--- title: "BOD versus COD and a BOD and COD calculation" subject: "Environment and Pollution Control" chapter: 4 key: "semester_7/environment_and_pollution_control:bod-cod-difference-numerical" source: https://www.bctengineer.com/q/semester_7/environment_and_pollution_control/bod-cod-difference-numerical site: BCT Engineering --- # BOD versus COD and a BOD and COD calculation Environment and Pollution Control · Chapter 4: Water pollution - Asked in: year not given, 3+5 marks - Video: [BOD and COD Numericals / Numericals Based on Biological Oxygen Demand and Chemical Oxygen Demand](https://www.youtube.com/watch?v=LFTeF4jKfeI) — Rapid Chemistry - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/environment_and_pollution_control/bod-cod-difference-numerical - Whole chapter: https://www.bctengineer.com/md/semester_7/environment_and_pollution_control/chapter-4 ## Question (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. ## 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. ## Answer #### (a) BOD and COD | Basis | BOD | COD | |---|---|---| | Meaning | Oxygen used by microorganisms to oxidise biodegradable organic matter | Oxygen equivalent of organic matter oxidised by a strong chemical oxidant | | Oxidant | Bacteria (biological) | Potassium dichromate in boiling acid (chemical) | | Time | 5 days at 20 $^\circ$C | 2-3 hours | | Measures | Only biodegradable matter | Biodegradable plus non-biodegradable matter | | Value | Lower | Higher (COD $\ge$ BOD) | | Use | Pollution strength, design of biological treatment | Quick control test; toxic wastes | #### (b) BOD5 and ultimate BOD Dilution factor $P = \dfrac{\text{volume of sample}}{\text{volume of mixture}} = \dfrac{10}{300} = 0.0333$. $$ \text{BOD}_5 = \frac{D_1 - D_2}{P} = \frac{8.6 - 4.1}{0.0333} = 135\ \text{mg/L} $$ First-order law: $\text{BOD}_t = L_0\,(1 - e^{-kt})$, so $$ L_0 = \frac{\text{BOD}_5}{1 - e^{-0.23\times5}} = \frac{135}{1 - e^{-1.15}} = \frac{135}{1 - 0.3166} = 197.6\ \text{mg/L} $$ #### (c) COD $$ \text{COD} = \frac{(A - B)\times N \times 8000}{V_{sample}} $$ where $A$ = blank (24.0 mL), $B$ = sample titre (14.0 mL), 8000 is the equivalent weight of oxygen in mg/eq. $$ \text{COD} = \frac{(24.0 - 14.0)\times0.1\times8000}{20} = 400\ \text{mg/L} $$ $$ \frac{\text{BOD}_5}{\text{COD}} = \frac{135}{400} = 0.34 $$ 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: BOD$_5$ = 135 mg/L; ultimate BOD $L_0$ = 198 mg/L; COD = 400 mg/L; BOD$_5$/COD = 0.34.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which statement is correct? A. COD uses bacteria and takes 5 days B. BOD measures only biodegradable matter, while COD is usually higher C. COD is always lower than BOD D. BOD uses dichromate Answer: B. BOD is biological and counts only biodegradable matter; COD uses a chemical oxidant and counts more, so COD >= BOD. (Key point: BOD vs COD) 2. With dilution P = 10/300 and DO drop 8.6 - 4.1 = 4.5 mg/L, BOD5 is: A. 13.5 mg/L B. 135 mg/L C. 1350 mg/L D. 4.5 mg/L Answer: B. BOD5 = 4.5/0.0333 = 135 mg/L. (Key point: BOD5 calculation) 3. Ultimate BOD is obtained from: A. L0 = BOD5 x (1 - e^(-5k)) B. L0 = BOD5/(1 - e^(-5k)) C. L0 = BOD5 x e^(-5k) D. L0 = BOD5 + k Answer: B. From BODt = L0(1 - e^(-kt)), giving L0 = 135/0.6834 = about 198 mg/L. (Key point: Ultimate BOD) 4. COD = (24.0 - 14.0) x 0.1 x 8000/20 and BOD5/COD are: A. 400 mg/L and 0.34 B. 40 mg/L and 3.4 C. 400 mg/L and 2.96 D. 800 mg/L and 0.17 Answer: A. COD = 400 mg/L; 135/400 = 0.34, suggesting much non-readily biodegradable load. 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