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Original post by GeorgeL3
For the Aspirin practical make sure you look up it's melting point first.
The answer you give should be a few degrees lower due to impurities (but not 60°C lower like mine turned out :-S ).

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Yep. To the asker, have a look at the role of the apparatus and why you've set it up that way as well.
guys is there a thread for the paper 6 int'l variant?

If not, on the paper 6 for 2013 jan, on question1, theres a part where they add XS NaOH to a Fe 2+ containing solution, and a green PPT persists. so its an expected result.
in the marking scheme you are expected to say:

not amphoteric / no furtherdeprotonation occurs / no ligand substitution occurs.

----
I dont get the part where it says "not amphoteric" - what does that mean, to what unit does that refer to anyway?

any help would be appreciated!
Reply 802
Original post by yahyazhar
guys is there a thread for the paper 6 int'l variant?

If not, on the paper 6 for 2013 jan, on question1, theres a part where they add XS NaOH to a Fe 2+ containing solution, and a green PPT persists. so its an expected result.
in the marking scheme you are expected to say:

not amphoteric / no furtherdeprotonation occurs / no ligand substitution occurs.

----
I dont get the part where it says "not amphoteric" - what does that mean, to what unit does that refer to anyway?

any help would be appreciated!

considering xs means excess :tongue:

Its because nothing further occurs when the excess naoh is added. with limited naoh a green ppt is the result. with excess the green precipitate remains so the addition of the excess naoh has no effect.

for eg with naoh and chromium in limited naoh a green ppt forms but with excess naoh a green solution forms slowly, so further reactions occur (ligand exchange).
Original post by jojo1995
considering xs means excess :tongue:

Its because nothing further occurs when the excess naoh is added. with limited naoh a green ppt is the result. with excess the green precipitate remains so the addition of the excess naoh has no effect.

for eg with naoh and chromium in limited naoh a green ppt forms but with excess naoh a green solution forms slowly, so further reactions occur (ligand exchange).


What does it mean by non-amphoteric though ?

Is it trying to suggest OH- can't act as a base and acid... (where as NH3 can, I believe?)

Also if the ligand exchange was with an amphoteric one... apart from colour change, what would the significant be? Faster rate? :s

Thanks
Original post by yahyazhar
guys is there a thread for the paper 6 int'l variant?

If not, on the paper 6 for 2013 jan, on question1, theres a part where they add XS NaOH to a Fe 2+ containing solution, and a green PPT persists. so its an expected result.
in the marking scheme you are expected to say:

not amphoteric / no furtherdeprotonation occurs / no ligand substitution occurs.

----
I dont get the part where it says "not amphoteric" - what does that mean, to what unit does that refer to anyway?

any help would be appreciated!


Are you an international student? :smile:

We don't do paper 6 in the UK........
Reply 805
Original post by posthumus
What does it mean by non-amphoteric though ?

Is it trying to suggest OH- can't act as a base and acid... (where as NH3 can, I believe?)

Also if the ligand exchange was with an amphoteric one... apart from colour change, what would the significant be? Faster rate? :s

Thanks


chromium and zinc form amphoteric hydroxides that can react with either acids or bases - they can react with naoh further for eg when cr3+ reacts with limited naoh, it forms cr(oh)3(h20)3.... when excess naoh is added to this, cr(oh)6 is formed - the green solution. the only other 1st row TM that behaves this way is zn. solubility.... the precipitates of both zn and cr dissolve in excess naoh ... im not sure about rate tbh - i dont think its really to do with rate though
Original post by jojo1995
chromium and zinc form amphoteric hydroxides that can react with either acids or bases - they can react with naoh further for eg when cr3+ reacts with limited naoh, it forms cr(oh)3(h20)3.... when excess naoh is added to this, cr(oh)6 is formed - the green solution. the only other 1st row TM that behaves this way is zn. solubility.... the precipitates of both zn and cr dissolve in excess naoh ... im not sure about rate tbh - i dont think its really to do with rate though


I see so
Cr is reduced from +3 ----> 0 with limited NaOH
& with addition to that is also oxidized from 0 ----> +3 with excess NaOH

Is there a specific reason why Cr or Zn ? :smile:

I asked about rate because you said something about green solution appearing slowly :tongue:

Thanks for the reply :biggrin:
Reply 807
Original post by posthumus
I see so
Cr is reduced from +3 ----> 0 with limited NaOH
& with addition to that is also oxidized from 0 ----> +3 with excess NaOH

Is there a specific reason why Cr or Zn ? :smile: i really don't know, i tried to google it and even made a thread about it, but i still dont know :frown:

I asked about rate because you said something about green solution appearing slowly :tongue:

Thanks for the reply :biggrin:



no redox occurs, the charge of cr always remains at 3+
the reactions are simply deprotonation (w/ limited naoh ) and ligand exchange (w/ excess naoh).

haha sorry about that ... i just thought id slip it in :P
you are welcome !
Contact process, V2O5 used as a catalyst, reacts to form vanadium (iv) oxide, is this V2O4 or VO2? I've seen it put both ways, which one is correct(rather, which one will get me the marks in the exam)?


SO2 + V2O5 → SO3 + 2VO2
2VO2 +½O2 → V2O5
or
SO2 + V2O5 → SO3 + V2O4
V2O4 + ½O2 → V2O5

---

Oh, does anyone happen to have a definitive list of metal aqua-ion colours and the colours when you add OH- etc. One moment [Cr(H2O)6]3+ is violet, then I have my teacher telling me it's green, the test tube in front of me telling me it's closer to black than anything else. >.> Driving me mad and it's only a trivial bit of memorising a few colours but I don't know which colours to remember. Again, I don't care if it's actually red, if saying it's blue get's me the mark in the exam that's what counts. :>
(edited 10 years ago)
Original post by 15weekwonder
Contact process, V2O5 used as a catalyst, reacts to form vanadium (iv) oxide, is this V2O4 or VO2? I've seen it put both ways, which one is correct(rather, which one will get me the marks in the exam)?


SO2 + V2O5 → SO3 + 2VO2
2VO2 +½O2 → V2O5
or
SO2 + V2O5 → SO3 + V2O4
V2O4 + ½O2 → V2O5

---

Oh, does anyone happen to have a definitive list of metal aqua-ion colours and the colours when you add OH- etc. One moment [Cr(H2O)6]3+ is violet, then I have my teacher telling me it's green, the test tube in front of me telling me it's closer to black than anything else. >.> Driving me mad and it's only a trivial bit of memorising a few colours but I don't know which colours to remember. Again, I don't care if it's actually red, if saying it's blue get's me the mark in the exam that's what counts. :>


They both look right to me :smile: I think you would get all your marks, as they both can happen... I don't know if they're equally as feasible however :tongue:

I got this from another thread :smile: :

[Fe(H20)6)]2+


with OH- [Fe(h20)4(OH)2] - green gelatinous ppt
with XS OH- [Fe(H20)(OH)2] - green gelatinous ppt
with NH3- [Fe(H20)4(OH)2] - green gelatinous ppt
with XS NH3- [Fe(H20)4(OH)2] - green gelatinous ppt
with carbonate [FeCO3] - green


[Cu(H20)6]2+

with OH- [Cu(H20)4(OH)2] pale blue
with XS OH- [Cu(H20)4(OH)2] pale blue
with NH3- [Cu(H20)4(OH)2] pale blue

with XS NH3- [Cu(NH3)4(H20)6]2+ deep blue


[Co(H20)6]2+ pink


with OH- [Co(H20)4(OH)2] blue
with XS OH- [Co(H20)4(OH)2] blue
with NH3 [Co(H20)4(OH)2] blue

with XS NH3 [Co(NH3)6]2+ pale yellow/brown
with carbonate [CoCo3] pink
How can you tell if NH3 is going to result in deprotonation or ligand exchange?

Also does anyone know if we need to know the uses of phenol (including the production of thermosetting and thermoplastic plastics?) Thank you. :smile:
Original post by LeaX
How can you tell if NH3 is going to result in deprotonation or ligand exchange?

Also does anyone know if we need to know the uses of phenol (including the production of thermosetting and thermoplastic plastics?) Thank you. :smile:


Well they would have to tell you in the questions what conditions it's being carried out under.

Don't need to know the uses of phenol, but it's good to know them anyway. Just briefly, it's one of the most common starting chemical for organic synthesis, apart from benzene of course and as you were saying, thermoplastics.
Original post by James A
Well they would have to tell you in the questions what conditions it's being carried out under.

Don't need to know the uses of phenol, but it's good to know them anyway. Just briefly, it's one of the most common starting chemical for organic synthesis, apart from benzene of course and as you were saying, thermoplastics.

But in the example above, excess ammonia produced [Fe(H20)4(OH)2] whereas with copper it produced [Cu(NH3)4(H20)6]2+ and with cobalt it produced [Co(NH3)6]2+. Is there a reasoning behind this or is it just something we need to memorise?

Ooh good, thank you. If it wasn't for this thread I'd be revising so much useless stuff which is in my textbook haha.
Original post by LeaX
But in the example above, excess ammonia produced [Fe(H20)4(OH)2] whereas with copper it produced [Cu(NH3)4(H20)6]2+ and with cobalt it produced [Co(NH3)6]2+. Is there a reasoning behind this or is it just something we need to memorise?

Ooh good, thank you. If it wasn't for this thread I'd be revising so much useless stuff which is in my textbook haha.


The george facer book says that with the adding of excess ammonia, to only the cobalt, nickel, copper, zinc and silver hydroxide complexes, you get ammines, which have (NH3)4(H20)2 attatched to the transition metal ion.

If in doubt, check the syllabus.
Given that the E-values for two half equations give an overall of +1.74 and that all conditions are standard. Am I right in thinking that we can predict the reaction will go to completion?
Original post by HarryMWilliams
Given that the E-values for two half equations give an overall of +1.74 and that all conditions are standard. Am I right in thinking that we can predict the reaction will go to completion?


Remember E values are only predictions, not actual assumptions.

Yes, but remember that the activation energy of the reaction could be too high to prevent this from happening. Or the reaction is not carried out under standard conditions....
Original post by HarryMWilliams
Given that the E-values for two half equations give an overall of +1.74 and that all conditions are standard. Am I right in thinking that we can predict the reaction will go to completion?


It's a feasible reaction which means it CAN happen :smile: However you have to keep in mind that a high activation energy and also a very slow rate... may mean the reaction would never go to completion. It may even be so slow that it would be hard to observe even the reaction taking place.

Hope that makes sense :tongue: :colondollar:
Original post by James A
Remember E values are only predictions, not actual assumptions.

Yes, but remember that the activation energy of the reaction could be too high to prevent this from happening. Or the reaction is not carried out under standard conditions....


Original post by posthumus
It's a feasible reaction which means it CAN happen :smile: However you have to keep in mind that a high activation energy and also a very slow rate... may mean the reaction would never go to completion. It may even be so slow that it would be hard to observe even the reaction taking place.

Hope that makes sense :tongue: :colondollar:


Thanks both.

As I thought! :biggrin:
Can anyone explain what is the difference between E, Edelta and Ecell ?
I'm pretty sure we don't need to know about the reactions of Cobalt as a transition metal/ion, right? (I can't find it on the specification)

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