Nat_LPS
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Hello. Been given a titration to do some calculations on, could someone check it? sitting clueless in chemistry hurts. I have no idea if what i'm doing makes sense, so any help would be awesome
here is is:

Potassium iodate(V), KIO3, is used in the standardisation of aqueous sodium thiosulphate.
Iodine is liberated from excess acidified aqueous potassium iodide in:
IO3^- + 6H^+ + 5I^- --> 3I2 + 3H2O
and the liberated iodine is titrated against sodium thiosulphate.

Q1) Write down the balanced equation for the reaction between iodine and sodium thiosulphate.

Q2) A mass of 0.1500g of potassium iodate(V), KIO3, was dissolved in water and made up to 250.0cm^3 of solution.

25.00cm^3 of this solution was added to excess aqueous potassium iodide and aqueous acid and the liberated iodine just reacted with 22.60cm^3 of aqueous sodium thiosulphate.

Calculate the concentration (in mol dm^-3) of the potassium iodate(V) solution and hence the concentration of the aqueous sodium thiosulphate.

note: Not sure about the first paragraph, we're adding iodate to iodide, to get iodine, (which will then be titrated against thiosulphate), how does this standardise thiosulphate? The excess iodide reacts hear, so what did the previous amount before excess react with?


Alright, so for question uno, I know thiosulphate is oxidised so iodine is I'm guessing reduced, from those i got:

I2 + 2S2O3^2- --> S4O6^2- + 2I^- (eq. 1)

I think this is above equation is one used for the calc in Q2, a long with the given:

IO3^- + 6H^+ + 5I^- --> 3I2 + 3H2O (eq. 2)

maybe to see how many moles of iodine in (eq. 2) is available to react in (eq 1) - the titration

Q2) I get a bit lost here,

moles iodate in 250cm^3 = = (0.15/214) = 0.0007009...
conc iodate = (0.15/214)/.25 = 0.0028... In 25cm^3 = .../10 = 0.00028...
(or moles in 250cm^3 x4)

moles iodate in 25cm^3 = moles in 250cm^3 / 10 = 0.00007009...

3x this = 0.00021028...moles of iodine reacted with 22.60cm^3 thiosulphate

from (eq. 1) would need to do moles iodine x 6 = moles thiosulphate = 0.00126168...

conc thiosulphate = mol/vol = 0.00126168.../ 0.0226 = 0.0558mol dm^-3

none of this feels familiar aha.
any hint? Thanks. Hope it's clear enough.
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Borek
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(Original post by Nat_LPS)
note: Not sure about the first paragraph, we're adding iodate to iodide, to get iodine, (which will then be titrated against thiosulphate), how does this standardise thiosulphate? The excess iodide reacts hear, so what did the previous amount before excess react with?
We use produced iodine for standardization. Iodide itself is not reacting with thiosulphate.

See if this: http://www.titrations.info/iodometric-titration won't help you understand what is going on (this is just the entry point, there are more pages about the iodometry).

I2 + 2S2O3^2- --> S4O6^2- + 2I^- (eq. 1)
OK

moles iodate in 250cm^3 = = (0.15/214) = 0.0007009...
OK

conc iodate = (0.15/214)/.25 = 0.0028...
OK

In 25cm^3 = .../10 = 0.00028...
Nope, something is wrong here. Concentration doesn't change just because you took 25 mL. Amount of iodate does.

moles iodate in 25cm^3 = moles in 250cm^3 / 10 = 0.00007009...
so that's OK again.

3x this = 0.00021028...moles of iodine reacted with 22.60cm^3 thiosulphate
OK

from (eq. 1) would need to do moles iodine x 6 = moles thiosulphate = 0.00126168...
No, not 6. 2 only. 3 was already taken care of above, when you calculated number of moles of iodine.
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Nat_LPS
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(Original post by Borek)
We use produced iodine for standardization. Iodide itself is not reacting with thiosulphate.

See if this: http://www.titrations.info/iodometric-titration won't help you understand what is going on (this is just the entry point, there are more pages about the iodometry).



OK



OK



OK



Nope, something is wrong here. Concentration doesn't change just because you took 25 mL. Amount of iodate does.



so that's OK again.



OK



No, not 6. 2 only. 3 was already taken care of above, when you calculated number of moles of iodine.
That's much clearer. Thanks a lot. and for the link.
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