Question

In a biochemical oxygen demand (BOD) test, 15 ml of wastewater sample was diluted with distilled water to completely fill a 300 ml BOD bottle and incubated at 20 \(^{\circ}\text{C}\) for 5 days. The dissolved oxygen (DO) level before and after incubation are 9.2 mg liter\(^{-1}\) and 4.4 mg liter\(^{-1}\), respectively. The BOD of the sample, in mg liter\(^{-1}\), is . (answer in integer)

Hint:

Scale the observed drop in dissolved oxygen up by the dilution factor of the BOD bottle to get the BOD of the original sample.

Answer 1

Correct Answer: 96

Step 1: Find the dilution factor.
Only a small volume of the raw wastewater is placed in the bottle, and the rest is made up with clean distilled water so the oxygen sensors can be read properly. The dilution factor tells us how many times the sample was diluted: \[ P = \dfrac{\text{volume of BOD bottle}}{\text{volume of sample}} = \dfrac{300}{15} = 20 \]

Step 2: Find the oxygen used up during incubation.
The drop in dissolved oxygen over the 5 day incubation shows how much oxygen the microorganisms in the sample consumed while breaking down the organic matter: \[ \Delta DO = DO_i - DO_f = 9.2 - 4.4 = 4.8\ \text{mg/liter} \]

Step 3: Scale this back up to the strength of the undiluted sample.
Since the sample itself was diluted 20 times before this drop was measured, the true BOD of the full strength wastewater is 20 times the observed drop: \[ BOD = \Delta DO \times P = 4.8 \times 20 = 96\ \text{mg/liter} \]

Final Answer:
The BOD of the wastewater sample is 96 mg per liter. \[ \boxed{96\ \text{mg/liter}} \]