Starch solution as indicator

When starch solution is added to a solution containing iodine, a dark blue complex [starch-I2] is formed. The formation of such complex is reversible and an equilibrium is established.
starch + I2 = [starch-I2]

When Na2S2O3 solution is added to the reaction mixture, iodine reacts. Due to the decrease in concentration of I2, the equilibrium position of the above equilibrium shifts to the left. At the equivalence point, it is expected that all I2 is released from the [starch-I2] complex and is then reacted. This causes the disappearance of the dark blue colour and is determined as the end point.

However, if the concentration of starch is too high, the binding force between starch and I2 in the [starch-I2] complex is very great . In other words, it is difficult for the complex to release I2, and the complex still exists in the reaction mixture even at the theoretical equivalence point. This causes a great error.

Therefore, starch indicator should be added when concentration of I2 is low enough. When the iodine solution turns yellow, it indicates that the concentration of I2 is low.

呢個係預科其中一個實驗遇到既問題
2S2O3^2- + I2 -> S4O6^2- + 2I- starch as indicator

如果個 solution係鮮橙色／even啡色，starch同iodine就會 form左一隻 brown precipitate 出黎
而該 precipitate唔會 redissolve返落個 titrant入面，咁樣個 titration即使 determine到個 end point但唔可以搵到 exact amount of the analyte(iodine)。因為brown precipitate係一個由多個iodine molecule cooridinate with starch molecule既 complex，測出既 amount of iodine就少左。因為用既 thiosulpate份量都係少左。
如果係pale yellow就代表 iodine既 concentration 就係 very low，唔會 form brown precipitate, instead 會 form 左 purple complex。對 amount of analyte既影響亦會細好多，determine 出黎既 thiosulphate used就會準好多，亦都搵到 amount of iodine presents in the solution

但而家對於 formation of starch & iodine 依然都係一個 unknown，even查大學既 reference/ journal都唔會搵到該 purple complex既結構