Chemistry 2nd Paper · Classes 11-12 · Chapter 1 · 1.18 (Water Quality)

DO, BOD, COD and TDS.

River water can look clean and still be killing its fish. This note covers four ways to measure what the eye cannot see: DO, BOD, COD and TDS. One unit, ppm, ties them all together.

1. Dissolved oxygen, DO

The oxygen mixed into water is called dissolved oxygen, DO for short. In oxygen-saturated water at 15°C, DO is 10 mg/L, which is 10 ppm. At 20°C it is 9.2 ppm. There are two ways to measure it: an oxygen sensor probe, or the Winkler iodometric method (section 3).

Surface water needs DO of 5 mg/L or more; in a river estuary it should be above 6 mg/L. Organic pollutants pull DO down, because bacteria use up the oxygen to break them down. When DO drops, aerobic fish die. Anaerobic decay then takes over, making CH₄, H₂S, PH₃ and amines, and their foul smell spreads.

2. BOD: how much oxygen the bacteria eat

While breaking down degradable organic matter, bacteria use up oxygen from the DO. The O₂ they consume in five days at 20°C is the BOD. So BOD = initial DO − final DO. “BOD 5 mg/L” means 5 mg of oxygen used per litre in five days at 20°C.

One handy figure: oxidising each 1 mg of carbon needs 2.67 mg of O₂.

The lower the BOD, the better the water. 1-2 mg/L is very good; 3 mg/L is fairly good; the WHO limit is 6 mg/L; above 10 is bad; and 20 mg/L is very bad. The 6 is a limit, not a quality word, so below it goes by the name “WHO limit”.

The book's example: a sample has an initial DO of 8.5 ppm. After five days in the dark at 20°C the final DO is 5.5 ppm. So BOD = 8.5 − 5.5 = 3.0 ppm.

3. DO by the Winkler method

Here the O₂ in the water is first locked into a manganese compound, then released as iodine, then measured with thiosulphate. Three reactions, with starch as the indicator:

2Mn²⁺ + O₂ + 4OH⁻ → 2MnO(OH)₂
2MnO(OH)₂ + 8H⁺ + 4I⁻ → 2Mn²⁺ + 2I₂ + 6H₂O
I₂ + 2S₂O₃²⁻ → S₄O₆²⁻ + 2I⁻

Put the reactions together and the mole ratio is: 1 mol O₂ ≡ 2 mol MnO(OH)₂ ≡ 2 mol I₂ ≡ 4 mol Na₂S₂O₃.

There are five reagents: 1 M MnSO₄, NaI + NaOH, 6 M H₂SO₄, starch and 0.006 M Na₂S₂O₃. In a 350 mL bottle put 2 mL MnSO₄ and 2 mL alkaline NaI, then 4 mL H₂SO₄. Take 100 mL of the sample, add 0.5 mL starch, and titrate with 0.006 M Na₂S₂O₃ to get Y mL.

The arithmetic is short. Y mL of 0.006 M Na₂S₂O₃ is 0.048Y mg of O₂, because 4 mol Na₂S₂O₃ ≡ 1 mol O₂. In a 100 mL sample the initial DO is therefore 0.48Y ppm. If the same job takes Z mL after five days, the final DO is 0.48Z ppm and BOD = 0.48(Y − Z) ppm. Mind the limit: 0.48 holds only for the book's 100 mL sample and 0.006 M thiosulphate; change the volume or the concentration and the factor changes too.

4. COD: all the organic matter

BOD counts only the degradable matter. COD is the oxygen needed to oxidise both degradable and non-degradable organic matter, so COD is always larger than BOD. The WHO maximum for COD is 10 ppm, in units of mg/L or ppm.

Set the two values side by side and you can read what is in the water: BOD points to the degradable dead organic matter, and COD brings in the non-degradable organic compounds as well.

The general oxidation equation, in which nitrogen leaves as NH₃:

CₙHₐO_bN_c + (n + a/4 − b/2 − 3c/4) O₂ → nCO₂ + cNH₃ + (a/2 − 3c/2) H₂O
CH₃CONH₂ (C₂H₅ON) + 2O₂ → 2CO₂ + NH₃ + H₂O

COD is found by a redox titration with dilute H₂SO₄ and M/60 K₂Cr₂O₇, in which Cr³⁺ forms. Add 5 mL of 85% H₃PO₄ and diphenylamine as the indicator. At the end point the colour is deep violet.

5. TDS and the WHO limits

TDS is all the dissolved solids in water. It includes Ca²⁺, Na⁺, K⁺, PO₄³⁻, NO₃⁻, Cl⁻ and pesticides.

TDS is measured gravimetrically or with a conductivity meter. By EPA the ideal TDS is 500 ppm; up to 1000 ppm is good; above that the water is unusable.

Table 1.10 gives the WHO standards: pH 6.5-8.5; DO 5.0-6.0; BOD 6.0 ppm; COD 10.0 ppm; TDS 500 ppm; hardness Ca²⁺ 100 ppm and Mg²⁺ 150 ppm; NaCl 500 ppm. Match each measure to its limit yourself.

For NaCl, WHO says water with more than 500 ppm is unfit for drinking.

6. ppm: one unit for all of it

Every value in this chapter is in ppm. 1 ppm = 1 mg/L (mg L⁻¹). To turn molarity into ppm: ppm = M × molar mass (g/mol) × 1000. The 1000 is there because molarity gives g/L, and 1 g = 1000 mg. The same thing in other units: 1 ppm = 1 mg/L = 1 g/m³ = 1 µg/mL.

Molar masses you will need (g mol⁻¹): HCl 36.5, HNO₃ 63, H₂SO₄ 98, NaOH 40, KOH 56, NaCl 58.5.

Example: 2.0 × 10⁻⁴ M O₂. 2.0 × 10⁻⁴ × 32 = 6.4 × 10⁻³ g/L = 6.4 mg/L = 6.4 ppm.

A few more: 0.01 M HCl = 0.01 × 36.5 × 1000 = 365 ppm, whether the solution is 250 mL or 1 L, since concentration does not depend on volume. 0.01 M HNO₃ = 630 ppm. 0.05 M H₂SO₄ = 0.05 × 98 × 1000 = 4900 ppm. From percent (w/v) to ppm, multiply by 10⁴: 10% = 10⁵ ppm, and 2.5% NaOH = 2.5 × 10⁴ ppm. The other way, 877.5 ppm NaCl = 877.5/(58.5 × 1000) = 0.015 M.

7. Check yourself

Three questions in the exam's style. Start with the first.

The second wants you to recall the steps.

In the last one, watch the unit and the number together.

The whole note on eight cards.

Source: Chemistry 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Environmental Chemistry”, section 1.18 (Water Quality).