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

I will do a titration to standardise a solution of sodium thiosulfate with potassium permanganate solution and potassium iodide solution.

My book says that when the potassium permanganate solution begins to lighten, add a few drops of starch indicator and titrate until the solution turns clear.

This seems to suggest to me that starch turns $$\ce{I_2}$$ dark but not $$\ce{KI}$$ as some of the $$\ce{KI}$$ solution would be still in the flask. Is this the case?

• Yes, that's right. – Ivan Neretin Sep 15 '20 at 18:50
• @Ivan Great, so just to confirm, starch indicator does not turn KI solution dark ? – Kantura Sep 15 '20 at 18:53
• Yes, starch is an indicator for I2 and not for KI. – Ivan Neretin Sep 15 '20 at 18:58

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.

• Cool , I didn't know that. But what about KI ? Does that turn dark with starch ? – Kantura Sep 15 '20 at 18:51
• @Maurice Much as I respect your expertise and commitment, I can't help but notice that you've chosen an excessively long and convoluted way to say "yes, that's right". – Ivan Neretin Sep 15 '20 at 18:56
• @ Kantura. Starch and KI do not react with one another, and do not produce any color. – Maurice Sep 15 '20 at 19:26
• @ Ivan. You may be right. But I cant't prevent me from thinking that Kantura did not know how to measure the surface of a peak. And now he knows. – Maurice Sep 15 '20 at 19:28
• Thanks all , my question was answered and I welcome and appreciate the deeper understanding – Kantura Sep 15 '20 at 19:45

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:

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).