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Dissolved oxygen by 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.

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

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

DO (mg/l)=V×N×8×1000Vs\text{DO (mg/l)} = \frac{V\times N\times8\times1000}{V_s}

where VV is the ml of thiosulphate, NN its normality and VsV_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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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