Wastewater analysis – compute BOD5 from diluted bottle data In a 5-day BOD test at 20 °C, a 5% dilution is used (15 mL sample + 285 mL dilution water in a 300 mL bottle). After incubation, the DO of the sample bottle is 3.80 mg/L and the DO of the dilution-water blank bottle is 8.80 mg/L. The dissolved oxygen originally present in the undiluted sample is 0.80 mg/L. Determine the BOD5 of the sample (mg/L).

Introduction:
Biochemical Oxygen Demand (BOD5 at 20 °C) quantifies oxygen consumed by microbial oxidation of biodegradable organics in a water sample. When dilution is used, the BOD is computed from oxygen depletion in the diluted sample corrected for the depletion observed in the dilution-water blank and then scaled by the dilution fraction.

Given Data / Assumptions:

• Bottle volume = 300 mL; sample volume Vs = 15 mL → dilution fraction P = 0.05.
• Final DO after 5 days: sample bottle = 3.80 mg/L; blank bottle = 8.80 mg/L.
• Initial DO of undiluted sample = 0.80 mg/L (contributes to initial DO of the mixture).
• No seeded correction other than the blank; uniform mixing assumed.

Concept / Approach:

The net oxygen used by the sample in the bottle equals the total depletion in the sample bottle minus the depletion attributable to the dilution water. The initial DO of the sample contributes P * 0.80 mg/L to the mixed bottle at time zero and must be included when computing depletion. BOD5 of the undiluted sample is then the net oxygen used divided by P.

Step-by-Step Solution:

Verification / Alternative check:

If one ignores the small 0.05 × 0.80 = 0.04 mg/L initial oxygen from the sample, the result would be close to 100 mg/L; including it yields the correct refined value of 92 mg/L, matching standard dilution–blank correction.

Why Other Options Are Wrong:

100 mg/L omits the initial sample oxygen and blank fraction correction; 108 and 116 mg/L double-count or misapply the blank; 84 mg/L underestimates depletion.

Common Pitfalls:

Not scaling the blank correction by the dilution water fraction; forgetting to divide the net oxygen used by the dilution fraction P; mixing mg/L and mg in bottle computations.

Final Answer:

92 mg/L