Worksheet Grade 12 Science

Effluent Toxicity and Dilution at the Marledge Outfall

A grade 12 science worksheet on whole effluent toxicity: students find the instream waste concentration, interpolate an LC50, and convert it to toxic units.

Grade: Grade 12
Subject: Science
Material: Worksheet
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

HS-LS2-7 NGSS

Design, evaluate, and refine a solution for reducing the impacts of human activities on the environment and biodiversity.

How this material aligns: Items 10 and 11 have students set the effluent ammonia limit that keeps the receiving stream below its aquatic-life criterion, which is refining a discharge solution to protect stream life.

All materials for HS-LS2-7 →

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: Items 1 through 4 and 10 through 12 carry cubic feet per second, million gallons per day, and milligrams per liter through a mixing calculation, choosing units consistently at every step.

All materials for HSN-Q.A.1 →

HSA-CED.A.1 Common Core Math

Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.

How this material aligns: Item 11 asks students to write and solve a one-variable mass-balance equation for the effluent concentration that meets a stated downstream criterion.

All materials for HSA-CED.A.1 →

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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 calculate the instream waste concentration and dilution ratio for a permitted discharge.
Students will interpolate an LC50 from a survival table and express it as acute toxic units.
Students will use a mass balance to find the effluent concentration that meets a downstream criterion.

Vocabulary

Key words: whole effluent toxicity, instream waste concentration, LC50, toxic unit, mass balance, mixing zone

Teacher overview

Marledge is a fictional town whose water reclamation facility discharges to the Quillmere River. Its permit is written around two questions: how much river water is available to dilute the effluent at low flow, and whether the effluent is toxic to aquatic life at that dilution. This worksheet walks students through both, using a 96-hour survival test and a chemical mass balance for ammonia nitrogen.

One error recurs in this kind of calculation. Students compare the LC50 to 100 percent effluent and decide the discharge is safe because the fish survived the weaker test concentrations. The comparison that matters is between the toxicity of the effluent and the fraction of effluent actually present in the stream after mixing, so the dilution has to be calculated first.

Effluent Toxicity and Dilution at the Marledge Outfall — Grade 12 Science

Name: ____________________________ Date: ______________

Student Worksheet

Use the following values for the fictional Marledge facility and the Quillmere River. Show your work and label every unit.

Effluent flow at the outfall: 4.0 MGD
Upstream river flow during the critical low-flow period: 24.0 cfs
Conversion: 1 MGD = 1.547 cfs
Effluent ammonia nitrogen: 12 mg/L; upstream ammonia nitrogen: 0.2 mg/L
Effluent dissolved oxygen: 6.5 mg/L; upstream dissolved oxygen: 8.4 mg/L
Mass balance: downstream concentration = (effluent flow x effluent concentration + upstream flow x upstream concentration) / total flow

96-hour survival test on fathead minnows, run on the same effluent:

Effluent concentration	Survival	Mortality
6.25 percent	100 percent	0 percent
12.5 percent	95 percent	5 percent
25 percent	90 percent	10 percent
50 percent	45 percent	55 percent
100 percent	10 percent	90 percent

Convert the effluent flow of 4.0 MGD to cubic feet per second. Round to two decimal places.
Calculate the total downstream flow in cubic feet per second, using your answer to item 1 and the upstream flow of 24.0 cfs.
Calculate the instream waste concentration, which is the effluent flow divided by the total downstream flow, expressed as a percent. Round to one decimal place.
Calculate the dilution ratio, which is the total downstream flow divided by the effluent flow. Round to one decimal place.
The LC50 is the effluent concentration at which half the test fish die. Between which two tested concentrations does the LC50 fall? Use the survival table.
Interpolate between those two rows to estimate the LC50 as a percent effluent. Round to one decimal place.
Acute toxic units are calculated as 100 divided by the LC50 expressed in percent. Calculate the toxic units in the undiluted effluent. Round to two decimal places.
Multiply the toxic units from item 7 by the instream waste concentration from item 3, written as a decimal, to find the toxic units present in the river below the mixing zone. Round to two decimal places.
The fictional permit is met when the river below the mixing zone holds fewer than 1.0 toxic unit. Does this discharge meet that condition? Answer yes or no.
Use the mass balance to calculate the ammonia nitrogen concentration in the river below the mixing zone. Round to one decimal place.
The state criterion for ammonia nitrogen in the Quillmere River is 1.9 mg/L below the mixing zone. Write and solve an equation for the effluent ammonia nitrogen concentration that would put the river exactly at that criterion. Round to one decimal place.
Use the mass balance to calculate the dissolved oxygen concentration in the river below the mixing zone. Round to one decimal place.
Permits are written around a low-flow condition rather than an average flow. Explain what would happen to your answer in item 3 if the river were running at 120 cfs instead, and why the low-flow case is the one that governs.
A chemical-specific test looks for one substance at a time. Explain one thing a whole effluent toxicity test can detect that a list of chemical tests could miss.

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: 6.19 cfs.
4.0 MGD x 1.547 cfs per MGD = 6.188 cfs, which rounds to 6.19 cfs. Carry 6.188 through the later items to avoid rounding drift.
Answer: 30.19 cfs.
6.188 cfs + 24.0 cfs = 30.188 cfs, which rounds to 30.19 cfs.
Answer: 20.5 percent.
6.188 / 30.188 = 0.20498. Multiplied by 100, that is 20.5 percent effluent in the river below the mixing zone.
Answer: 4.9; the dilution ratio is 4.9 to 1.
30.188 / 6.188 = 4.878, which rounds to 4.9. Every part of effluent is joined by about 3.9 parts of river water.
Answer: Between 25 percent and 50 percent effluent.
At 25 percent effluent, mortality was 10 percent, which is below half. At 50 percent effluent, mortality was 55 percent, which is above half. The concentration that kills half the fish therefore falls between those two rows.
Answer: 47.2 percent effluent.
Mortality rises from 10 percent to 55 percent across a 25-point span of concentration, a change of 45 points of mortality. Reaching 50 percent mortality takes 40 of those 45 points, so the LC50 sits 40/45 of the way up: 25 + (40/45) x 25 = 25 + 22.2 = 47.2 percent effluent.
Answer: 2.12 toxic units.
100 / 47.2 = 2.1186, which rounds to 2.12 acute toxic units in the undiluted effluent.
Answer: 0.43 toxic units.
2.1186 x 0.205 = 0.4343, which rounds to 0.43 toxic units in the river below the mixing zone.
Answer: Yes.
0.43 toxic units is below the 1.0 toxic unit condition, so the diluted effluent meets it. The undiluted effluent is toxic at 2.12 toxic units, which is why the dilution has to be calculated before any judgment is made.
Answer: 2.6 mg/L.
(6.188 x 12) + (24.0 x 0.2) = 74.256 + 4.8 = 79.056. Dividing by the total flow: 79.056 / 30.188 = 2.619 mg/L, which rounds to 2.6 mg/L.
Answer: 8.5 mg/L.
Let C be the effluent concentration. (6.188C + 4.8) / 30.188 = 1.9. Multiplying both sides by 30.188 gives 6.188C + 4.8 = 57.357. Subtracting: 6.188C = 52.557, so C = 8.494, which rounds to 8.5 mg/L. That is a reduction of about 29 percent from the current 12 mg/L.
Answer: 8.0 mg/L.
(6.188 x 6.5) + (24.0 x 8.4) = 40.222 + 201.6 = 241.822. Dividing by the total flow: 241.822 / 30.188 = 8.011 mg/L, which rounds to 8.0 mg/L. That stays above the 5 mg/L most fish need.
Answer: The instream waste concentration would drop to about 4.9 percent.
At 120 cfs the total flow is 126.188 cfs, so 6.188 / 126.188 = 0.049, or 4.9 percent effluent. More river water means more dilution and less exposure. A permit written around that generous case would leave fish unprotected during the dry weeks when the river runs low, so the low-flow condition is the one that governs.
Answer: It measures the combined effect of everything in the effluent, including substances nobody tested for.
Chemical tests only find what an analyst decides to look for, and they do not show how substances interact. Two chemicals that are each below their limits can still harm fish together. A survival test on the actual effluent captures that combined effect and catches contaminants no one listed.

Differentiation

Support

Give students a filled-in flow diagram for the fictional Marledge outfall showing the two incoming flows and the combined flow, with boxes for effluent flow, upstream flow, and total flow. Complete items 1 and 2 together as a class so every student starts item 3 with correct flows. For item 6, supply a partially worked interpolation with the 45-point mortality span already identified.

Extension

Have students recalculate items 3, 8, and 10 for a wet-weather case in which the fictional facility discharges 6.5 MGD into a river running at 68 cfs, then write two sentences on whether higher flow at the plant always means more risk to the stream. Ask them to explain why the answer depends on how effluent concentration changes with flow.

ELL supports

Provide a bilingual term card for the six vocabulary terms with a one-line definition and a small sketch of the mixing zone. Allow students to answer items 13 and 14 in a labeled diagram plus two short sentences rather than a paragraph. Pair students for items 5 and 6 so the interpolation language can be talked through before it is written.

Extensions

Ask students to research how their own state reports permitted discharges, then compare the reporting categories they find with the two questions the fictional Marledge permit asks. Students can also design a survival test with a sixth concentration between 25 and 50 percent effluent and explain how that added row would sharpen the LC50 estimate they made in item 6.