Chemistry - Do both iodine and potassium iodide turn dark in the presence of starch?

Solution 1:

Starch is a long ribbon or a long filament made of a great number of glucose units attached to one another like the wagons in a train. But this filament is wound in a helicoidal way. And the inner part of this helix is a sort of long hole. And the dimension of this hole is just big enough to allow the iodine molecule $\ce{I2}$ to enter and stay there, maintained by van der Wals forces. But when the iodine molecule is inserted in the starch helix, it produces a "starch-iodine" complex which has an intense blue color.

Solution 2:

I want OP to understand that iodine ($\ce{I2}$) is not the one giving the starch solution deep blue color. It is rather triiodide ion ($\ce{I3-}$), which would form in presence of excess iodide ion ($\ce{I-}$):

$$ \ce{I2 + I- <=> I3-} \tag1$$

This is an important aspect in iodometry titrations, because $\ce{I3-}$ is very water soluble compared to partially or not particularly water soluble $\ce{I2}$, which is one of the two major disadvantages of using $\ce{I2}$ as a titrant. The second disadvantage of using $\ce{I2}$ as a titrant is $\ce{I2}$ is somewhat volatile. Consequently, there is a significant amounts of dissolved iodine from the solution by evaporation (sublimation?). Therefore, adding enough iodide ($\ce{I-}$) to iodine ($\ce{I2}$) solution would make it to overcome both of these disadvantages by the reaction in equation $(1)$ (i.e. in the presence of $\ce{I-}$, $\ce{I2}$ reacts to form $\ce{I3-}$, which is highly soluble, and most importantly, not volatile). In addition, $\ce{I3-}$ ions give the needed deep blue color with the indicator, the starch solution.

For example, the color changes in one perticular iodometric titration is depicted in following image:

Color change in iodometric titration

Keep in mind that when you have known concentration of potassium permanganate solution and excess potassium iodide solution together in acidic medium, it is actually become a $\ce{KI3/KI}$ solution. The major chemical species present in the solution is $\ce{I3-}$:

$$ \ce{MnO4- + 8 H+ + 5 e- <=> Mn^2+ + 4 H2O } \tag2$$ $$ \ce{2I- <=> I2 + 2e-} \tag3$$ Sum of $2 \times (2)$ and $5 \times (3)$ gives: $$ \ce{2 MnO4- + 16 H+ + 10I- -> 2Mn^2+ + 5I2 + 8 H2O } \tag4$$ The excess of $\ce{I-}$ react with freshly formed $\ce{I2}$ to give: $$ \ce{I2 + I- <=> I3-} \tag1$$

Dilute triiodide solutions are yellow (as shown in $(b)$ in the image), more concentrated solutions are brown, and even more concentrated solutions are violet (as shown in $(a)$ in the image). If you add starch solution at the beginning, excess $\ce{I3-}$ would destroy the starch structure. That's why you need to titrate dark color to yellow color with thiosulfate solution first before the addition of starch. That time, $\ce{I3-}$ concentration is dilute enough, yet give a dark blue color by making the $\ce{I3-}$-starch complex (see the insert in the image at right hand side). The end point would be dark blue to very pale pink becace of the presence of $\ce{Mn^2+}$ ions (not colorless as shown in the image).