chem(titration)

Q.1
Iodine was formed between the reaction of iodide ions and iodate(V) ions in an acidic medium. The iodine formed then reacted with the excess iodide ions to give deep brown triiodide ions. Therefore, the solution is deep brown at the beginning of the titration.
I2(aq) + I^-(aq) ≒ I3^-(aq) ...... (*)

On addition of sodium thiosulphate solution, the thiosulphate ions removed the iodine ions in the solution.
I2(aq) + 2S2O3^2-(aq) → 2I^-(aq) + S4O6^2-(aq)
The removal of iodine led to that the equilibrium position of reaction (*) shifted to the right, and thus the concentration of triiodide ions decreased. As a result, the solution became paler in colour.

When the solution became yellow, the concentration of triiodide ions was low. On addition of starch indicator, the solution became dark blue due to the formation of iodine-starch complex, which is dark blue in colour.
I2(aq) + starch(aq) ≒ [I2-starch] ...... (#)

Further addition of thiosulphate ions continued to remove iodine in the solution, and this shifted the equilibrium position of reaction (#) to the left. At the end point of the titration, all iodine was released from the iodine-starch complex. Due to the absence of iodine, triiodide ions and iodine-starch complex in the solution, the solution changes to colourless.


Q.2
This is because iodine is volatile.


Q.3
At the beginning of the titration, the concentration of iodine is high. If starch is added at that moment, the binding force between iodine and starch in the iodine-starch complex is very strong. This makes the complex not completely dissociate at the equivalent point. This causes a fatal error of the experiment.