BOD calculation with dilution: A 2% diluted sewage sample is incubated for 5 days at 20°C and shows an oxygen depletion of 5 ppm. What is the BOD of the original sewage?

Introduction / Context:

BOD (Biochemical Oxygen Demand) quantifies the oxygen required to biologically oxidize biodegradable organic matter. High-strength sewage is often diluted for the 5-day, 20°C test (BOD5) to keep DO within measurement range.

Given Data / Assumptions:

• Dilution fraction f = 2% = 0.02 of original sewage.
• Observed DO depletion in the bottle after 5 days = 5 mg/L (ppm).
• Nitrification inhibition is assumed or negligible; we report BOD5 of the original sample.

Concept / Approach:

The BOD of the original undiluted sample equals the measured depletion divided by the dilution fraction.

Step-by-Step Solution:

Verification / Alternative check:

Sanity check: A 2% sample causing 5 mg/L depletion implies full-strength BOD ≈ 50 * 5 = 250 mg/L, a plausible municipal value.

Why Other Options Are Wrong:

• 200 or 225 ppm: Underestimates; does not reflect the 50× dilution factor.
• None of these: Incorrect because 250 ppm is directly obtained from the calculation.

Common Pitfalls:

• Using 2 (not 0.02) as the dilution factor—remember it is a fraction, not percent value.
• Confusing ppm with % or mg/L; here 1 ppm ≈ 1 mg/L.

Final Answer:

250 ppm