A-Level Chemistry Sodium Thiosulfate and Iodine Titrations

[@jensychiang8352]

You have no idea how much this saved me. Thank you so much!

[@tomiolasoji9712]

I'm so happy I found this. This was so well explained, thank so much you for doing this!

[@beauchampchemistry9031]

Very welcome- glad you found it useful

[@athithyan3879]

I want you to know i loved this vid, saved me for my practical thank you SOOOO much

[@beauchampchemistry9031]

So pleased it was helpful for you! 👍

[@MishaAisyah]

thank you so much Mrs Lucas for the explanation!

[@beauchampchemistry9031]

Very welcome! Thanks for watching

[@ophira4081]

Thanks for such a amazing video . I wish I could have teachers like u in our university 😢 By the way it's amazing 👏

[@beauchampchemistry9031]

Thank you for your kind comment- so glad you found it useful!

[@essie180]

This is so well explained, thankyou so much !!

[@Anant_1939]

Nice explanation!!

[@yousifmajeed9316]

These are very good and clear explanations.

[@multi-talentwithgrace1761]

Thanks so much....I can't expressed how I feel bcos being an empty headed student in this topic just changed to a sensible student by watching ur video

[@jeond9762]

Thank you so much mrs lucas

[@kevinkularatne327]

Nicely Explained! Thank you!

[@beauchampchemistry9031]

Very welcome- thanks for watching!

[@ayasamer7731]

Thank u soo much❤ Finally I understand iodometric 😍❤ We want more practices to be able to calculate precisely

[@angelkorbyn3038]

I guess I am pretty off topic but does anybody know of a good place to watch new movies online ?

[@thinkdontshoot8065]

@@angelkorbyn3038 bruh

[@oluwakemiodedina2858]

@@angelkorbyn3038 there are much better places to ask this

[@jenicagong5193]

Thank you so much for such a clear explanation. This video has been extremely helpful

[@GangGang-dx7te]

Could you help me how to solve this problem ? Ma'am.... One vitamin C tablet is dissolved in an acidic solution. Add 2 g of solid KI and 50.00 mL of the 0.01795 M KIO3 solution. The solution is titrated with a 0.1258 M thiosulfate solution and reaches the endpoint at 6.12 mL of titrant. What are the moles of triiodide consumed by the vitamin C tablet?

[@adrianaalejandrahernandezp8040]

Greetings from México, thanks for this video now I understand

[@beauchampchemistry9031]

You are so very welcome- thanks for watching

[@mattearthling1691]

Thanks, really helped glad I found this.

[@agborjames3375]

Thank you so much for the insight madam

[@anfaljibril3494]

Step 7 is incorrect as the mass of KIO3 should be over mass of the impure KIO3

[@tamaradifc7566]

Yes. 0.847/1 x 100

[@obey8100]

THANK YOU FOR THE EXPLANATION

[@beauchampchemistry9031]

Very welcome!

[@sherifhabbak8868]

I have a commercial Povidone-Iodine solution that I would like to test to know the amount of free iodine in this solution. how do I prepare the Iodate solution? should I jump to equation 2 right away and start titrating with sodium thiosulphate without the previous reaction?

[@amjedelnima2852]

Great explanation!

[@beauchampchemistry9031]

Thanks so much!

[@zahirahhannani7362]

I have a question regarding on redox titration experiment. Can you explain what is the purpose of standardisation on aknown concentration sodium thiosulphate solution?

[@rume3137]

In the begging, we accept the concentration of sodium sulphate as 0.100 M but are we sure that this is the real concentration value of the sodium sulphate? For example; if the real value is 0.089, the number of moles will change and it will effect the results.My opinion is like that.

[@bakacheyounes9325]

thank you so much

[@aishworyaparajuli3949]

do we need to know sodium thiosulfate redox calculations for aqa? thanks

[@Frederick-765]

What is the order of reaction

[@OwnsYourFace204]

Why is H+ added?

[@thomasc886]

the H+ takes place in the reaction by combining with the oxygen from the iodate ions to make water, and also, although it isn't shown, the anions (negative ions) that would have been bonded to the hydrogens in whatever acid they were part of would replace the iodide anions on the left side of the equation by bonding to the cations (positive ions) that they had come bonded to, most often sodium or potassium, though it could of course be others.

[@nainakaur9680]

Thank you

[@willhales4553]

I'm confused- in step 4 where 1.875x10^-3 is divided by 3 how do you get 3.958x10^-4? Isn't it 6.25x10^-4? Am I missing something?

[@beauchampchemistry9031]

Thanks Will- I have absolutely no idea how I got this wrong!! I shall blame lockdown brain! Thanks for spotting!! As long as everyone understands the process then hopefully it will still be useful!

[@kamilakucerova1620]

​@@beauchampchemistry9031 First of all - thank you for this video, it really helped me! I think there's a mistake in the step 2 instead of step 4. I've tried to solve this task on my own and when I divided 2.375^-3 by 2, I got 1.1875^-3 instead of 1.875^-3. Then when I used 1.1875^-3 in step 4, I got 3.958x10^-4 as in the video. Is it right? :)

[@ehandpiece]

WoW! Your smart.. i don't understand this stuff and i am a MAN! I would not have known it was wrong. Weird? Thanks.:)

[@RonnieHeuvel]

@@kamilakucerova1620 What you did is correct. She didn't put 1 before the 875 in the video. But the numbers don't really matter if you just want to understand the process.

[@673king6]

[@agborjames3375]

Please step 7 is 0.847/1* 100

[@sirus6167]

Why is purity 1g/0.847 and not the other way around

[@sirus6167]

yeah just rewatched it and yeah that's a mistake. It's (8.47/1) x 100

[@uchenduchigozie3488]

Thanks

[@golddrogershankskaido8614]

i love it....

[@golddrogershankskaido8614]

make me more understand

[@patrickk7946]

Video was very useful but the mcvities bant is too dead