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Offline abdillah

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qualitative analysis
« on: 31/10/2007 05:58:12 »
Ok here's a question:

For the GCE or even GCSE 'O' levels examinations (Cambridge), under qualitative analysis:-

to test for the presence of anions such as chlorides, iodides and sulphates, it is mentioned that we should acidify it with dilute nitric acid, before we proceed to add silver nitrate, lead(II) nitrate and barium nitrate respectively.

chloride : add dil. nitric acid, then add silver nitrate  --> white ppt. AgCl formed
iodide   : add dil. nitric acid, then add lead(II) nitrate --> yellow ppt. PbCl2 formed
sulphate : add dil. nitric acid, then add barium nitrate  --> white ppt., BaSO4 formed

my colleague said it is to remove other anions such as carbonates and hydroxides that might interfere with the reaction, is that true?

Is there any other underlying reason?


 

Offline lightarrow

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« Reply #1 on: 31/10/2007 09:59:02 »
Ok here's a question:

For the GCE or even GCSE 'O' levels examinations (Cambridge), under qualitative analysis:-

to test for the presence of anions such as chlorides, iodides and sulphates, it is mentioned that we should acidify it with dilute nitric acid, before we proceed to add silver nitrate, lead(II) nitrate and barium nitrate respectively.

chloride : add dil. nitric acid, then add silver nitrate  --> white ppt. AgCl formed
iodide   : add dil. nitric acid, then add lead(II) nitrate --> yellow ppt. PbCl2 formed
It's PbI2 actually, but I'm sure you didn't mean to write PbCl2.
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sulphate : add dil. nitric acid, then add barium nitrate  --> white ppt., BaSO4 formed

my colleague said it is to remove other anions such as carbonates and hydroxides that might interfere with the reaction, is that true?

Is there any other underlying reason?
In presence of high conc. of OH- or CO32-, the acid you add could be not enough to acidify the solution; it must be acidified just to destroy those ions, since Ag+, Pb2+ precipitates with both ions, and Ba2+ precipitates with CO32-.
 

Offline Bored chemist

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« Reply #2 on: 31/10/2007 20:34:20 »
"my colleague said it is to remove other anions such as carbonates and hydroxides that might interfere with the reaction, is that true?"
Yes that's right; also it removes sulphide, cyanide (though not at GCSE) and some other things too.
 

Offline abdillah

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qualitative analysis
« Reply #3 on: 01/11/2007 06:00:04 »
Thanks peeps...

ok for my next question. Aluminium, lead (II) and zinc, when they form a precipitate with sodium hydroxide, it will dissolve in excess. My question here is, is it because they form sodium aluminate, sodium leadate (not sure of the naming) and lastly sodium zincate? Is it because of their nature as amphoteric hydroxides?

Al(OH)3 + NaOH --> NaAl(OH)4 Is this equation correct?

Or am I entirely wrong about their nature being amphoteric as the reason for these reactions? Do we call these salts complexes? How do you define a complex? :) sorry a lot of questions here actually.

My next question here is aluminium and lead(II) hydroxides ppts. are insoluble in excess aqueous ammonia but zinc is soluble. Why is that so? Here copper (II) hydroxide ppt. also dissolves in excess aqueous ammonia to give a dark blue solution. Are these two reactions of complexes and less similar to the one above?

Oh... and for ammonium compounds, when we heat it, ammonia gas will be given out also right? So what is the purpose of adding sodium hydroxide? as a catalyst?

Since I'm at it, I might as well ask this last question. How does the mechanism of testing for nitrates work? Add sodium hydroxide, then aluminium foil and warm, ammonia will be formed. hmmm...

I'm looking for the answers in the internet as well but I can't find it... Hope you guys can help me out... Thanks a mil
« Last Edit: 01/11/2007 06:06:54 by abdillah »
 

Offline lightarrow

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qualitative analysis
« Reply #4 on: 01/11/2007 12:37:43 »
Thanks peeps...

ok for my next question. Aluminium, lead (II) and zinc, when they form a precipitate with sodium hydroxide, it will dissolve in excess. My question here is, is it because they form sodium aluminate, sodium leadate (not sure of the naming) and lastly sodium zincate? Is it because of their nature as amphoteric hydroxides?

Al(OH)3 + NaOH --> NaAl(OH)4 Is this equation correct?

Yes but you don't need to write it that way, because you have similar reactions with any other strong basis:

1.Al(OH)3 +   OH- <--> [Al(OH)4]-   
2.Pb(OH)2 + 2OH- <--> [Pb(OH)4]2-
3.Zn(OH)2 + 2OH- <--> [Zn(OH)4]2-

Similar reactions with:

4.Cr(OH)3  +  OH- <--> [Cr(OH)4]-
5.Sn(OH)2 + 2OH- <--> [Sn(OH)4]2-
6.Sn(OH)4 + 2OH- <--> [Sn(OH)6]2-
7.Sb(OH)3 +   OH- <--> [Sb(OH)4]- 

I can give you the approx. pH at which those reactions are completely displaced towards the right (total dissolution of the hydroxide):

1. pH ~ 12
2. pH ~ 13
3. pH ~ 13.5
4. pH ~ 13.5
5. pH ~ 13
6. pH ~ 9.5
7. pH ~ 12

Remember however that some hydroxides, especially Al(OH)3 (and probably others but I don't know exactly), get old and are not soluble anylonger in excess OH- after a while.

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Or am I entirely wrong about their nature being amphoteric as the reason for these reactions?
You're not wrong.
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Do we call these salts complexes?
Yes, those anions I've written up are all complexes.
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How do you define a complex?
A chemical species where there are atoms covalently bond to a number of other species greater than that atom's oxidation number(O.N.). Example: [Al(OH)4]-. Aluminum here has O.N. = +3 and it's covalently bond with 4 species (the 4 OH-); 4 > 3 so it's a complex...

Other examples: Al2O3 in α-alumina (every atom of Al is bond with 6 O atoms); B2O3; solid AlCl3; LBX3 where B = boron, X = F, Cl, Br, I and L = ammonia, amines, diethylic ether, alogenide...

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My next question here is aluminium and lead(II) hydroxides ppts. are insoluble in excess aqueous ammonia but zinc is soluble. Why is that so? Here copper (II) hydroxide ppt. also dissolves in excess aqueous ammonia to give a dark blue solution. Are these two reactions of complexes and less similar to the one above?

Zn(OH)2 + 4NH3 <--> [Zn(NH3)4]2+ +2OH-  (colorless).

Cu(OH)2 + 4NH3 <--> [Cu(NH3)4]2+ +2OH-   (deep blue).

Quote
Oh... and for ammonium compounds, when we heat it, ammonia gas will be given out also right? So what is the purpose of adding sodium hydroxide? as a catalyst?
It won't work with all ammonium compounds, e.g. with (NH4)2SO4.

Edit: it won't work if you only heat it, without adding sodium hydroxide.

The purpose of adding sodium hydroxide is to displace the equilibrium towards the reaction:

NH4+ + OH- --> NH3↑ + H2O

Quote
Since I'm at it, I might as well ask this last question. How does the mechanism of testing for nitrates work? Add sodium hydroxide, then aluminium foil and warm, ammonia will be formed. hmmm...

I'm looking for the answers in the internet as well but I can't find it... Hope you guys can help me out... Thanks a mil

The metal (usually it's used "Devarda's alloy" which contains Al, Zn and Cu) must be in eccess and in these conditions it reduces NO3- to NH3:

8Al + 3NO3- + 5OH- + 18H2O --> 3NH3↑ + 8[Al(OH)4]-

Warming helps ammonia coming out of the solution.
« Last Edit: 05/11/2007 12:45:38 by lightarrow »
 

Offline vinuuraj

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qualitative analysis
« Reply #5 on: 07/11/2007 16:18:46 »
chloride : add dil. nitric acid, then add silver nitrate  --> white ppt. AgCl formed
iodide   : add dil. nitric acid, then add lead(II) nitrate --> yellow ppt. PbCl2 formed
sulphate : add dil. nitric acid, then add barium nitrate  --> white ppt., BaSO4 formed
 

Offline lightarrow

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qualitative analysis
« Reply #6 on: 07/11/2007 16:24:36 »
chloride : add dil. nitric acid, then add silver nitrate  --> white ppt. AgCl formed
iodide   : add dil. nitric acid, then add lead(II) nitrate --> yellow ppt. PbCl2 formed
sulphate : add dil. nitric acid, then add barium nitrate  --> white ppt., BaSO4 formed

PbI2
 

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qualitative analysis
« Reply #6 on: 07/11/2007 16:24:36 »

 

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