To find out the amount of dissolved oxygen in a water sample

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Experiment 9

Aim

To find out the amount of dissolved oxygen in a water sample.

Procedure

  1. A water sample was collected by a 250cm3 volumetric flask. The flask was filled completely with water without trapping any air bubbles.
  2. 1 cm3 of manganese(II) sulphate solution was added to the sample using a pipette. The solution was discharged well below the surface.
  3. 1 cm3 of alkaline potassium iodide solution was introduced similarly. It was sure that no air becomes entrapped. The bottle was inverted to distribute the precipitate uniformly.
  4. When the precipitate has settled at least 3 cm  below the stopper, 1 cm3 of concentrated sulphuric(VI) acid was introduced well below the surface. The stopper was replaced and carefully mixed until the precipitate disappears. A magnetic stirrer was used.
  5. The mixture was allowed to stand for 5 min and then 100 cm3 of the acidified sample was withdrawn into a 250 cm3 conical flask.
  6. 0.0125M sodium thiosulphate was titrated until the iodine color becomes faint. Then 1 cm3 of starch solution was added, and continue adding the thiosulphate solution until the blue color disappears.
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Result

Titration Result

The first try encountered error, and only the result of the second try was used.

Calculation

I2+2S2O32-  2I-+S4O62-

2Mn(OH)3(s) + 2I-(aq) + 6H+(aq)  2Mn2+(aq) + I2(aq) + 6H2O(l)

8OH-(aq) + 4Mn2+(aq) + O2(aq) + 2H2O(l)  4Mn(OH)3(s)

No. of mole of sodium thiosulphate used: 6 × 10-6 × 0.0125 = 7.5 × 10-8

No. of mole of I2 in 100cm3 sample: 7.5 × 10-8 ÷2 = 3.75 × 10-8

No. of mole of Mn(OH)3 in 100cm3 sample: 3.75 × 10-8 × 2 = 7.5 × 10-8

No. of mole of dissolved O2 in 100cm3 sample: 7.5 × 10-8 ÷4 = 1.875 × 10-8

No. of mole of dissolved O2 per dm3 : 1.875 × 10-8 ÷100 × 106 = 1.875 × 10-4

Mass of dissolved O2 per dm3 : 1.875 × 10-4 × 16 × 2 = 0.006g

DO value of the water sample is: ...

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