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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Michael on December 28, 2004, 12:24:00 AM

Title: Identification tests
Post by: Michael on December 28, 2004, 12:24:00 AM
Hello again i have another problem,

It deals with identification tests of chlorides, bromides and iodides.

The materials are:
Silver nitrate solution
Dilute HNO3
NaBr
NaCl
KI

From what i have researched i learnt that:

Chlorides give a white curdy ppt of silver chloride which dissolves in ammonia

Bromides give a creamy ppt of silver bromide which dissolve slightly in cold ammonia soultion.

Iodides gives a yellow ppt of silver iodide which does not dissolve in ammonia solution.

In each case a few drops of dilute HNO3 was added first followed by the silver nitrate.

Why was HNO3 added in each case.

What are the reactions taking place (equations)?

Why was "cold" ammonia solution needed?

What changes would be observered in each case other than the ones stated above like in colour or other wise. if any?

And for the quest for more knowledge what effects will concentrated sulphuric acid have on the bromides, chlorides and iodides (equations)?

I am limited to picturing what these reactions are like because i haven't the oppertunity to do them. So your help will be greatly appreciated.

Thanks in advance.
Title: Re:Identification tests
Post by: AWK on December 28, 2004, 04:20:45 AM
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Why was HNO3 added in each case.
Many anions form precipitates with AgNO3. This is for check we are dealing with halides

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What are the reactions taking place (equations)?
Ag(+) + X(-) = AgX(s)

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Why was "cold" ammonia solution needed?
Only concentrated ammonia dissolves AgBr. Solubility of NH3 in water strongly depends on temperature.

Quote
What changes would be observered in each case other than the ones stated above like in colour or other wise. if any?
AgCl(s) + 2NH3 = [Ag(NH3)2]Cl - ppt disappears, colorless solution
AgBr(s) + 2NH3 (concentrated) = partiall dissolution of ppt

Quote
And for the quest for more knowledge what effects will concentrated sulphuric acid have on the bromides, chlorides and iodides (equations)?

2NaCl + H2SO4 = 2HCl + Na2SO4
For NaBr and NaI with conc. H2SO4 the same reaction take place follloved by
2HX + H2SO4 = X2 + SO2 + 2H2O