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Worksheet Grade 12 Science

Nitrification Stoichiometry and Alkalinity at Cawdry

A grade 12 science worksheet deriving oxygen demand and alkalinity destruction per gram of nitrogen from a balanced equation, then applying both to a plant's daily load.

Objectives

  1. Students will derive the oxygen and alkalinity ratios of nitrification from a balanced chemical equation.
  2. Students will apply the pounds formula to convert a nitrogen concentration into daily oxygen and alkalinity demand.
  3. Students will explain why nitrification lowers pH and why cold water slows it.

Vocabulary

Alkalinity
Alkalinity is water's ability to resist becoming acidic, acting like a buffer that keeps pH from swinging wildly when acids are added. Water with good alkalinity helps protect fish and other aquatic life from sudden pH changes.

Also: nitrification, stoichiometry, denitrification, buffering capacity, pounds formula

Teacher overview

Cawdry is a fictional water reclamation facility whose permit requires it to convert ammonia to nitrate before discharging. That conversion is carried out by bacteria, and the chemistry sets two hard operating limits: the oxygen the blowers must supply, and the alkalinity the water must have on hand. This worksheet has students derive both ratios from a balanced equation rather than look them up, then apply them to the plant's daily load.

The misconception worth naming is that nitrification removes nitrogen. It does not. Nitrification changes the form of the nitrogen from ammonia to nitrate and leaves the mass of nitrogen unchanged, while consuming oxygen and destroying alkalinity along the way. Removing nitrogen from the water takes a second, anoxic step, which is why items 11 and 12 bring denitrification into the alkalinity budget.

Nitrification Stoichiometry and Alkalinity at Cawdry — Grade 12 Science
Name: ____________________________ Date: ______________

Student Worksheet

Use these values and relationships for the fictional Cawdry facility. Show every step, including units.

  • Balanced nitrification equation: NH4+ + 2 O2 gives NO3- + 2 H+ + H2O
  • Atomic and formula masses: nitrogen 14 g/mol, oxygen gas 32 g/mol, calcium carbonate 100 g/mol
  • Two moles of hydrogen ion consume one mole of alkalinity expressed as calcium carbonate
  • Plant flow: 6.0 MGD
  • Influent ammonia nitrogen: 25 mg/L
  • Influent alkalinity: 220 mg/L as calcium carbonate
  • Operators want at least 50 mg/L of alkalinity remaining after nitrification
  • Pounds formula: pounds per day = concentration (mg/L) x flow (MGD) x 8.34
  • Denitrification returns half a mole of alkalinity as calcium carbonate per mole of nitrogen, which is half of what nitrification destroys

  1. According to the balanced equation, how many moles of oxygen gas are consumed for every mole of ammonium oxidized?

  2. Using the masses given, calculate the grams of oxygen gas consumed per gram of nitrogen oxidized. Round to two decimal places.

  3. Using the masses given and the two-to-one relationship between hydrogen ion and calcium carbonate, calculate the grams of alkalinity as calcium carbonate destroyed per gram of nitrogen oxidized. Round to two decimal places.

  4. Calculate the nitrogen load reaching the plant, in pounds per day, from the influent ammonia nitrogen and the plant flow.

  5. Using your ratio from item 2 and your load from item 4, calculate the oxygen the blowers must supply for nitrification, in pounds per day. Round to the nearest pound.

  6. Using your ratio from item 3, calculate the alkalinity destroyed by nitrification, in milligrams per liter. Round to one decimal place.

  7. Convert your answer to item 6 into pounds per day of alkalinity destroyed. Round to the nearest pound.

  8. Subtract the alkalinity destroyed from the influent alkalinity of 220 mg/L to find the alkalinity remaining after nitrification. Round to one decimal place.

  9. Compare your answer to item 8 with the 50 mg/L the operators want to keep. How many milligrams per liter short is the plant?

  10. Calculate the supplemental alkalinity the plant must add, in pounds per day as calcium carbonate, to close the shortfall from item 9. Round to the nearest pound.

  11. Suppose the plant adds an anoxic zone that denitrifies all of the nitrate it produces. Using the half ratio given above, calculate the alkalinity returned to the water, in milligrams per liter. Round to one decimal place.

  12. Add the alkalinity returned in item 11 to the alkalinity remaining in item 8 to find the alkalinity leaving a plant that both nitrifies and denitrifies. Round to one decimal place.

  13. Explain, using the balanced equation, why nitrification drives pH down and why a plant with low influent alkalinity is at greater risk of a pH crash than one with high alkalinity.

  14. Nitrifying bacteria grow slowly, and their growth rate falls sharply as water temperature drops. Explain what a plant operator would have to change in winter to keep the same degree of nitrification.

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: 2 moles of oxygen gas.
    The balanced equation shows NH4+ reacting with 2 O2, so each mole of ammonium consumes two moles of oxygen gas.

  2. Answer: 4.57 g oxygen per g nitrogen.
    Two moles of oxygen gas weigh 2 x 32 = 64 grams. One mole of nitrogen weighs 14 grams. 64 / 14 = 4.5714, which rounds to 4.57 grams of oxygen per gram of nitrogen.

  3. Answer: 7.14 g calcium carbonate per g nitrogen.
    Each mole of nitrogen oxidized releases 2 moles of hydrogen ion, and 2 moles of hydrogen ion consume 1 mole of calcium carbonate, which weighs 100 grams. 100 / 14 = 7.1428, which rounds to 7.14 grams per gram of nitrogen.

  4. Answer: 1,251 pounds per day.
    25 mg/L x 6.0 MGD x 8.34 = 150 x 8.34 = 1,251 pounds of nitrogen per day.

  5. Answer: 5,717 pounds per day.
    1,251 pounds of nitrogen per day x 4.57 pounds of oxygen per pound of nitrogen = 5,717.07, which rounds to 5,717 pounds of oxygen per day. The ratio derived in grams per gram carries over to pounds per pound because both sides scale the same way.

  6. Answer: 178.5 mg/L.
    7.14 mg of calcium carbonate destroyed per mg of nitrogen x 25 mg/L of nitrogen = 178.5 mg/L of alkalinity destroyed.

  7. Answer: 8,932 pounds per day.
    178.5 mg/L x 6.0 MGD x 8.34 = 178.5 x 50.04 = 8,932.14, which rounds to 8,932 pounds per day. Multiplying the item 4 load by 7.14 gives the same result.

  8. Answer: 41.5 mg/L.
    220 mg/L - 178.5 mg/L = 41.5 mg/L of alkalinity remaining after nitrification.

  9. Answer: 8.5 mg/L short.
    50 mg/L - 41.5 mg/L = 8.5 mg/L. The plant finishes below the cushion its operators want, which leaves pH vulnerable to any increase in ammonia load.

  10. Answer: 425 pounds per day.
    8.5 mg/L x 6.0 MGD x 8.34 = 8.5 x 50.04 = 425.34, which rounds to 425 pounds per day of alkalinity as calcium carbonate.

  11. Answer: 89.3 mg/L.
    Denitrification returns half of the 7.14 ratio, which is 3.57 mg of calcium carbonate per mg of nitrogen. 3.57 x 25 mg/L = 89.25 mg/L, which rounds to 89.3 mg/L.

  12. Answer: 130.8 mg/L.
    41.5 mg/L remaining + 89.25 mg/L returned = 130.75 mg/L, which rounds to 130.8 mg/L. Adding denitrification does more for the alkalinity budget than the chemical feed in item 10, and it also removes nitrogen rather than only converting it.

  13. Answer: Nitrification produces two hydrogen ions per nitrogen, and alkalinity is what absorbs them.
    Every ammonium ion oxidized releases 2 H+, which is an acid load. Alkalinity, mostly bicarbonate, neutralizes those hydrogen ions and holds pH steady. A plant with high influent alkalinity has a large reserve to spend, while a plant starting near the amount that nitrification will destroy runs out of buffer partway through the basin. Once the buffer is gone the pH falls quickly, and nitrifying bacteria slow down below roughly neutral pH, so the process stalls itself.

  14. Answer: The operator must hold the bacteria in the system longer, usually by carrying more solids in the aeration basin.
    Slower growth in cold water means the nitrifiers reproduce more slowly than in summer. If the plant washes them out at the summer rate, the population shrinks and ammonia begins passing through. Keeping a larger inventory of solids, and therefore a longer average residence time for the bacteria, compensates for the slower growth. Doing so also raises oxygen demand, so blower output has to rise with it.

Differentiation

Support

Give students the balanced equation with the mole ratios already annotated above each species, and work items 1 through 3 as a class so every student begins item 4 with the correct 4.57 and 7.14 ratios. Supply a two-column organizer with "concentration in mg/L" on the left and "pounds per day" on the right so the fictional Cawdry conversions in items 4, 7, and 10 follow one visible pattern.

Extension

Ask students to recompute items 4 through 8 for a summer condition in which the fictional plant treats 7.5 MGD at an influent ammonia nitrogen of 31 mg/L, then write two sentences comparing the alkalinity shortfall with the one they found for the base case. Students who finish can estimate how much of the year the plant would need a chemical feed if influent alkalinity varied between 180 and 260 mg/L.

ELL supports

Provide a labeled diagram of the fictional Cawdry aeration basin with arrows for ammonia in, oxygen in, nitrate out, and hydrogen ion produced, and put the six vocabulary terms on the diagram where they apply. Give sentence frames for items 13 and 14, such as "Nitrification lowers pH because ___, and alkalinity matters because ___." Allow the explanation items to be answered in three short sentences rather than a paragraph.

Extensions

Students can extend the fictional Cawdry case into a cost comparison by pricing supplemental alkalinity per pound and comparing a year of chemical feed against the capital cost of an anoxic zone, stating clearly which figures they invented for the exercise. A second option is to have students look up how their state reports ammonia limits for treatment plants and describe how a limit written as a monthly average differs from one written as a daily maximum.