Question

In Winkler's method why is some amount of Na2S2O3 added before addition of starch solution turning the deep straw coloured solution pale straw?

Is it to generate I− ions needed for formation of I3− in the I2 containing solution? This I3− inturn reacts with the starch to form blue-black complex.

However once we didn't add Na2S2O3 and the addition of starch solution did turn the solution it blue-black.

Answer 1

In the first step, manganese(II) sulfate (at 48% of the total volume) is added to an environmental water sample. Next, potassium iodide (15% in potassium hydroxide 70%) is added to create a pinkish-brown precipitate. In the alkaline solution, dissolved oxygen will oxidize manganese(II) ions to the tetravalent state.

2 MnSO4(s) + O2(aq) → 2 MnO(OH)2(s)
MnO(OH)2 appears as a brown precipitate. There is some confusion about whether the oxidised manganese is tetravalent or trivalent. Some sources claim that Mn(OH)3 is the brown precipitate, but hydrated MnO2 may also give the brown colour.

4 Mn(OH)2(s) + O2(aq) + 2 H2O → 4 Mn(OH)3(s)
The second part of the Winkler test reduces (acidifies) the solution. The precipitate will dissolve back into solution. The acid facilitates the conversion by the brown, Manganese-containing precipitate of the Iodide ion into elemental Iodine.

The Mn(SO4)2 formed by the acid converts the iodide ions into iodine, itself being reduced back to manganese(II) ions in an acidic medium.

Mn(SO4)2 + 2 I−(aq) → Mn2+(aq) + I2(aq) + 2 SO42−(aq)
Thiosulfate is used, with a starch indicator, to titrate the iodine.

2 S2O32−(aq) + I2 → S4O62−(aq) + 2 I−(aq)

From the above stoichiometric equations, we can find that:

1 mole of O2 → 2 moles of MnO(OH)2 → 2 mole of I2 → 4 mole of S2O32−
Therefore, after determining the number of moles of iodine produced, we can work out the number of moles of oxygen molecules present in the original water sample. The oxygen content is usually presented as mg/dm3.