BOD test by dilution method

Importance of BOD value

BOD tells the amount of biodegradable organic matter in wastewater, i.e. its strength and polluting power. It is needed to

• design the treatment units and assess their performance (percentage BOD removal),
• decide the degree of treatment before disposal into a river, from the oxygen demand on the stream,
• check effluent against standards, and
• know the oxygen the sample would take from receiving water (cause of fish kill and odours). A high BOD shows heavily polluted wastewater (raw sewage 200-400 mg/L; treated effluent should be below 30-50 mg/L).

Principle

A diluted sample is incubated for 5 days at 20°C in the dark in a standard bottle, and the oxygen consumed by the microorganisms in oxidising the organic matter is measured by the dissolved oxygen (DO) before and after incubation.

Apparatus and reagents

BOD bottles (300 mL, glass-stoppered), incubator at 20 ± 1°C, pipettes, DO meter or Winkler titration set. Dilution water is prepared from distilled water, with 1 mL each per litre of phosphate buffer, $MgSO_4$, $CaCl_2$ and $FeCl_3$ solutions (nutrients for bacteria), and aerated to saturate with oxygen. Seed (settled domestic sewage) is added if the sample has no bacteria.

Procedure

1. Select dilutions. Sewage needs far more oxygen than water can hold, so it is diluted in such a way that, after 5 days, at least 2 mg/L of DO is used up and at least 1 mg/L of DO remains. Take 1-5% for strong wastes (raw sewage).
2. Prepare the diluted samples: pour a measured volume of sample (e.g. 5 mL) into each of the BOD bottles and fill to the neck with dilution water without trapping air bubbles. Make a blank with only dilution water.
3. Initial DO ($D_1$): measure the DO of one set of bottles immediately (Winkler azide method or probe).
4. Incubate the second set, with stoppers sealed by a water seal, for 5 days at 20°C in the dark (to stop algae making oxygen).
5. Final DO ($D_2$): after 5 days measure the DO of the incubated bottles.
6. Calculate

$$BOD_5=\frac{D_1-D_2}{P}\ \text{mg/L},\qquad P=\frac{\text{volume of sample}}{\text{volume of diluted mixture}}$$

For example, 5 mL of sample in a 300 mL bottle gives $P=5/300=0.0167$. With $D_1=9$ and $D_2=5$ mg/L, $BOD_5=4/0.0167=240$ mg/L. For seeded samples, a seed correction is applied, and the blank is used to check the dilution water (its depletion should be below 0.2 mg/L).

The ultimate BOD can be found by $BOD_t = L(1-10^{-kt})$ if $k$ is known.