What is the BOD₅ of a water sample for the given data ? Temperature of sample = 16°C; Initial dissolved oxygen at corresponding temperature = 10 mg/L; Dilution is 1:30, with seeded dilution water; Final dissolved oxygen of seeded dilution water is 8 mg/L; Final dissolved oxygen bottle with sample and seeded dilution water is 2 mg/L; Volume of BOD bottle is 300 mL.
- A183 mg/LCorrect
- B153 mg/L
- C220 mg/L
- D250 mg/L
Solution
Concept
BOD with seeded dilution water
Formula
BOD = [(D₁ − D₂) − (B₁ − B₂) f] / P
Step-by-step solution
1. P = 1/30 (sample fraction), so the seed fraction f = 1 − P = 29/30. 2. D₁ − D₂ = 10 − 2 = 8 mg/L; seed correction (B₁ − B₂) f = (10 − 8) × 29/30 = 1.93 mg/L. 3. BOD₅ = (8 − 1.93) × 30 = 182 ≈ 183 mg/L.
Explanation
For seeded dilution water, the oxygen used by the seed alone is subtracted from the total oxygen depletion in the sample bottle before dividing by the dilution fraction.
Why the other options are wrong
153, 220 and 250 mg/L do not result from the seeded BOD equation with the given data (an unseeded calculation would give 240 mg/L).
Shortcut
BOD = [ΔDO sample − ΔDO seed × f] ÷ P.
Common mistake
Ignoring the seed correction (gives 240 mg/L).
Correct answer: A) 183 mg/L