Water Quality Parameters — pH, DO, BOD, COD Explained

1. Classification of Water Quality Parameters

2. Physical Parameters

2.1 Temperature

Temperature affects all chemical and biological processes in water — higher temperature reduces dissolved oxygen solubility, accelerates biochemical reactions, and increases taste/odour problems. Ideal drinking water: 10–15°C. Temperature affects BOD rate constant k_d and must be corrected in design.

2.2 Turbidity

2.3 Colour

True colour (after filtration) vs apparent colour (including suspended matter). Measured in Hazen units (HU) or Platinum Cobalt units. IS 10500 limit: 5 HU (desirable); 15 HU (permissible).

2.4 Total Dissolved Solids (TDS)

3. Chemical Parameters

3.1 pH

3.2 Hardness

3.3 Alkalinity

3.4 Chlorides

IS 10500 limit: 250 mg/L (desirable); 1000 mg/L (permissible). High chlorides: salty taste; indicator of sewage contamination (chlorides are conservative in water). Seawater intrusion in coastal aquifers raises chloride levels.

3.5 Fluoride

IS 10500: 1.0 mg/L (desirable); 1.5 mg/L (permissible). Fluoride < 0.6 mg/L → dental caries; 1.5–4 mg/L → dental fluorosis; > 4 mg/L → skeletal fluorosis. Defluoridation required when > 1.5 mg/L (Nalgonda technique, activated alumina).

3.6 Nitrates

IS 10500: 45 mg/L as NO₃⁻ (permissible). Excess nitrates (from agricultural fertilisers, sewage): methaemoglobinaemia (blue baby syndrome) in infants under 6 months. Indicates agricultural runoff or inadequate sewage treatment in catchment.

4. Dissolved Oxygen (DO)

4.1 DO Significance

4.2 DO Deficit

5. Biochemical Oxygen Demand (BOD)

BOD is the amount of dissolved oxygen consumed by microorganisms (primarily bacteria) while aerobically decomposing organic matter in a water sample at a specified temperature (20°C) over a specified time (5 days for BOD₅).

5.1 BOD Kinetics — First Order Model

5.2 Typical Values

5.3 Determination of k_d and L₀ from Test Data

5.4 Relationship Between BOD₅ and Ultimate BOD

6. BOD Temperature Correction

6.1 DO Saturation Temperature Correction

DO saturation (DO_sat) decreases with increasing temperature. The empirical formula for fresh water:

7. Chemical Oxygen Demand (COD)

7.1 COD/BOD Ratio

7.2 Theoretical Oxygen Demand (ThOD)

8. Nitrogen and Phosphorus

8.1 Nitrogen Forms in Water

8.2 Nitrogenous BOD (NBOD)

8.3 Phosphorus and Eutrophication

Phosphorus (as PO₄³⁻) is the limiting nutrient for algal growth in most freshwater lakes. Excess phosphorus from sewage effluent, agricultural runoff, and detergents triggers eutrophication — explosive algal blooms that deplete DO when they decompose, kill fish, and produce toxins. IS 10500 limit for phosphate: 0.3 mg/L. Removal in wastewater treatment: biological excess phosphorus removal (BEPR) or chemical precipitation with alum/lime/ferric salts.

9. Biological Parameters — Coliforms

Coliform bacteria are used as indicators of faecal contamination in water — their presence suggests the water has been contaminated by human or animal faeces and may contain pathogenic organisms (Vibrio cholerae, Salmonella typhi, hepatitis A virus, Giardia, Cryptosporidium).

9.1 MPN Method

The Most Probable Number (MPN) method uses multiple tube fermentation with serial dilutions to statistically estimate the number of coliform organisms per 100 mL. A 5-tube × 3-dilution MPN test gives coliform counts interpreted from MPN tables. Higher MPN = more contamination.

10. DO Sag Curve — Streeter-Phelps Equation

When sewage (or any oxygen-demanding waste) is discharged into a river, bacteria in the water consume oxygen to decompose the organic matter (deoxygenation), while the river re-aerates from the atmosphere (reaeration). The interplay creates a characteristic “sag” in the dissolved oxygen profile downstream — the Streeter-Phelps DO sag curve.

10.1 Streeter-Phelps Equation

10.2 Critical Point (Minimum DO)

10.3 Mixing of River and Sewage

10.4 Typical k Values

11. IS 10500:2012 — Drinking Water Standards

12. Worked Examples (GATE CE Level)

Example 1 — BOD at Time t (GATE CE 2022 type)

Problem: The ultimate BOD of a sewage sample is 280 mg/L and the BOD rate constant k_d = 0.20 day⁻¹ (base-e). Find (a) the 5-day BOD and (b) the BOD remaining after 5 days.

Answer: BOD₅ = 177 mg/L; BOD remaining = 103 mg/L

Example 2 — Determination of Ultimate BOD (GATE CE 2021 type)

Problem: A BOD test gives: BOD at 2 days = 120 mg/L; BOD at 5 days = 180 mg/L. Find the ultimate BOD L₀ and the rate constant k_d.

Answer: L₀ ≈ 199 mg/L; k_d ≈ 0.46 day⁻¹

Example 3 — DO Deficit and Mixing (GATE CE type)

Problem: A river (flow = 10 m³/s, DO = 8 mg/L, BOD = 2 mg/L) receives sewage (flow = 2 m³/s, DO = 1 mg/L, BOD = 200 mg/L). DO saturation = 9.1 mg/L at 20°C. Find (a) the mixed stream DO, (b) the initial DO deficit D₀, and (c) the mixed stream ultimate BOD (assume BOD₅ = ultimate BOD for this problem).

Answer: DO_mixed = 6.83 mg/L; D₀ = 2.27 mg/L; L₀ = 35.0 mg/L

Example 4 — Streeter-Phelps Critical Deficit (GATE CE 2020 type)

Problem: For the mixed stream in Example 3, k_d = 0.20 day⁻¹ and k_r = 0.40 day⁻¹. Find (a) the critical time t_c and (b) the critical DO deficit D_c. Is the minimum DO adequate for aquatic life? (DO_sat = 9.1 mg/L)

Answer: t_c = 3.13 days; D_c = 9.36 mg/L (exceeds DO_sat) → stream goes anaerobic.

Example 5 — Hardness Calculation (GATE CE type)

Problem: A water sample contains 80 mg/L Ca²⁺ and 24 mg/L Mg²⁺. Calculate the total hardness as mg/L CaCO₃. Atomic weights: Ca = 40, Mg = 24, C = 12, O = 16.

Answer: Total hardness = 300 mg/L as CaCO₃

Example 6 — BOD Temperature Correction (GATE CE type)

Problem: The BOD rate constant at 20°C is k_d,20 = 0.23 day⁻¹. Find k_d at 15°C and 30°C using θ = 1.047. Then find BOD₅ at 15°C if L₀ = 250 mg/L.

Answer: k_d,15 = 0.183 day⁻¹; k_d,30 = 0.365 day⁻¹; BOD₅ at 15°C = 149.7 mg/L

13. Common Mistakes

Mistake 1 — Confusing y_t (BOD Exerted) with L_t (BOD Remaining)

Error: Stating that BOD₅ = L₀ × e^–k_dt instead of L₀(1 – e^–k_dt).
Root Cause: Two quantities are defined: y_t = BOD exerted (oxygen consumed) = L₀(1 – e^–k_dt); L_t = BOD remaining (unoxidised organic matter) = L₀ e^–k_dt. The question may ask for either; students often apply the wrong formula.
Fix: y_t starts at 0 and increases toward L₀. L_t starts at L₀ and decreases toward 0. y_t + L_t = L₀ always. BOD₅ test measures y₅ (exerted) — use (1 – e^–k_dt).

Mistake 2 — Using k_d (Base-e) and K₁ (Base-10) Interchangeably

Error: Writing BOD_t = L₀(1 – 10^–k_dt) when k_d is the base-e rate constant (or vice versa).
Root Cause: Some textbooks use natural logarithm (base-e, k_d); others use base-10 (K₁). The two are related by k_d = 2.303 K₁. Using K₁ in the base-e formula overestimates BOD by a large amount.
Fix: Identify which base is used: if the formula is y = L₀(1 – e^–kt), use k in base-e units. If y = L₀(1 – 10^–Kt), use K in base-10 units. k_d (base-e) is typically 0.15–0.30 day⁻¹; K₁ (base-10) ≈ 0.065–0.13 day⁻¹. Always verify consistency.

Mistake 3 — Taking D_c as the Minimum DO Instead of the Maximum Deficit

Error: Reporting D_c = 5 mg/L as “minimum DO = 5 mg/L” when it actually means the DO has dropped 5 mg/L below saturation.
Root Cause: D_c is a deficit (how much DO has been removed from saturation), not the actual DO value. Minimum DO = DO_sat – D_c.
Fix: After finding D_c, always compute: minimum DO = DO_sat – D_c. If D_c > DO_sat: the minimum DO would be negative — the stream actually goes anaerobic before the critical point is reached.

Mistake 4 — Using BOD₅ Directly as Ultimate BOD in Streeter-Phelps Equation

Error: Substituting BOD₅ (5-day BOD) for L₀ (ultimate BOD) in the Streeter-Phelps equation without converting.
Root Cause: The Streeter-Phelps equation requires the ultimate BOD L₀ because the deoxygenation term k_dL₀ represents total oxygen demand. BOD₅ is only a fraction of L₀ (typically 50–70%).
Fix: Convert BOD₅ to L₀: L₀ = BOD₅/(1 – e^–5k_d). Then use L₀ in the Streeter-Phelps equation. If the problem gives L₀ directly, use it without conversion.

Mistake 5 — Hardness Calculation: Using MW Instead of Equivalent Weight for Conversion

Error: Computing Ca²⁺ hardness as 80 mg/L × (100/40) = 200 mg/L CaCO₃, using molecular weights instead of equivalent weights.
Root Cause: The conversion factor is 50/(equivalent weight of ion), not 100/(atomic weight). CaCO₃ equivalent weight = 50; Ca²⁺ equivalent weight = 20; factor = 50/20 = 2.5. Using MW ratio 100/40 = 2.5 happens to give the same answer for Ca²⁺, but this is coincidental — for Mg²⁺ the MW ratio (100/24 ≈ 4.17) equals the equivalent weight ratio (50/12 ≈ 4.17), which also coincides. The correct basis is always equivalent weights (MW/valence).
Fix: Hardness as CaCO₃ = (mg/L of ion) × (50/equivalent weight of ion). Equivalent weight = atomic weight/valence. Ca²⁺: 40/2 = 20; Mg²⁺: 24/2 = 12; SO₄²⁻: 96/2 = 48; HCO₃⁻: 61/1 = 61.

14. Frequently Asked Questions

Q1. What is the physical meaning of the BOD rate constant k_d, and what factors affect its value?

The BOD rate constant k_d represents the rate at which microorganisms oxidise the biodegradable organic matter — specifically, the fraction of remaining ultimate BOD that is consumed per unit time (day⁻¹ in first-order kinetics). A high k_d means fast decomposition; a low k_d means slow. Physically, k_d reflects the combined effect of microbial population density, the biodegradability of the organic matter (easily degradable sugars vs recalcitrant compounds), temperature (higher T → faster kinetics → higher k_d), pH, and nutrient availability. Laboratory BOD bottle tests give k_d values of 0.05–0.30 day⁻¹ for domestic sewage at 20°C. In actual rivers, the rate is often higher (0.10–0.30 day⁻¹) because the microbial population is enriched and there is physical mixing. Temperature correction (k_d,T = k_d,20 × 1.047^T–20) is essential for design — BOD kinetics are significantly slower in winter (15°C) Indian rivers compared to summer, affecting the DO sag profile and treatment requirements.

Q2. Why is the COD test preferred over BOD₅ for industrial effluent monitoring?

The 5-day BOD test has three significant practical disadvantages for industrial effluent monitoring: (1) time — it takes 5 days to get a result, which is too slow for real-time process control at a treatment plant; (2) inhibition — many industrial effluents contain toxic substances (heavy metals, solvents, biocides) that inhibit the microbial activity in the BOD test, giving artificially low BOD readings that don’t reflect the true organic load; (3) relevance — for effluents with high non-biodegradable content (textile, pharmaceutical, petrochemical), BOD₅ gives little information about the total organic pollution load. COD overcomes all three: it is completed in 2 hours, it oxidises all organic matter regardless of biodegradability, and it is not affected by toxicants. In practice, a good correlation between COD and BOD is established for a specific effluent type, and COD is then used for routine monitoring with periodic BOD checks for calibration. Indian environmental regulations (IS 2490, Environment Protection Rules Schedule 6) specify both BOD and COD limits in most effluent discharge standards precisely because neither alone is sufficient.

Q3. How is the Streeter-Phelps model used in practice for river management in India?

The Streeter-Phelps model is used to predict the dissolved oxygen profile downstream of sewage discharge points and to set effluent quality standards that will maintain the river’s DO above minimum acceptable levels (typically 4–5 mg/L for fisheries use). In India, the Central Pollution Control Board (CPCB) and state PCBs use DO modelling for rivers like the Ganga, Yamuna, Cauvery, and Krishna to designate stretches as Class A (clean, drinking water source), Class B (bathing), Class C (drinking with treatment), Class D (fisheries), or Class E (irrigation). When a new industrial facility or municipality wants to discharge treated effluent to a river, engineers run the Streeter-Phelps model (or more sophisticated 1D/2D stream quality models) for critical low-flow conditions (7-day, 10-year minimum flow) to ensure the discharge will not push the river DO below the class standard. The parameters k_d and k_r are calibrated from field data (dissolved oxygen surveys and stream gauging). The Namami Gange programme has required extensive DO modelling along the 2,500 km Ganga river to identify critical stretches and prioritise sewage treatment infrastructure investments.

Q4. What is the significance of the carbonate hardness system (temporary vs permanent hardness) for water treatment in India?

Temporary hardness (due to calcium and magnesium bicarbonates) can be removed by lime softening: Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃↓ + 2H₂O. This is the basis of the lime-soda softening process used in water treatment plants serving hard water areas (much of peninsular India, particularly groundwater from granitic and basaltic aquifers). Permanent hardness (sulphates, chlorides of Ca and Mg) requires either excess lime-soda treatment or ion exchange softening (zeolite process, sodium cycle cation exchange). In domestic boilers and industrial steam generators, even moderate hardness (>100 mg/L) causes scaling on heat exchange surfaces, reducing efficiency and eventually causing tube failure. In textiles, high hardness interferes with dyeing and soaping. In hospitals and pharmaceutical manufacturing, water hardness must be below 50 mg/L for autoclaves and processes. The IS 10500 desirable limit of 200 mg/L is set to prevent significant scale formation in household plumbing, while the permissible limit of 600 mg/L is set to avoid significant taste impacts — very hard water has a distinctly bitter, salty taste that makes it unacceptable to consumers even if not strictly harmful.