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What is the role of iodine here? And why only sodium thiosulphate is used for titration? Why do we add $\ce{Na2CO3}$ here. Also why do we add acetic acid here?

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closed as off-topic by Jan, getafix, Todd Minehardt, Klaus-Dieter Warzecha, paracetamol Jan 21 '17 at 9:56

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There is a pretty good explanation of this procedure here.
It is stated that:

Iodometric determination of copper is based on the oxidation of iodides to iodine by copper(II) ions, which get reduced to $\ce{Cu+}$.

The initial reaction, not the titration, is given as:

$\displaystyle \ce{2Cu^2+ + 4I- → 2CuI(s) + I2}$

Then the titration reaction is:

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

Sodium thiosulphate makes for an inexpensive, safe and effective reducing agent to react with $\ce{I2}$.

Since starch turns to a dark blue in the presence of $\ce{I2}$, then back to white when the $\ce{I2}$ is depleted, starch makes a good indicator for the endpoint of this reaction.

The oxidation reaction of iodides to iodine by copper(II) ions is most effective under slightly acidic conditions, with a pH of around 4-5. This is the reason for the addition of acetic acid and sodium carbonate. Whenever a procedure calls for the addition of a weak acid and a weak base, there is a pretty good chance that the purpose is to form a buffer solution at some desired pH. At least in this case, that is exactly the purpose. You will notice in the reference above that they use ammonia and acetic acid instead. Sodium carbonate is preferred in some situations for safety reasons.

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    $\begingroup$ The presence of iodine is detected because of the equilibrium $$\ce{I- + $n$ I2 <=> [I_{$2n+1$}]^-}$$ which forms a complex with starch; just to be more precise. Otherwise, this is of course correct. $\endgroup$ – Martin - マーチン Jan 21 '17 at 10:02

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