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Standardization of sodium thiosulfate using potassium dichromate
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Michael Ahler
Standardization of sodium thiosulfate using potassium dichromate
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.
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.
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.
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.
no I am pretty sure thats the process.because later we add potussium dichromate solution.this is the process of producing iodine gas from potussium iodide only with potussium dichromate.More
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.
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.
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.
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.
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