Iodine / Thiosulfate Redox Titration Demonstration
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- เผยแพร่เมื่อ 15 ก.ค. 2024
- How can the reaction of iodine and thiosulfate ions be used as a titration?
00:00 - Introduction
00:06 - Reaction between Iodine and Sodium Thiosulfate
00:46 - Detecting the End-point Using Starch as an Indicator
02:35 - 2-step procedure to produce iodine and then titrate against thiosulfate ions.
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I am an A/ L student in sri lanaka .This vedio was very helpful to my redox lesson.Thank you so much ♥️ from Sri Lanka
Yes, This is very helpful for A/L exam in Sri Lanka 🇱🇰 thank u very much 😊
Quite relatable mate, it's an entirely different feeling when you have actually seen the reaction compared to just knowing theory.
This video is PERFECT, thank you so much for recording it and sharing it here. I finally understand it!
It really is impossible to study chemistry just by looking at formulas. Seeing it with my own eyes made the reaction crystal clear for me. Can't wait to do this myself! :D
Thank you sir. This is what I wanted to see
Neat, thanks. Really handy for Winkler method endpoint.
This was awesome. Thanks a lot sir!!
I'm A/L student from sri lanka...thank you very much...
Thank you very much
Thank you so much
Thank you very much...(A/L student Sri Lanka)
Thanks so much sir
Thanks sir
Really helpful thx
Glad it helped
ncert chem brought me here :) . Thanks
what is there in conical flask in 2nd case
Waec 2022 chemistry practical Q1
Redox titration
Given 25.4g of I2 n 6.0g KI per dm3
24.82g Na2SO3.5H2O
Starch solution
Please do the titration and possible calculations involved
hadir
If sodium thiosulfate is 0.1N ,If we use10%ki than this is right method?
thanks for the video. I had a question and that is when I add starch, the dark blue color doesn't get colorless no matter how much sodium thiosulfate I titer into it. what can I do to solve this problem?
You probably added the starch too early leading for the starch to bind too tightly to the iodine
What happens when excess sodium thiosulfate is added? Does the solution in the conical flask remain colourless?
Yes, it will do. But if you carry out the titration correctly you'll never get to see this!
@@KEGSChemistry Thank you Sir!
Hi, I did this few days ago and there was quite alot of precipitate and I think it is iodine, but I dont know how to explain why this is happening? and I saw pink-red gas forming and I cant come to any conclusion. I hope you answer me.
Hi Shad, if you could let me know at which point this happened and the exact solutions you were using then I may be able to help.
@@KEGSChemistry Hi, the precipitate and the pink gas was when i first added 2mL 5M HCl and 5mL 0.1M KI and 20mL diluted chlorine (we diluted 4mL chlorine to 250mL with deionized water). The precipitate was totally gone after i titrated til it became light yellow and there was nothing after the titration was totally done(it became colorless). Im not sure about if the gas was gone, because we didnt even notice it and the teacher pointed it out afterawrds. I already handed my report and explained that the precipitate might be iodine (since it is black), and this happened because there wasnt enough iodide to make triiodide, which is more soluble. And the gas might be explained with iodines volayility. I really hope you can help me, because i can fix on it before getting the grade. (We needed 20.8mL 0.02mL Na2S2O3 in average to finish the titration, the mass precentage of NaOCl in house chlorine was found to be 4.6% but on the bottle it saids around 4%). Thank you very much.
@@shadl2676 Thanks for the clairification. Based on the numbers you have given me I think your conclusion of there not being enough KI present is accurate. After mixing your Chlorine solution and KI you would have a 0.17% (wt/v) concentration of Iodine, but the solubility limit in pure water is 0.029% (wt/v) - i.e. much lower. This is enhanced in the presence of iodide, but in an aqueous iodine (Lugol's) solution you have doulbe the % (wt/v) of KI to form triiodide. Based on the volume of KI which you used, you will only have 0.086% (wt/v) of KI, which is actually lower than iodine and hence not enough to make iodine soluble. When you add thiosulfate, it regenerated iodide, which presumably then re-dissolved your iodine and led to it completely reacting with the thiosulfate. I guess this could have taken time hence you over-shot the end-point. But if gas left the beaker I would have expected you to register less chlorine than expected in your titration. If there's a next time, make sure you have a big excess of iodide ions. We tend to use 1 M KI to achieve this, but it is expensive so I'm sure you could get away with less.
@@KEGSChemistry Thank you very much for answering. It makes more sense now. About the gas, i expected to register less chlorine too but from the data and calculations i registered a bit more. As i said earlier i registered 4.55% while on the flask it said around 4% and i think they mean at least less than 4.5%, so i'm really flustered. As i mentioned earlier i wrote that the gas might have something to do with iodine's volatility, but do you think that the gas is iodine or could it be something else?. And i wonder about another thing, does thiosulfate react with only triiodide, or with iodine too?. Thank you again, i really appreciate it, none of my teachers was willing to help me, you saved me.
How to indicate the endpoint when the potassium iodide is in the burette during titration?
It's never done this way as you want excess iodide in order to generate a quantity of iodine that is directly proportional to the number of moles of your analyte. If you look at this video it explains more: th-cam.com/video/4YYmj_OtOV4/w-d-xo.html
Is it possible to perform this experiment by filling burette with iodine and conical flask with sodium thiosulfate?
I've never seen it done this way! It would make the colour change very hard to detect as iodine would bind to starch present and make it go black immediately after addition. It would then fade and come back when you added more. Thiosulfate in the burette is much better!
Hey! May I ask how did you manage to solve I2 in water?
Iodine is very insoluble in water (as you perhaps already know). The trick is to make it up in a solution of potassium iodide, which causes much more to dissolve as the I^- and I_2 form a triiodide ion (I_3^-). This ion is what gives the brown colour to the solution. If you want to make an aqueous iodine solution, try searching for a recipe for "Lugol's iodine" and you should find something sensible. Alternatively, this type of solution is probably fairly easy to buy, but you have to be careful with the label as to what the concentration of I_2(aq) actually is.
As we use excess KI to form the I_2 here, there will always be enough I^- ions to keep the I_2 soluble and the I_2 produced thus has the typical brown colour of the I_3^- ion.
@@KEGSChemistry all right thank you very much
@@KEGSChemistry How was the I_2 produced when KI was added to Fe^3+ solution?
@@SandeepKumar-jj7zi Fe3+ reduces I- to form I2
@@bh-rf9dd or oxidizes?
How do i write its observations and calculations in practical notebook any body help me ?
Hi there, please correct me if I'm wrong, but I think I can explain it:
The point of the whole thing is: you want to determine the amount of Fe3+ in the solution, so you first reduce it with 2I- which turns into I2, and then you reduce the I2 with 2Na2S2O3 to 2I-. Amount of Fe3+ in the original solution was equal to HALF the amount of Na2S2O3 you added in the end:
1) Fe3+ + 2I- ---> I2 + Fe2+ (I2 is dark brown)
2) I2 + 2Na2S2O3 ---> 2NaI + Na2S4O6 (solution turns transparent)
See? 1 mole of Fe3+, 2 moles of Na2S2O3.
You know how much of Na2S2O3 you've added, so you can calculate how much of Fe3+ there was originally.
Now, why do we make a pause in adding Na2S2O3 when the solution is light yellow and add starch? Why can't we just add Na2S2O3 until the solution turns transparent?
To answer this, let's look at step 2 again:
I2 + 2Na2S2O3 ---> 2NaI + Na2S4O6
As i said, solution turns transparent, but it does so gradually(!), thus NOT letting you determine precise amount of sodium thiosulfate you needed to reduce all the I2 to 2I-.
Well, starch solves this problem:
1) Fe3+ + 2I- ---> I2 + Fe2+ (I2 is dark brown)
2) 2I2 + 2Na2S2O3 ---> I2 + 2NaI + Na2S4O6 (that's the pause: small amount of remaining I2 makes the solution light yellow)
3) starch + remaining I2 ---> dark blue complex of starch and I2 (I2 is still reactive!)
4) [I2 + starch complex] + 2Na2S2O3 ---> starch + 2NaI + Na2S4O6 (solution turns transparent instantly(!) when all the iodine reacted, thus LETTING you determine precise amount of sodium thiosulfate you needed to reduce all the I2 to 2I-).
I'm three weeks late but I hope this helps!
Here because we can't do this at school
anyone here from the lesson lol
What order of reaction is this sir?
It's zeroth order as the concentration of iodine doesn't affect the rate of the reaction
21
But the starch is an indicator, you were supposed to add it to the iodine before titrating
As he points out at minute 1:14, the starch is added close to the end-point to avoid inaccurate results because of strong bindings to the iodine
Kindly sir help.me
Sorry this is so late, but if you want to learn about the calculations you could try watching this video here: th-cam.com/video/4YYmj_OtOV4/w-d-xo.html
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