In this process first we need to liberate iodine to react with sodium thiosulfate.

First, sodium bicarbonate is added to a iodate-free solution of potassium iodide. The mixture is shaken until the salts dissolve, then conc. hydrochloric acid is added. The solution is shaken again, then covered and keept in a dark place. I have two questions here:

  1. What is the role of sodium bicarbonate here?
  2. Can conc. sulfuric acid be used to provide the acidic media instead of conc. hydrochloric acid?

1 Answer 1

  1. I think your procedure is incorrect. Typically you would mix $\ce{KIO3}$ (limiting reagent) with access of $\ce{KI}$. Then you add $\ce{HCl}$ to have a reaction: $$\ce{KIO3 + 5KI + 6HCl -> 6KCl + 3I2 + 3H2O}$$ This is a way to generate a known amount of $\ce{I2}$.

    You would then dissolve $\ce{Na2S2O3}$ in a buffer (sodium bicarbonate). This is done because $\ce{Na2S2O3}$ is not stable under acidic conditions. Then you can titrate $\ce{I2}$ in presence of starch with $\ce{Na2S2O3/NaHCO3}$. $\ce{NaI}$ and $\ce{Na2S4O6}$ are formed in this reaction.

  2. In any case you cannot replace $\ce{HCl}$ with other acids. $\ce{H2SO4}$ and $\ce{HNO3}$ are oxidizers strong enough to turn some more $\ce{KI}$ into $\ce{I2}$ and mess up your measurement.

  • 1
    $\begingroup$ no I am pretty sure that's the process.because later we add potussium dichromate solution.this is the process of producing iodine gas from potussium iodide only with potussium dichromate. $\endgroup$ Jul 24, 2016 at 4:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.