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My class is currently performing a reaction to determine the oxidizing power (i.e. the amount of hypochlorite) of household bleach. The general process is as follows:

Standardize thiosulfate to be used as a titrant of $\ce{I2}$ solution.

  • An excess of I- ions are allowed to react with a known amount of $\ce{KIO3}$ in acidic solution.
  • The iodine formed from this reaction is titrated with thiosulfate using starch as an indicator of triiodide anion.
  • The relation between moles $\ce{KIO3}$ and thiosulfate is used to determine molarity of thiosulfate.

Analysis of oxidizing capacity of liquid bleach

  • Excess iodide ion solution added to bleach.

  • Iodide ions are oxidized to I2 after this solution is acidified simultaneously with the reduction of hypochlorite.

  • The iodine that is formed is then titrated with the above standardized thiosulfate solution.

  • The indicator (starch) is not added until the dark-brownish color of the iodine has changed to pale yellow. When it is added, the solution color changes to blue black.

  • The end point of titration is indicated when the solution becomes colorless.

  • (Note that if the starch is added too soon in the titration, the formation of the blue-black complex is not easily reversed, making the end point very slow and difficult to detect.

Pertinent reactions are given below

Reactions Pertaining to Titration

$\ce{HOCl + 2I- + H+ -> I2 + Cl- + H2O}$

$\ce{2S2O3^-- + I2 -> 2I- + S4O6^--}$

$\ce{HOCl + 2S2O3^-- + H+ -> Cl- + S4O6^-- + H2O}$ [the addition of the above two reactions]

$\ce{I3- + starch <-> starch*I3-(complex)}$ [blue-black in solution]

Reactions pertaining to the standardization of thiosulfate

$\ce{IO3- + 5I- +6H+ -> 3I2 + 3H2O}$

$\ce{3I2 + S2O3^-- -> 6I- + 3S4O6^--}$

The question

We were told that, during the standardization process, if the iodate-iodide solution is acidified with $\ce{H2SO4}$, it should be titrated immediately. My question is why? Using thiosulfate, this is not the case, correct? Why does this change when sulfuric acid is used in its stead?

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After acidification with $\ce{H2SO4}$, starch starts to hydrolyze to its component sugars. Lugol's iodine does not react with those sugars to form the blue complex.

Our amylose is gone, today,

Its trace we'll see no more,

For what we thought was $\ce{H2O}$

Was $\ce{H2SO4}$.

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