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BOD versus COD and a BOD and COD calculation

Short answer

BOD measures biodegradable organics using bacteria over 5 days; COD uses chemical oxidation, yielding a higher value that includes non‑biodegradable matter. For the sample: BOD₅ = 135 mg/L, ultimate BOD = 198 mg/L, COD = 400 mg/L, resulting in a BOD₅/COD ratio of 0.34.

(a) Differentiate between BOD and COD. (b) 10 mL of a wastewater sample is diluted to 300 mL with dilution water. The initial dissolved oxygen is 8.6 mg/L and after 5 days at 20 degrees C it is 4.1 mg/L. Calculate the 5-day BOD. If the reaction rate constant k = 0.23 per day (base e), find the ultimate BOD. (c) In a COD test, 20 mL of the same sample required 14.0 mL of 0.1 N ferrous ammonium sulphate (FAS), while the blank required 24.0 mL. Calculate the COD and the BOD5/COD ratio.

(a) BOD and COD

BasisBODCOD
MeaningOxygen used by microorganisms to oxidise biodegradable organic matterOxygen equivalent of organic matter oxidised by a strong chemical oxidant
OxidantBacteria (biological)Potassium dichromate in boiling acid (chemical)
Time5 days at 20 ∘^\circC2-3 hours
MeasuresOnly biodegradable matterBiodegradable plus non-biodegradable matter
ValueLowerHigher (COD ≥\ge BOD)
UsePollution strength, design of biological treatmentQuick control test; toxic wastes

(b) BOD5 and ultimate BOD

Dilution factor P=volume of samplevolume of mixture=10300=0.0333P = \dfrac{\text{volume of sample}}{\text{volume of mixture}} = \dfrac{10}{300} = 0.0333.

BOD5=D1−D2P=8.6−4.10.0333=135 mg/L\text{BOD}_5 = \frac{D_1 - D_2}{P} = \frac{8.6 - 4.1}{0.0333} = 135\ \text{mg/L}

First-order law: BODt=L0 (1−e−kt)\text{BOD}_t = L_0\,(1 - e^{-kt}), so

L0=BOD51−e−0.23×5=1351−e−1.15=1351−0.3166=197.6 mg/LL_0 = \frac{\text{BOD}_5}{1 - e^{-0.23\times5}} = \frac{135}{1 - e^{-1.15}} = \frac{135}{1 - 0.3166} = 197.6\ \text{mg/L}

(c) COD

COD=(A−B)×N×8000Vsample\text{COD} = \frac{(A - B)\times N \times 8000}{V_{sample}}

where AA = blank (24.0 mL), BB = sample titre (14.0 mL), 8000 is the equivalent weight of oxygen in mg/eq.

COD=(24.0−14.0)×0.1×800020=400 mg/L\text{COD} = \frac{(24.0 - 14.0)\times0.1\times8000}{20} = 400\ \text{mg/L} BOD5COD=135400=0.34\frac{\text{BOD}_5}{\text{COD}} = \frac{135}{400} = 0.34

A ratio of about 0.34 indicates that a considerable part of the organic load is not easily biodegradable (domestic sewage is about 0.4-0.8).

Answer: BOD5_5 = 135 mg/L; ultimate BOD L0L_0 = 198 mg/L; COD = 400 mg/L; BOD5_5/COD = 0.34.

✓ Answer checked by Claude Opus 5.5 (high effort)

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