1. CONCEPTUAL FRAMEWORK & CHEMICAL OXIDATION

A. Definition & Core Principle

Chemical Oxygen Demand (COD)

Chemical Oxygen Demand (COD) is the amount of oxygen required to chemically oxidize oxidizable substances present in a water or wastewater sample.

This includes:

• Biodegradable organic compounds

• Non-biodegradable or refractory organic compounds

• Some oxidizable inorganic substances

BOD vs COD

BOD measures oxygen consumed through biological oxidation.

COD measures oxygen demand through chemical oxidation using a strong oxidizing agent.

Therefore, COD is generally greater than BOD for the same wastewater.

Oxygen Demand Hierarchy

A useful conceptual relationship is:

ThOD ≥ COD ≥ BODᵤ ≥ BOD₅

Where:

• ThOD = Theoretical Oxygen Demand

• COD = Chemical Oxygen Demand

• BODᵤ = Ultimate BOD

• BOD₅ = 5-day BOD

These relationships are useful as a general conceptual framework, although the exact relationship depends on the wastewater composition and analytical method.

BOD vs COD — Test Duration

BOD₅

• Biological test

• 5 days

• 20°C

• Requires microorganisms

• Measures oxygen consumed biologically

COD

• Chemical test

• Usually completed in about 2–3 hours using the conventional dichromate digestion method

• Uses strong chemical oxidizing conditions

• Much faster than BOD₅

Engineering Significance

Because COD can be determined much faster than BOD₅, it is widely used for:

• Industrial wastewater monitoring

• Treatment-plant process control

• Rapid assessment of organic pollution load

OXYGEN DEMAND SPECTRUM

                         OXYGEN DEMAND
                              │
                              ▼
                             COD
                              │
            ┌─────────────────┴─────────────────┐
            │                                   │
            ▼                                   ▼
     Biodegradable                         Refractory /
       fraction                         poorly biodegradable
            │                                   │
            ▼                                   ▼
          BODᵤ                           Complex organics
            │
            ▼
          BOD₅

Examples of Refractory Organic Matter

Depending on the wastewater source, refractory compounds may include:

• Synthetic dyes

• Lignin

• Resins

• Certain pesticides

• Complex industrial solvents

• Other persistent organic compounds

B. LABORATORY COD DIGESTION CHEMISTRY

The conventional COD test uses a strong oxidizing system under acidic and elevated-temperature conditions.

1. Primary Oxidizing Agent

Potassium dichromate — K₂Cr₂O₇

It is used in the presence of concentrated sulphuric acid.

The conventional digestion is carried out at approximately:

150°C for about 2 hours

2. Catalyst

Silver sulphate — Ag₂SO₄

Silver sulphate acts as a catalyst and improves oxidation of certain organic compounds, particularly some straight-chain aliphatic compounds.

3. Chloride Interference Control

Mercuric sulphate — HgSO₄

Chloride ions can interfere with the dichromate COD determination.

HgSO₄ is used to complex chloride ions and reduce this interference.

Without adequate chloride masking:

Chloride interference ↑ → measured COD may be falsely high

4. Back-Titration

After digestion, the remaining unreacted dichromate is determined by titration using:

Ferrous Ammonium Sulphate (FAS)

with Ferroin indicator.

Basic Principle

Dichromate added initially
          ↓
Some dichromate oxidizes wastewater
          ↓
Remaining dichromate
          ↓
Back-titrated using FAS
          ↓
Difference between blank and sample
          ↓
COD

The dichromate reduction can be represented conceptually as:

Cr₂O₇²⁻ → Cr³⁺

The dichromate ion is reduced from hexavalent chromium to trivalent chromium.

2. BOD/COD RATIO & WASTEWATER TREATABILITY

BOD/COD Ratio

The ratio:

BOD₅ / COD

is commonly used as an indicator of the biodegradability of wastewater.

It helps engineers assess whether biological treatment is likely to be effective.

Basic Logic

BOD/COD ratio
      │
      ├── High ratio
      │      ↓
      │  Large biodegradable fraction
      │      ↓
      │  Biological treatment more suitable
      │
      └── Low ratio
             ↓
       Large refractory fraction
             ↓
       Biological treatment may be less effective

BOD/COD TREATABILITY MATRIX

BOD/COD RATIO

0.0 ─────────────── 0.2 ───────────────── 0.5 ───────────────── 1.0
│                    │                      │
▼                    ▼                      ▼

Very Low           Moderate              High
Biodegradability   Biodegradability       Biodegradability

Refractory /       Biological treatment   Conventional
toxic compounds   may require             biological treatment
                   pre-treatment

General Engineering Interpretation

Important

These ratio ranges are engineering rules of thumb, not universal regulatory limits.

Treatment selection must also consider:

• Toxicity

• Specific contaminants

• Nutrient concentrations

• pH

• Salinity

• Temperature

• Hydraulic loading

• Existing microbial population

• Required effluent quality

Refractory Organic Matter

Industries such as:

• Tanneries

• Paper and pulp mills

• Textile dyeing units

• Chemical industries

• Refineries

may generate wastewater containing complex organic compounds.

Such wastewater can have a low BOD₅/COD ratio.

For example:

BOD₅/COD < 0.2

suggests that only a relatively small fraction of the measured COD is readily biodegradable.

In such cases, direct biological treatment may be inadequate and suitable pre-treatment may be required.

3. CORE MATHEMATICAL FORMULAS FOR COD

                         COD CALCULATIONS
                              │
        ┌─────────────────────┼─────────────────────┐
        │                     │                     │
        ▼                     ▼                     ▼
  COD TITRATION            ThOD                TREATABILITY
        │                     │                     │
        ▼                     ▼                     ▼
COD = (Vb−Vs)N×8000/Vsample  Stoichiometry       BOD₅/COD
                                                    │
                                                    ▼
                                             Treatment selection

FORMULA 1 — COD FROM FAS TITRATION

The conventional dichromate COD method can be calculated using:

COD (mg/L) = [(Vb − Vs) × N_FAS × 8000] / Vsample

Where:

• Vb = FAS volume used for blank titration, mL

• Vs = FAS volume used for sample titration, mL

• N_FAS = Normality of FAS solution

• Vsample = Wastewater sample volume, mL

• 8000 = conversion factor based on the equivalent weight of oxygen

Why 8000?

Equivalent weight of oxygen:

32 / 4 = 8 g/eq

Converting grams to milligrams:

8 × 1000 = 8000 mg/eq

Therefore:

8000 = mg O₂ per equivalent

FORMULA 2 — THEORETICAL OXYGEN DEMAND (ThOD)

For a compound represented by:

CₙHₐOᵦN𝑐

complete oxidation can be represented as:

CₙHₐOᵦN𝑐 + (n + a/4 − b/2 − 3c/4)O₂ → nCO₂ + [(a − 3c)/2]H₂O + cNH₃

The oxygen requirement is:

O₂ moles required = n + a/4 − b/2 − 3c/4

Therefore:

ThOD (g O₂/g compound) = [O₂ moles required × 32] / Molar Mass

Where:

32 = molecular mass of O₂ in g/mol

FORMULA 3 — BIODEGRADABILITY INDEX

Biodegradability Index

BI = BOD₅ / COD

Interpretation

Higher BI:

More biodegradable material → Biological treatment more feasible

Lower BI:

More refractory material → Biological treatment becomes more difficult

REFRACTORY COD

The simplified calculation used in this chapter is:

Refractory COD = Total COD − Ultimate BOD

Therefore:

CODrefractory = CODtotal − BODᵤ

Where:

BODᵤ = BOD₅ / (1 − e⁻⁵ᵏ)

This calculation assumes that the BOD kinetics can be represented by the first-order model used in Unit 16.1.

FORMULA 4 — COD REMOVAL EFFICIENCY

% COD Removal = [(CODin − CODout) / CODin] × 100

Where:

• CODin = COD entering treatment

• CODout = COD leaving treatment

FORMULA 5 — DAILY COD MASS LOAD

COD Load (kg/day) = Q × COD × 10⁻³

Where:

• Q = wastewater flow, m³/day

• COD = concentration, mg/L

4. SOLVED NUMERICAL PROBLEMS

PROBLEM 1 — LABORATORY COD DETERMINATION BY TITRATION

Question

A 50 mL industrial wastewater sample is digested with potassium dichromate in acidic medium.

After digestion:

• FAS used for sample titration = 14.2 mL

• FAS used for blank titration = 26.8 mL

• FAS normality = 0.10 N

• Wastewater sample volume = 50 mL

Calculate the COD.

Solution

Step 1 — Identify the values

Vb = 26.8 mL

Vs = 14.2 mL

N_FAS = 0.10 N

Vsample = 50 mL

Step 2 — Apply COD formula

COD = [(Vb − Vs) × N_FAS × 8000] / Vsample

Substitute:

COD = [(26.8 − 14.2) × 0.10 × 8000] / 50

First:

26.8 − 14.2 = 12.6 mL

Therefore:

COD = (12.6 × 0.10 × 8000) / 50

COD = 10,080 / 50

COD = 201.6 mg/L

Answer

COD = 201.6 mg/L

PROBLEM 2 — ThOD OF GLUCOSE & ETHANOL

Question

Calculate the theoretical oxygen demand of a solution containing:

• Glucose = 300 mg/L

• Ethanol = 100 mg/L

Calculate the total ThOD.

PART A — GLUCOSE

Glucose formula:

C₆H₁₂O₆

Oxidation Reaction

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

Therefore:

1 mole glucose requires 6 moles O₂

Step 1 — Molecular Mass of Glucose

Molar mass = (6 × 12) + (12 × 1) + (6 × 16)

= 72 + 12 + 96

= 180 g/mol

Step 2 — Oxygen Required

6 × 32 = 192 g O₂/mol glucose

Therefore:

ThOD = 192 / 180

ThOD = 1.067 g O₂/g glucose

Step 3 — Oxygen Demand for 300 mg/L

ThOD = 300 × 1.067

ThOD = 320.1 mg/L

PART B — ETHANOL

Ethanol formula:

C₂H₅OH

Equivalent molecular formula:

C₂H₆O

Oxidation Reaction

C₂H₆O + 3O₂ → 2CO₂ + 3H₂O

Therefore:

1 mole ethanol requires 3 moles O₂

Step 1 — Molecular Mass

Molar mass = (2 × 12) + (6 × 1) + (1 × 16)

= 24 + 6 + 16

= 46 g/mol

Step 2 — Oxygen Required

3 × 32 = 96 g O₂/mol ethanol

Therefore:

ThOD = 96 / 46

ThOD = 2.087 g O₂/g ethanol

Step 3 — Oxygen Demand for 100 mg/L

ThOD = 100 × 2.087

ThOD = 208.7 mg/L

PART C — TOTAL ThOD

Total ThOD = ThODglucose + ThODethanol

= 320.1 + 208.7

= 528.8 mg/L

Answer

Total ThOD = 528.8 mg/L

PROBLEM 3 — BOD/COD RATIO & TREATABILITY

Question

A pharmaceutical industry generates wastewater having:

COD = 1,200 mg/L

BOD₅ = 180 mg/L

At 20°C:

k = 0.23 day⁻¹

Calculate:

1. BOD₅/COD ratio

2. Treatability status

3. Ultimate BOD

4. Refractory COD

Step 1 — Calculate BOD₅/COD

BI = BOD₅ / COD

BI = 180 / 1200

BI = 0.15

Interpretation

BI = 0.15 < 0.20

Therefore, the wastewater has low biodegradability according to the rule-of-thumb classification used here.

Step 2 — Treatment Implication

A low BOD₅/COD ratio suggests that a large portion of the COD may be refractory or poorly biodegradable.

Therefore:

Direct biological treatment may be inadequate.

Possible treatment strategy:

Industrial Wastewater
        ↓
Physico-Chemical / Oxidative Pre-treatment
        ↓
Breakdown of refractory compounds
        ↓
Biological Treatment
        ↓
Final Treatment / Polishing
        ↓
Discharge or Reuse

Possible pre-treatment processes include:

• Fenton oxidation

• Ozonation

• Other Advanced Oxidation Processes

• Adsorption

• Appropriate physico-chemical treatment

The actual treatment train must be selected based on the specific contaminants and treatability studies.

Step 3 — Calculate Ultimate BOD

Use:

BOD₅ = L₀(1 − e⁻⁵ᵏ)

Given:

BOD₅ = 180 mg/L

k = 0.23 day⁻¹

Therefore:

180 = L₀(1 − e⁻⁽⁰·²³ × ⁵⁾)

180 = L₀(1 − e⁻¹·¹⁵)

Since:

e⁻¹·¹⁵ ≈ 0.3166

Therefore:

180 = L₀ × 0.6834

Hence:

L₀ = 180 / 0.6834

L₀ ≈ 263.4 mg/L

Step 4 — Calculate Refractory COD

Using the simplified relationship:

CODrefractory = CODtotal − BODᵤ

Therefore:

CODrefractory = 1200 − 263.4

CODrefractory = 936.6 mg/L

Percentage of Total COD

Refractory fraction = 936.6 / 1200 × 100

≈ 78.05%

Answer

• BOD₅/COD = 0.15

• Biodegradability = Low

• Ultimate BOD ≈ 263.4 mg/L

• Refractory COD ≈ 936.6 mg/L

• Refractory fraction ≈ 78.1%

PROBLEM 4 — INDUSTRIAL COD MASS BALANCE

Question

A textile dyeing facility generates:

Flow = 1,500 m³/day

Influent COD = 850 mg/L

Treatment system:

Stage 1

Chemical coagulation removes 40% COD.

Stage 2

Biological aerated filter removes 70% of the remaining COD.

Calculate:

1. Final effluent COD

2. Daily COD discharged

Step 1 — Stage 1 COD

40% is removed.

Therefore, 60% remains.

CODstage 1 = 850 × (1 − 0.40)

CODstage 1 = 850 × 0.60

CODstage 1 = 510 mg/L

Step 2 — Stage 2 COD

The biological filter removes 70% of the remaining COD.

Therefore, 30% remains.

CODfinal = 510 × (1 − 0.70)

CODfinal = 510 × 0.30

CODfinal = 153 mg/L

Step 3 — Daily COD Mass Discharged

Use:

COD Load = Q × COD × 10⁻³

Therefore:

Load = 1500 × 153 × 10⁻³

Load = 229.5 kg/day

Answer

• Final COD = 153 mg/L

• Daily COD discharged = 229.5 kg/day

5. FORMULA SUMMARY & QUICK REFERENCE SHEET

6. EXAM-ORIENTED COD LOGIC

If FAS titration values are given

Use:

COD = [(Vb − Vs) × N × 8000] / Vsample

Remember:

Blank − Sample

not:

Sample − Blank

If chemical formula is given

Calculate:

1. Molecular mass

2. Moles of O₂ required

3. Oxygen mass required

4. ThOD

If BOD₅ and COD are given

Calculate:

BOD₅/COD

Then interpret biodegradability.

If BOD₅ and k are given

First calculate ultimate BOD:

BODᵤ = BOD₅ / (1 − e⁻⁵ᵏ)

Then, if required:

Refractory COD = Total COD − BODᵤ

If treatment removal percentages are given

Apply each removal percentage sequentially.

Example:

COD = 1000 mg/L

40% removal
       ↓
600 mg/L remains

70% removal of remaining COD
       ↓
180 mg/L remains

Never add:

40% + 70% = 110%

Instead:

1000 × 0.60 × 0.30 = 180 mg/L

If flow and COD are given

Use:

COD Load = Q × COD × 10⁻³ kg/day

7. BOD vs COD — FINAL COMPARISON

8. BOD/COD TREATABILITY MEMORY

                 BOD₅ / COD

                      │
        ┌─────────────┼─────────────┐
        │             │             │
        ▼             ▼             ▼

      < 0.2         0.2–0.5        ≥ 0.5
        │             │             │
        ▼             ▼             ▼

   Low biological   Moderate       High
   treatability     biological     biological
                    treatability   treatability
        │             │             │
        ▼             ▼             ▼

 Pre-treatment     Combined /     Conventional
 often required    adapted         biological
                   treatment       treatment

Easy Memory

High BOD/COD → Biological treatment is easier

Low BOD/COD → Refractory fraction is higher → Pre-treatment may be needed

9. MUST REMEMBER

COD

Chemical Oxygen Demand

Oxygen required to chemically oxidize oxidizable substances.

Main COD Oxidant

Potassium dichromate — K₂Cr₂O₇

Acid Medium

Sulphuric acid — H₂SO₄

Catalyst

Silver sulphate — Ag₂SO₄

Chloride Interference Control

Mercuric sulphate — HgSO₄

Back-Titration

Ferrous Ammonium Sulphate (FAS)

Indicator

Ferroin

COD Formula

COD = [(Vb − Vs) × N × 8000] / Vsample

Biodegradability Index

BI = BOD₅ / COD

ThOD

Based on the stoichiometric oxygen requirement for complete oxidation.

Mass Load

COD Load = Q × COD × 10⁻³ kg/day

10. ONE-MINUTE MEMORY MAP

                         COD
                          │
         ┌────────────────┼────────────────┐
         │                │                │
         ▼                ▼                ▼
     LAB TEST          TREATABILITY       LOAD
         │                │                │
         ▼                ▼                ▼
 K₂Cr₂O₇ + H₂SO₄     BOD₅ / COD       Q × COD × 10⁻³
         │                │
         ▼                ▼
 Ag₂SO₄ + HgSO₄      High → biological
         │           Low → pre-treatment
         ▼
      Digestion
         │
         ▼
    FAS titration
         │
         ▼
        COD

11. QUICK REVISION QUESTIONS

1. What is COD?

2. What is the main difference between BOD and COD?

3. Why is COD determination faster than BOD₅?

4. What is the main oxidizing agent used in the conventional COD test?

5. What is the role of sulphuric acid?

6. What is the role of silver sulphate?

7. Why is mercuric sulphate used?

8. What is FAS?

9. Why is a blank titration required?

10. Why is the blank FAS volume greater than the sample FAS volume?

11. What does the BOD₅/COD ratio indicate?

12. What does a very low BOD₅/COD ratio indicate?

13. What is ThOD?

14. How is ThOD calculated from a chemical formula?

15. What is refractory COD?

16. How is daily COD mass load calculated?

17. Why should treatment removal percentages be applied sequentially?

18. What happens to biological treatability when the BOD₅/COD ratio decreases?

CORE FORMULAS

1. COD = [(Vb − Vs) × N_FAS × 8000] / Vsample

2. ThOD = (Moles O₂ required × 32) / Molar Mass

3. BI = BOD₅ / COD

4. BODᵤ = BOD₅ / (1 − e⁻⁵ᵏ)

5. CODrefractory = CODtotal − BODᵤ

6. % COD Removal = [(CODin − CODout) / CODin] × 100

7. COD Load = Q × COD × 10⁻³ kg/day

These formulas cover the principal COD calculation patterns required for this unit.