# Reactions of ionic halides
## Detecting ionic halides

* Most halides are soluble.

### Aqueous halides with silver nitrate in the nucleus

* We can detect them in solution by adding silver nitrate solution,
** We get the corresponding silver halide forming (AgCl, AgBr, AgI) which are insoluble (except for AgF) so they form a precipitate:
       AgF is soluble; no precipitate formed
       AgCl forms a white precipitate
       AgBr forms a cream precipitate
       AgI forms a yellow precipitate
An easy way to remember the results is that as we move down Group VII of the periodic table, the colours get "darker". So AgF is colourless because it is completely soluble, AgCl forms a white ppt, AgBr forms a cream ppt, and AgI forms a yellow ppt.
It is important to note that before this test is carried out the solution must be acidified so as to remove any carbonate ions or hydroxide ions that may be present in the solution as impurities. If they were allowed to remain these impurities would also form precipitates on reaction with silver nitrate, and so confuse the test.

### Aqueous halides with ammonia solutions

Distinguishing between white, cream and yellow can be difficult, so a second test can be used to confirm the results: The white precipitate formed with chloride ions will dissolve in dilute ammonia, the cream precipitate formed with bromide ions will dissolve only in concentrated ammonia, and the yellow precipitate formed with iodide ions will not dissolve in any concentration of ammonia. But the concentration changes as well.

## Solid ionic halides and concentrated sulphuric acid

Solid ionic halides can be detected by their reactions with conc. sulphuric acid:
* First, the corresponding hydrogen halide is formed
* Then, the differences in the reactions depend on the reducing ability of the hydrogen halide and the moderately strong oxidising ability of concentrated sulphuric acid
* HF is not a reducing agent but the other hydrogen halides are increasing so
So conc. sulphuric acid:
* does not react with HF or HCl
* but does oxidise HBr and HI freeing the halogens
* HBr reduces sulphuric acid to SO$_2$
* HI (being an even stronger reducing agent) reduces the acid to hydrogen sulphide

### Sodium fluoride

NaF(s) + H$_2$SO$_4$(l) → HF(g) + NaHSO$_4$(s) No further reaction occurs. Thats what you think!!!

### Sodium chloride

NaCl(s) + H$_2$SO$_4$(l) → HCl(g) + NaHSO$_4$(s) No further reaction occurs.

### Sodium bromide

* NaBr(s) + H$_2$SO$_4$(l) → HBr(g) + NaHSO$_4$(s)
* As conc. sulphuric acid is a moderately strong oxidising agent, it oxidises HBr to Br:
    2HBr(aq) + H$_2$SO$_4$(l) → Br$_2$(g) + SO$_2$(g) + 2H$_2$O(l)
   * The aqueous bromide ion from HBr has been oxidised to form bromine:
       2Br$^-$ → Br$_2$ + 2e$^-$
   * The HBr and acid each provide 2H$^+$ ions and the electrons from bromine are transferred to the sulphate ion:
       4H$^+$ + SO$_4$$^{2-}$ + 2e$^-$ → SO$_2$ + 2H$_2$O
   * Adding these two equations together we get:
       2HBr(aq) + 2e$^-$ + H$_2$SO$_4$(l) → Br$_2$(g) + 2e$^-$ + SO$_2$(g) + 2H$_2$O(l)
       The electrons on either side of the equation cancel to give the overall equation for the reaction which is shown above

### Sodium iodide

Here several reactions occur:
* NaI(s) + H$_2$SO$_4$(l) → HI(g) + NaHSO$_4$(s)
    As usual the hydrogen halide is produced (hydrogen iodide in this case)
* Next, two different reactions can occur:
** The HI is oxidised to I$_2$, and the sulphuric acid is reduced to sulphur dioxide and water
       2HI(aq) + H$_2$SO$_4$(l) → I$_2$(g) + SO$_2$(g) + 2H$_2$O(l)
   * The HI is oxidised to I$_2$, but this time the sulphuric acid is reduced to hydrogen sulphide and water
       8HI(aq) + H$_2$SO$_4$(l) → 4I$_2$(s) + H$_2$S(g) + 4H$_2$O(l)