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Stoichiometry of the Titration of Diluted Bleach Solution. (Eqn. 1)   1 OCl− + 3 I− + 2...

Stoichiometry of the Titration of Diluted Bleach Solution.

(Eqn. 1)   1 OCl− + 3 I− + 2 H+ → 1 I3− + 1 Cl− + 1 H2O

(Eqn. 2)   2 S2O32− + 1 I3− + 0 H+ → 3 I− + 1 S4O62− + 0 H2O
What is the molar ratio of OCl− to S2O32− in the overall net ionic equation?
[OCl−(net) / S2O32−(net)]

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Answer #1

(Eqn. 1)   1 OCl− + 3 I− + 2 H+ → 1 I3− + 1 Cl− + 1H2O

(Eqn. 2)   2 S2O32− + 1 I3− + 0 H+ → 3 I− + 1S4O62− + 0 H2O

Overall reaction: 1OCl- + 2H+ +2S2O32- -> 1Cl- + 1 S4O62- + 1 H2O

Here I3-, I- got cancelled by their stochiometry.

Here 1 mole OCl- reacts with 2 miles of S2O32-.

So the molar ratio is OCl−(net) / S2O32−(net)=1/2.

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