Math problem set Grade 6 Science

Larkspur Basin Chlorine Residual: Dose, Demand, and Decay

A grade 6 problem set on a fictional system's chlorine tests, where students find chlorine demand, the mean and range of residuals, and compare them to the safe range.

Objectives

  1. Students will find chlorine demand as the dose minus the measured residual.
  2. Students will find the range, mean, and median of chlorine residual readings and compare them to a safe range.

Vocabulary

Chlorine residual
Chlorine residual is the small amount of chlorine still left in the water after it leaves the treatment plant. It keeps protecting the water from germs all the way to your faucet.
Disinfection
Disinfection is the step that kills or disables germs in water so it is safe to drink. Water plants usually disinfect with chlorine, ultraviolet light, or ozone.

Also: chlorine demand

Teacher overview

Water plants add chlorine to kill germs. Some of the chlorine gets used up doing that job, which is the chlorine demand, and the amount left over is the residual. Operators want a small residual to stay in the water all the way to the last house, so germs cannot grow back. This set follows the fictional Larkspur Basin system as operators test the residual at points along one pipe. A common error is mixing up the dose with the residual: the dose is what goes in, the residual is what is left, and demand is the difference between them.

Answer Key

  1. Answer: 0.5 mg/L.
    Dose 2.5 mg/L minus residual 2.0 mg/L = 0.5 mg/L of chlorine demand.

  2. Answer: 1.2 mg/L.
    Highest reading 2.0 minus lowest reading 0.8 = 1.2 mg/L.

  3. Answer: 1.4 mg/L.
    2.0 plus 1.7 plus 1.4 plus 1.1 plus 0.8 = 7.0. 7.0 divided by 5 = 1.4 mg/L.

  4. Answer: 1.4 mg/L.
    In order the readings are 0.8, 1.1, 1.4, 1.7, 2.0. The middle value is 1.4 mg/L.

  5. Answer: Yes.
    Every reading falls between 0.2 and 2.0 mg/L, so all five are inside the safe range.

  6. Answer: 2.0 mg/L.
    4.0 mg/L MRDL minus the 2.0 mg/L residual = 2.0 mg/L below the limit.

  7. Answer: 0.3 mg/L.
    From 2.0 to 1.7 is 0.3, from 1.7 to 1.4 is 0.3, from 1.4 to 1.1 is 0.3, and from 1.1 to 0.8 is 0.3. Each drop is 0.3 mg/L.

  8. Answer: 0.5 mg/L.
    The last reading is 0.8 mg/L, and the steady drop is 0.3 mg/L, so 0.8 minus 0.3 = 0.5 mg/L.

Differentiation

Support

Give students the five readings already listed in order for the median, and the sum 7.0 already found for the mean, so they only divide. Write the demand rule "dose minus residual" on a card, and let students use a calculator and a number line for the subtraction steps.

Extension

Ask students to find how many more test points the residual could pass in this fictional system, dropping 0.3 mg/L each time from 0.8 mg/L, before it would fall below the safe minimum of 0.2 mg/L, and to show their reasoning.

ELL supports

Pre-teach "dose," "residual," and "demand" with a picture of chlorine going in, some being used up, and a small amount staying in the water. Provide the frame "Demand is the dose minus the ___," and let students point to the five readings in the list at the top of the problems as they answer.