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This problem is taken from the Indonesian Chemical Olympiad.

A sample containing barium peroxide was added to a glass full of sulfuric acid. It produced an insoluble product and its filtrate. The filtrate was reacted with potassium iodide in acidic conditions. The product was titrated with $11.8 \,\mathrm{mL}$ of $0.1 \,\mathrm{M}$ sodium thiosulfate.

  • Write the reaction above.
  • What is the mass of the barium peroxide in the sample?

Actually I know if the first reaction will make H2O2, but when this peroxide reacted with potassium iodide, I'm confused what is the product.

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    $\begingroup$ @lambda23 start considering the following: first obtained $\ce{H2O2}$, then yellow $\ce{I2}$, then titration yielding $\ce{S4O6 2-}$ and $\ce{I-}$, both colorless. $\endgroup$ – mannaia Apr 3 '14 at 14:47
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In sulfuric acid, bariumperoxide generates hydrogen peroxide and barium sulfate (= precipitate).

$\ce{BaO2 + H2SO4 -> H2O2 + BaSO4 v}$

Under acidic conditions, iodide is oxidized to iodine.

$\ce{H2O2 + 2I- + 2H3O+ -> I2 + 4H2O}$

With an excess of iodide, triodide is formed.

$\ce{I- + I2 -> I3-}$

The amount of iodine (in triodide) is determined by titration with thiosulfate:

$\ce{I3- + 2 S2O3^2- -> S4O6^2- + 3 I-}$

I leave the math for you :)

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