Question

A BOD\(_5\) test is conducted. 25 mL wastewater with ultimate BOD of 75 mg/L is diluted to 300 mL. Initial DO = 6.5 mg/L. DO at 7th day = 1.25 mg/L. Find BOD\(_5\) of wastewater sample (mg/L, rounded off to two decimals).

Hint:

When dilution water is used, always multiply measured oxygen depletion by dilution factor. For BOD\(_5\), interpolate if exact data is missing.

Answer 1

Correct Answer: 54.4

Step 1: Dilution ratio.
Dilution factor = \( \frac{V_\text{total}}{V_\text{wastewater}} = \frac{300}{25} = 12 \).

Step 2: Oxygen consumed in 7 days.
\[ DO_\text{consumed} = 6.5 - 1.25 = 5.25 \, \text{mg/L} \]
Step 3: Correct for dilution.
BOD\(_7\) of sample = \( 5.25 \times 12 = 63 \, \text{mg/L} \).

Step 4: Estimate BOD\(_5\).
Since BOD follows first-order kinetics, BOD\(_5\) ≈ \( 0.99 \times BOD\(_7\) \).
\[ BOD_5 = 0.99 \times 63 = 62.5 \, \text{mg/L} \] Final Answer: \[ \boxed{62.50 \, \text{mg/L}} \]