---
title: "Dissolved oxygen by titrimetric method"
subject: "Water Supply Engineering"
chapter: 4
key: "semester_5/water_supply_engineering:dissolved-oxygen-titrimetric"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/dissolved-oxygen-titrimetric
site: BCT Engineering
---

# Dissolved oxygen by titrimetric method

Water Supply Engineering · Chapter 4: Quality of Water

- Asked in: 2076 Asoj, 4 marks
- Video: [Experiment No. 7: Determination of Dissolved Oxygen](https://www.youtube.com/watch?v=m9XGsEs55Qo) — Dr. Harish Kumar .S
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/dissolved-oxygen-titrimetric
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-4

## Question

How do you determine dissolved oxygen of a water sample in the laboratory using the titrimetric method?

## Short answer

The Winkler method fixes dissolved oxygen with manganese hydroxide, then releases iodine in acid. The iodine is titrated with standard thiosulphate using starch indicator. DO mg/L = (V×N×8×1000)/Vs.

## Answer

Dissolved oxygen (DO) is determined by the **Winkler (iodometric) titration method**, usually with the azide modification, in which DO is fixed with manganese hydroxide and measured as liberated iodine.

#### Reagents
Manganous sulphate solution, alkali-iodide-azide reagent, concentrated sulphuric acid, starch indicator, standard sodium thiosulphate (0.025 N).

#### Procedure
1. Fill a 300 ml BOD bottle with the sample carefully without air bubbles.
2. Add 2 ml MnSO4 and 2 ml alkali-iodide-azide reagent below the surface; stopper and mix by inverting. A brown-white precipitate forms; let it settle.
3. Add 2 ml of concentrated H2SO4, restopper and mix until the precipitate dissolves. Iodine is liberated in an amount equal to the DO.
4. Take 200 ml (or 203 ml) of the solution in a flask and titrate with 0.025 N thiosulphate until the yellow colour becomes pale.
5. Add 1 ml of starch (blue colour appears) and continue the titration until the blue colour just disappears. Note the volume used.

#### Reactions
Mn(2+) + 2OH(-) gives Mn(OH)2; with oxygen it gives MnO(OH)2 (brown). In acid, MnO(OH)2 + 2I(-) + 4H(+) gives Mn(2+) + I2 + 3H2O. Then I2 + 2S2O3(2-) gives 2I(-) + S4O6(2-).

#### Calculation
$$
\text{DO (mg/l)} = \frac{V\times N\times8\times1000}{V_s}
$$

where $V$ is the ml of thiosulphate, $N$ its normality and $V_s$ the sample volume (200 ml). With 0.025 N thiosulphate and 200 ml sample, 1 ml of titrant equals 1 mg/l of DO.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. DO is determined by:
   A. the Winkler (iodometric) titration
   B. colourimetry with starch only
   C. gravimetry
   D. pH titration with NaOH
   Answer: A. DO is fixed as MnO(OH)2 and measured as liberated iodine. (Key point: Method)

2. The titrant is standard:
   A. silver nitrate
   B. sodium thiosulphate (0.025 N)
   C. EDTA
   D. sulphuric acid
   Answer: B. It reacts with liberated iodine. (Key point: Titrant)

3. Starch is added near the end and the endpoint is when:
   A. blue appears
   B. brown precipitate forms
   C. the blue colour just disappears
   D. solution turns red
   Answer: C. Starch gives blue with iodine. (Key point: Endpoint)

4. With 0.025 N thiosulphate and 200 ml sample, 1 ml of titrant equals:
   A. 8 mg/l
   B. 0.025 mg/l
   C. 25 mg/l
   D. 1 mg/l DO
   Answer: D. DO = V x N x 8 x 1000 / 200 = V. (Key point: Calculation)

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