chemist sets an orthophosphate feed rate in pounds, gallons, and mL per minute, then ranks 20 lead sample results.

Objectives

  1. Students will calculate an orthophosphate feed rate in pounds per day, gallons per day, and milliliters per minute.
  2. Students will rank a set of lead sampling results and identify the 90th percentile value.
  3. Students will compare a 90th percentile result against a stated action level and explain why that statistic is used.

Vocabulary

Orthophosphate
A chemical that utilities add to water in very small amounts. It coats the inside of pipes so lead cannot easily get into the water.
Corrosion control
Treatment that keeps water from slowly eating away at the inside of pipes. It helps stop metals like lead from getting into drinking water.
Action level
A trigger number in lead and copper testing. If too many homes test above this level, the utility must take steps like improving treatment and telling the public.

Also: feed rate, 90th percentile

Teacher overview

Corrosion control is chemistry that a plant runs continuously, and its results appear weeks later in samples drawn at a customer's kitchen tap. A water quality chemist sets an orthophosphate dose that forms a protective film inside old pipes, then watches whether the lead results from sampled homes come down. This worksheet follows that work at the fictional Corbenwood Water System, from a target dose in milligrams per liter through the feed pump setting to a set of 20 first-draw results. The misconception this worksheet corrects is treating the average of the sampling results as the compliance number; the rule is written on the 90th percentile, so a system with mostly low homes and a handful of very high ones can average well and still exceed the action level.

Corbenwood Corrosion Control: Orthophosphate Feed and Lead Results — Grade 11 Careers & CTE
Name: ____________________________ Date: ______________

Student Worksheet

The fictional Corbenwood Water System treats 2.4 million gallons per day. The chemist is starting an orthophosphate corrosion control program at a target dose of 1.2 milligrams per liter. The liquid product is 32 percent active orthophosphate by weight and weighs 11.2 pounds per gallon. Use the pounds formula, which is pounds per day equals dose in milligrams per liter times flow in million gallons per day times 8.34. Use 3,785 milliliters per gallon and 1,440 minutes per day.

  1. Find the pounds of active orthophosphate the system must feed each day at the 1.2 milligram per liter dose. Round to the nearest hundredth.
  2. Find the pounds of liquid product per day required to deliver that active weight. Round to the nearest hundredth.
  3. Find the gallons of liquid product per day. Round to the nearest hundredth.
  4. Find the chemical feed pump setting in milliliters per minute. Round to the nearest tenth.

Twelve weeks later the system collects 20 first-draw lead samples from homes with lead service lines. The results, in parts per billion, are 5.9, 2.0, 12.4, 3.4, 22.5, 1.5, 8.4, 4.6, 16.2, 2.8, 7.1, 1.2, 10.8, 6.5, 3.1, 9.6, 2.3, 7.8, 4.0, and 5.2. For a set of 20 samples, the 90th percentile is the 18th value once the results are ordered from lowest to highest.

  1. Order the 20 results and give the 90th percentile value, in parts per billion.
  2. The current lead action level is 15 parts per billion. State how far the 90th percentile from item 5 sits below it, in parts per billion.
  3. Under the Lead and Copper Rule Improvements the action level drops to 10 parts per billion in 2027. State how far the same 90th percentile would sit above that level, in parts per billion.
  4. Convert the 90th percentile from item 5 to milligrams per liter, using 1 milligram per liter equals 1,000 parts per billion.
  5. The chemist decides to raise the dose to 1.5 milligrams per liter. Find the new liquid product feed rate in gallons per day, rounded to the nearest hundredth.
  6. The product costs $2.85 per gallon. Using your rounded answer from item 9 and a 365-day year, find the annual chemical cost. Round to the nearest cent.
  7. The mean of the 20 results is about 6.9 parts per billion. Explain in two or three sentences why the rule is written on the 90th percentile rather than on the mean.
  8. Orthophosphate works only within a certain pH range. Explain in two or three sentences why the chemist tracks pH and alkalinity alongside the phosphate dose rather than treating the dose as the whole program.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: 24.02 pounds per day.
    1.2 milligrams per liter times 2.4 million gallons per day times 8.34 equals 24.0192 pounds per day. First 1.2 times 2.4 equals 2.88, then 2.88 times 8.34 equals 24.0192, which rounds to 24.02 pounds per day.
  2. Answer: 75.06 pounds per day.
    The product is 32 percent active, so 24.0192 divided by 0.32 equals 75.06 pounds of liquid product per day.
  3. Answer: 6.70 gallons per day.
    75.06 pounds per day divided by 11.2 pounds per gallon equals 6.7018, which rounds to 6.70 gallons per day.
  4. Answer: 17.6 milliliters per minute.
    6.7018 gallons per day times 3,785 milliliters per gallon equals 25,366.3 milliliters per day. Dividing by 1,440 minutes per day gives 17.615, which rounds to 17.6 milliliters per minute.
  5. Answer: 12.4 parts per billion.
    Ordered, the results are 1.2, 1.5, 2.0, 2.3, 2.8, 3.1, 3.4, 4.0, 4.6, 5.2, 5.9, 6.5, 7.1, 7.8, 8.4, 9.6, 10.8, 12.4, 16.2, and 22.5. The 18th value is 12.4 parts per billion.
  6. Answer: 2.6 parts per billion below the action level.
    15 minus 12.4 equals 2.6 parts per billion of margin under the current action level.
  7. Answer: 2.4 parts per billion above the 2027 level.
    12.4 minus 10 equals 2.4 parts per billion. The same result that passes today would exceed the level taking effect in 2027, which is why the chemist is optimizing the program now rather than later.
  8. Answer: 0.0124 milligrams per liter.
    12.4 parts per billion divided by 1,000 parts per billion per milligram per liter equals 0.0124 milligrams per liter. Reporting more digits than this would claim precision the laboratory method does not provide.
  9. Answer: 8.38 gallons per day.
    1.5 times 2.4 times 8.34 equals 30.024 pounds of active orthophosphate per day. Dividing by 0.32 gives 93.825 pounds of product, and dividing by 11.2 pounds per gallon gives 8.3772, which rounds to 8.38 gallons per day.
  10. Answer: $8,717.30.
    8.38 gallons per day times 365 days equals 3,058.7 gallons per year. Multiplying by $2.85 per gallon gives $8,717.295, which rounds to $8,717.30.
  11. Answer: The mean can mask a small number of very high homes, which is where the health risk is concentrated.
    Two results in this set, 16.2 and 22.5 parts per billion, are far above the rest, yet they pull the mean up only to about 6.9 because 18 lower values dilute them. A rule written on the mean would let a system pass even if a large share of homes were high, as long as enough homes were low. The 90th percentile looks near the top of the distribution, so a system passes only when at least 90 percent of sampled homes are at or below the action level; in this set the two homes above 15 parts per billion are exactly the top tenth, so the system passes even though those two households are above the level.
  12. Answer: The phosphate film only forms and holds within a workable pH and alkalinity range, so the dose alone does not guarantee protection.
    Orthophosphate builds a low-solubility scale on the interior pipe wall, and both the formation and the stability of that scale depend on the water's pH and its buffering capacity. If pH drifts low or alkalinity is too thin to hold it steady, the same feed rate can produce a much weaker film, and lead results will not come down as expected. The chemist therefore treats the dose, the pH, and the alkalinity as one recipe and adjusts them together.

Differentiation

Support

Give students the fictional system's active pounds per day from item 1 already computed so the product and gallon conversions in items 2 and 3 can proceed independently, and provide the 20 sample results pre-sorted for item 5 so the percentile step is not blocked by an ordering error.

Extension

Ask students to find the dose in milligrams per liter that the fictional system would need in order to feed exactly 10.00 gallons of product per day, working the feed-rate chain in reverse, and to state the assumption about product strength that step relies on.

ELL supports

Pre-teach "orthophosphate," "feed rate," "first draw," and "action level" with a labeled diagram of the fictional plant showing the chemical day tank, the feed pump, and a sampled kitchen tap at the far end. Provide the sentence frame "The 90th percentile is ___ than the mean here because ___" for item 11.

Extensions

Have students look up how their own utility describes its corrosion control treatment in its annual water quality report, then write a short paragraph comparing the chemical named there with the orthophosphate program at the fictional system.