Math problem set Grade 9 Math

Marbury Phosphorus Removal: Alum Dose and Nutrient Recovery

A grade 9 set using the pounds formula to size chemical dosing for phosphorus removal at a fictional water resource recovery facility, then to weigh nutrient recovery.

Grade: Grade 9
Subject: Math
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

HSN-Q.A.1 Common Core Math

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

How this material aligns: Problems 1, 2, and 5 apply the pounds formula, tracking milligrams per liter, million gallons per day, and the 8.34 constant into pounds per day as this standard requires.

All materials for HSN-Q.A.1 →

HSN-Q.A.3 Common Core Math

Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.

How this material aligns: Problem 3 states no rounding, so students must choose how precisely to report the removed load, then hold that precision through the recovery figure in problem 7, the accuracy judgment this standard names.

All materials for HSN-Q.A.3 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will apply the pounds formula to find phosphorus loads and chemical feed rates in pounds per day.
Students will calculate removal efficiency and the mass of phosphorus recovered as fertilizer.

Vocabulary

Key words: pounds formula, phosphorus, removal efficiency, nutrient recovery

Teacher overview

Marbury is a fictional water resource recovery facility, the modern name for a wastewater plant that recovers useful materials. Phosphorus is the focus here. In a river it feeds algae blooms that starve fish of oxygen, so it is a pollutant, yet the same phosphorus is a valuable fertilizer, so it is also a resource. This set has students size the chemical dose that removes phosphorus and then weigh how much can be recovered.

A common misconception is that a concentration in milligrams per liter is already a weight of chemical. It is not. To find pounds per day, students must combine the concentration with the flow and the 8.34 constant. Skipping the flow leaves the answer in the wrong units and far too small.

Marbury Phosphorus Removal: Alum Dose and Nutrient Recovery — Grade 9 Math

Name: ____________________________ Date: ______________

Problems

Use these facts about the fictional Marbury facility for every problem.

The plant treats 3.0 MGD.
Pounds formula: loading (lbs/day) = concentration (mg/L) × flow (MGD) × 8.34.
Removal efficiency (%) = (influent − effluent) ÷ influent × 100.
Influent total phosphorus is 6.0 mg/L. The permit limit for effluent is 1.0 mg/L.

How many pounds of phosphorus enter the plant each day? Use the influent concentration of 6.0 mg/L. Round to the nearest tenth of a pound.
How many pounds of phosphorus leave the plant each day if the effluent just meets the 1.0 mg/L limit? Round to the nearest tenth of a pound.
How many pounds of phosphorus does the plant remove each day? Subtract your problem 2 answer from your problem 1 answer.
What is the phosphorus removal efficiency, as a percent? Use the influent and effluent concentrations. Round to the nearest tenth of a percent.
The plant adds alum to precipitate the phosphorus at a dose of 90 mg/L. How many pounds of alum does it feed each day? Round to the nearest tenth of a pound.
Alum costs $0.30 per pound. What is the plant's daily alum cost? Round to the nearest cent.
In a proposed upgrade, Marbury would remove this phosphorus biologically instead of with alum, and a recovery system on the digester side stream would capture 40 percent of the phosphorus removed in problem 3 as struvite fertilizer. How many pounds of phosphorus would it recover each day? Round to the nearest tenth of a pound.
In two or three sentences, explain how phosphorus can be both a pollutant and a resource, using the numbers from this set to support your answer.

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

Answer Key

Answer: 150.1 lbs/day.
Loading = 6.0 × 3.0 × 8.34 = 150.12, which rounds to 150.1 lbs/day.
Answer: 25.0 lbs/day.
Loading = 1.0 × 3.0 × 8.34 = 25.02, which rounds to 25.0 lbs/day.
Answer: 125.1 lbs/day.
150.12 − 25.02 = 125.10, which rounds to 125.1 lbs/day.
Answer: 83.3 percent.
Removal % = (6.0 − 1.0) ÷ 6.0 × 100 = 5.0 ÷ 6.0 × 100 = 83.33, which rounds to 83.3%.
Answer: 2,251.8 lbs/day.
Alum feed = 90 × 3.0 × 8.34 = 2,251.8 lbs/day.
Answer: $675.54.
Daily cost = 2,251.8 lbs × $0.30 = $675.54.
Answer: 50.0 lbs/day.
Recovered = 0.40 × 125.10 = 50.04, which rounds to 50.0 lbs/day.
Answer: Phosphorus is a pollutant because the 150.1 pounds arriving each day would feed algae and lower oxygen if released to a river. It is a resource because the plant removes about 125.1 pounds daily and could recover roughly 50 pounds of it as struvite fertilizer, returning that nutrient to farms instead of wasting it.

Differentiation

Support

Provide a worked pounds-formula example with different numbers before problem 1, showing each unit as it cancels. Give students a template that lists the three factors (concentration, flow, 8.34) in separate boxes to multiply in order.

Extension

Ask students to find the yearly alum cost and the yearly mass of phosphorus recovered, then estimate the fertilizer value if recovered struvite is worth a price they choose and state as an assumption. Have them compare that value with the alum cost the upgrade would avoid.

ELL supports

Pre-teach "phosphorus," "dose," and "recovery" with a picture showing algae in a river next to a bag of fertilizer, so the two roles of the nutrient are visible. Provide a sentence frame for problem 8: "Phosphorus is a pollutant because ___, and a resource because ___."