In a BOD test, 5 mL of wastewater (DO=0.0) is mixed with 295 mL of diluting water (DO= 9.0 
mg/L). After 5 days’ inoculation of the 
mixture, DO = 5.2 mg/L. What is the BOD5of 
the wastewater? Assume that all oxygen 
used represents the oxygen demand of the 
wastewater.

The Biochemical Oxygen Demand (BOD) is a measure of the amount of oxygen consumed by microorganisms while decomposing organic matter in water. In a BOD test, we can calculate the 5-day BOD (BOD₅) using the dilution method.

### Given:
- Volume of wastewater = 5 mL
- Volume of diluting water = 295 mL
- Initial dissolved oxygen (DO) of wastewater = 0.0 mg/L
- Initial DO of diluting water = 9.0 mg/L
- DO after 5 days = 5.2 mg/L

To calculate the BOD₅ of the wastewater, we will use the formula:

\[
\text{BOD}_5 = \left(\frac{V_w + V_d}{V_w}\right) \times (DO_i - DO_f)
\]

Where:
- \(V_w\) = Volume of wastewater (mL)
- \(V_d\) = Volume of diluting water (mL)
- \(DO_i\) = Initial dissolved oxygen (mg/L)
- \(DO_f\) = Final dissolved oxygen after 5 days (mg/L)

### Step 1: Calculate total volume of the mixture
\[
V_{\text{total}} = V_w + V_d = 5 + 295 = 300 \, \text{mL}
\]

### Step 2: Calculate the BOD₅
\[
\text{BOD}_5 = \left(\frac{300}{5}\right) \times (9.0 - 5.2)
\]
\[
\text{BOD}_5 = 60 \times 3.8 = 228 \, \text{mg/L}
\]

### Conclusion:
The 5-day Biochemical Oxygen Demand (BOD₅) of the wastewater is **228 mg/L**.

---

Would you like more details on any of the steps, or do you have any questions?

Here are 5 related questions you might find interesting:
1. How does BOD help in assessing water pollution?
2. What is the significance of the 5-day incubation period in the BOD test?
3. How would the BOD calculation change if the volume of wastewater increased?
4. What other factors besides organic matter could affect BOD levels?
5. How can BOD tests be used to improve wastewater treatment processes?

**Tip**: Always ensure the sample bottles for BOD tests are airtight to avoid the absorption of oxygen from the atmosphere during the incubation period.

In a BOD test, 5 mL of wastewater (DO=0.0) is mixed with 295 mL of diluting water (DO= 9.0 mg/L). After 5 days’ inoculation of the mixture, DO = 5.2 mg/L. What is the BOD5 of the wastewater? Assume that all oxygen used represents the oxygen demand of the wastewater.

Learn how to calculate the 5-day Biochemical Oxygen Demand (BOD5) of wastewater using the dilution method. The problem involves mixing wastewater and diluting water, with dissolved oxygen levels monitored over a 5-day incubation period. Suitable for university-level students studying environmental science or chemistry.

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