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In a certain titration experiment,

1) when $\ce{KIO3}$ solution is added to $\ce{KI}$, the $\ce{I-}$ ion is oxidised to the triiodide complex, $\ce{I3-} $

$\ce{IO3- + 8 I- -> 3 I3- + 3H2O}\tag{1} $

2) Ascorbic acid immediately reduces the triiodide complex to the colourless $\ce{I}^- $ ion.

$\ce{C6H8O6 + I3^- + H_2O -> C6H8O7 + 3I^- + 2H+}\tag{2} $

3) When all the ascorbic acid is reacted with $\ce{I3}^- $ , the excess $\ce{I3}^- $ will react with the starch solution to form a deep blue $\ce{I3}^- $ starch complex.

$\ce{I3^- \text{ + Starch} -> I3^-}\text{- starch complex}\tag{3}$

At the experiment I have the moles of $\ce{KIO3}$ let’s say $0.1$ moles.

If I want to find the moles of $\ce{C6H8O6}$ can I say that the mole fraction is -> $ \ce{IO3^-} : \ce{I3-} : \ce{C6H8O7} $ is $1:3:1$ ? Meaning the moles of $\ce{C6H8O7} $ is $0.1$ moles

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  • $\begingroup$ Try using equivalents concept. It is more effective $\endgroup$ – user600016 Nov 7 '18 at 5:41
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Combining equation (1) and three times equation (2) gives an overall reaction

$\ce{IO3- + 8 I- -> 3 I3- + 3H2O}\tag{1} $
$\ce{3C6H8O6 + 3I3^- + 3H_2O -> 3C6H8O7 + 9I^- + 6H+}\tag{2a} $

$\ce{IO3- + 3C6H8O6 -> 3C6H8O7 + I^- + 6H+}\tag{4}$

So $\ce{IO3- \text{ to} C6H8O6}$ is 1::3.

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