[EndlessDelusion]

Hi, I'm new here and having a little trouble with some coursework. If you could help, great! Basicly, to bring you upto speed, I've been planning an A2 investigation (if you're from anywhere but england, that just means an education stage between Highschool and Uni) and it requires an individual investigation.

Questions are in bold for your viewing pleasure. Sorry it's so long.


Basicly to find out how much oxygen is in a sample:
1) Reacting manganese (II) hydroxide with water containing oxygen creates Manganese (III) hydroxideoxide.
2) React the Manganese (III) Hydroxide with Hydrochloric acid and Pottasium iodide, which willl create a whole heap of things, including Iodine.
3) titrate the iodine with sodium thiosulphate and you can find when there are no more Iodine molecules (as they turn into ions).

In depth that means:
O₂ (aq) + 4Mn(OH)₂ (s) + 2H₂O (l)  4Mn(OH)₃ (s)

2Mn(OH)₃ (s) + 6HCl (aq) + 2KI (aq) 2MnCl₂(aq) + 2KCl (aq) + I₂ aq)+ 6H₂O (l)

Then a titration:
I₂ (aq) + 2Na₂S₂O₃(aq) Na₂S₄O₆ (aq) + 2NaI (aq) (Basicly when all the iodine has gone, that’s the end point)


First off, does that sound like a plausable experiment? This is the big thing that needs addressing before ANYTHING else happens. I mean, would the reactions happen at all? It took some time to work this out!

Heres were it gets a little tricky.
I have at my resources:
1 mole of Manganese (II) Hydroxide in a  40:60 ratio with dilute water
1 mole of Hydrochloric acid in a 30:70 ratio with dilute water with 2 grams of Potassium Iodide
0.0125 mol dm³ sodium thiosulphate solution (2Na₂S₂O₃)

So I’m just wondering..  Doesn’t the 40:60 ratio just mean that the solution is  reducing in molarity.. or am I just being a plank? (sorry if that's a bit numptyish.. It's been months since we revised anything on molar equations)!
So that means the manganese(II)Hydroxide is actually 0.4 mols? and HCL 0.3?

I have a question about actually working out how much oxygen is actually in a sample.. but that's long and laborious and best not push my luck on a new forum!

Sorry for the wall of text on 2 simple questions.. but hopefully some of you can answer a few questions.. At least this isn't last minute before actually doing the coursework!
thankyou for reading - cheers!

[Borek]

I am not sure about this part with Mn(III) - but otherwise the idea seems more or less logical. Even if there is no Mn(III) but Mn(IV) as the intermediate, it doesn't matter - what matters is molar ratio of iodine produced per mole of oxygen, and that won't change regardless of intermediates. You may as well think of it as O₂ + I^- -> I₂ (plus some water and OH^- or H⁺ added for balancing) - and treat manganese just like catalyst.

No idea how to read these 40:60 or 30:70 things. 1M HCl diluted 30:70 with DI water would be obvious, but as stated it doesn't make much sense to me.

[EndlessDelusion]

Ah thankyou Borek..
And what I ment was ratios with the 30:70 etc..  I should have made that clearer- so it technicly woould just mean it's being diluted down.
And I was going to ask about the ratio of I^- from the O₂ and how to work it out, but that's something to do later.

Thanks for a clear description and taking the time to reply!