[cliverlong]

Hi,

I am trying to learn, but more importantly understand, the titration of oxidizing agent, iodine and thiosulphate

Now, many, many books give detailed instructions on how to perform the procedure but precious little why it is performed that way. The procedure seems quite elaborate and I can't explain why it has to be done that way and no simpler approach would work.

Now what I have determined

The purpose of the exercise is to determine the concentration of the oxidizing agent in solution (I suppose this is a good thing)

Now,

Sodium thiosulphate is the reducing agent

Iodide/ Iodine is the indicator

Problem. I just keep getting "blocked" at the step I add thiosulphate from the burette

Iodine is gaining electrons to change to I^- so is being REDUCED

Therefore thiosulphate is the reducing agent. So why doesn't the thiosulphate react with the oxidizing agent? Why does it react with I₂? Surely the "important" reaction is reducing agent with oxidizing agent? I keep going round in circles at this point.

Can anyone throw any light on why the titration is run the way it is without restating how the titration is performed (which I have documented).


Thanks

Clive

[Borek]

I₂ is both oxidizer AND indicator (together with the starch).

[wilson]

Thiosulphate is reducing, iodine is oxidizing, iodide is reducing, and the oxidizer is of course, oxidizing.
I think the iodine is reduced by thiosulphate to form iodide, and the iodide is subsequently oxidized by your oxidizing agent to form iodine again--detected by using starch.

1. A known concentration of thiosulphate and iodine are reacted to produce a known amount of iodide ions.
2. The oxidizer is then titrated to the known amount of iodide ions. Acidification is required to introduce protons. Iodine is formed again.
A starch indicator is added near the end point (when the solution is brown). A blue ppt. is formed due to the complexation of iodine and iodide ions. The end point is detected when the blue ppt. fades.

What do you think?

[Borek]

A starch indicator is added near the end point (when the solution is brown). A blue ppt. is formed due to the complexation of iodine and iodide ions. The end point is detected when the blue ppt. fades.

Think: if blue ppt is formed due to the complexation of iodine and iodide ions, do you have to ad starch to see the ppt?

Besides, you seem to assume there is a separate oxidizer in the solution. There is no such thing, it is iodine that is being determined and that is oxidator. Could be it was prepared earlier using some oxidizer, but at the time of titration it no longer matters.

[wilson]

A starch indicator is added near the end point (when the solution is brown). A blue ppt. is formed due to the complexation of iodine and iodide ions. The end point is detected when the blue ppt. fades.

Think: if blue ppt is formed due to the complexation of iodine and iodide ions, do you have to ad starch to see the ppt?

Besides, you seem to assume there is a separate oxidizer in the solution. There is no such thing, it is iodine that is being determined and that is oxidator. Could be it was prepared earlier using some oxidizer, but at the time of titration it no longer matters.

1. Yes, a the starch indicator is required. I2 + I- -->I3- And I3- forms a complex with the starch to form a blue ppt. Sorry about any misunderstanding.

2. I get what you mean. I agree with you-- the iodide is the one that is determined here. For a moment I thought, what a "special" titration this is! When I read "the purpose of the exercise is to determine the concentration of the oxidizing agent in solution", I just treated the oxidizer as some other oxidizer (not I2).

I must have been low on coffee.

[cliverlong]

Besides, you seem to assume there is a separate oxidizer in the solution. There is no such thing, it is iodine that is being determined and that is oxidator. Could be it was prepared earlier using some oxidizer, but at the time of titration it no longer matters.

I start to see where you are coming from. The roles of the two oxidizing agents are mixed up in my question.

I will copy the question into this thread. My reading of the questions was definitely that it was the concentration of the acidified permanganate that was being determined.

I'm too tired to type it in now.

I'll come back to this in a couple of days and maybe it will all fall into place.

Clive