In the BOD test, the calculated BOD for a sample with IDO 8.5 mg/L, FDO 5.1 mg/L, sample used 9 mL, and total sample volume 300 mL is which value?

In the BOD test, the calculated BOD for a sample with IDO 8.5 mg/L, FDO 5.1 mg/L, sample used 9 mL, and total sample volume 300 mL is which value?

• A. 102 mg/L
• B. 113 mg/L
• C. 128 mg/L
• D. 85 mg/L

Answer: (B)

BOD is the amount of oxygen consumed by microorganisms during incubation, shown by the drop in dissolved oxygen in the bottle. The calculation uses how much oxygen is used (the difference between initial DO and final DO) and how diluted the sample is in the bottle (the dilution factor). The oxygen drop is 8.5 mg/L minus 5.1 mg/L, which equals 3.4 mg/L. The dilution factor is the total bottle volume divided by the sample volume: 300 mL / 9 mL = 33.333... Multiply the DO drop by the dilution factor: 3.4 × 33.333... ≈ 113 mg/L. So the calculated BOD is about 113 mg/L.

BOD is the amount of oxygen consumed by microorganisms during incubation, shown by the drop in dissolved oxygen in the bottle. The calculation uses how much oxygen is used (the difference between initial DO and final DO) and how diluted the sample is in the bottle (the dilution factor).

The oxygen drop is 8.5 mg/L minus 5.1 mg/L, which equals 3.4 mg/L.

The dilution factor is the total bottle volume divided by the sample volume: 300 mL / 9 mL = 33.333...

Multiply the DO drop by the dilution factor: 3.4 × 33.333... ≈ 113 mg/L.

So the calculated BOD is about 113 mg/L.