Wastewater BOD calculation — dilution method at 20°C for 5 days (BOD5) In a BOD test, 5 mL of wastewater is diluted with 295 mL of aerated dilution water (total volume 300 mL). The diluted sample has initial dissolved oxygen (DO) = 7.8 mg/L and, after 5 days at 20°C, DO = 4.4 mg/L. What is the BOD5 of the original wastewater (mg/L)?

Introduction / Context:
The biochemical oxygen demand over 5 days at 20°C (BOD5) is a standard measure of biodegradable organic pollution. When the sample is diluted, the observed oxygen depletion in the diluted bottle must be scaled by the dilution factor to recover the BOD of the original wastewater.

Given Data / Assumptions:

• Sample volume V_s = 5 mL; dilution water volume = 295 mL; total volume V_t = 300 mL.
• Initial DO = 7.8 mg/L; final DO after 5 days = 4.4 mg/L.
• Seeding/blank corrections are negligible (not stated).

Concept / Approach:
Oxygen depletion in the diluted bottle is ΔDO_diluted = DO_initial − DO_final. The dilution factor is DF = V_t / V_s. Then the BOD of the undiluted wastewater is BOD5 = ΔDO_diluted * DF.

Step-by-Step Solution:
Compute ΔDO_diluted = 7.8 − 4.4 = 3.4 mg/L.Compute DF = 300 / 5 = 60.BOD5 = 3.4 * 60 = 204 mg/L.

Verification / Alternative check:
The result lies in a typical range for municipal wastewater; if a seed correction existed, it would be subtracted before applying the dilution factor, but none is provided here.

Why Other Options Are Wrong:
196, 200, 208 mg/L: Close distractors but not equal to the exact product 3.4 * 60 = 204 mg/L.
180 mg/L: Larger error; does not match the data.

Common Pitfalls:

• Using the ratio V_s / V_t instead of V_t / V_s when scaling.
• Forgetting to apply seed/blank corrections when they are provided.

Final Answer:
204 mg/L