5.3.2 Tests for ions

(a) qualitative analysis of ions on a test-tube scale: processes and techniques needed to identify the following ions in an unknown compound:

(i) anions: CO₃^2–, Cl ^–, Br^–, I^–, SO₄^2–

(ii) cations: NH₄ ⁺; Cu²⁺, Fe²⁺, Fe³⁺, Mn²⁺, Cr³⁺

If you haven't worked it out yet, quantitative analysis involves "amounts of stuff" (quantities) - in other words mole calculations (yay!)

Qualitative analysis doesn't involve amounts, it investigates what is in substance by looking at how it reacts (its qualities, if you like)

A lot of these twests will be familiar from GCSE and AS and some from the earlier part of this Module.

I'm going to repeat them anyway, because I'm like that.

Ammonium (NH₄⁺)

The key to identifying Ammonium ions is that they easily break down into Ammonia gas and we can then test for that.

• • Add Sodium Hydroxide

• Warm the mixture

  • Ammonium ions + Hydroxide ions → Ammonia + Water

NH₄⁺ (aq) + OH^- (aq) → NH₃ (g) + H₂O (l)

• • Ammonia gas makes damp red litmus paper turn blue

  • If you still were uncertain you could hold a bottle of Concentrated HCl over the mixture and look for white smoke

NH_{3 (g)} + HCl _(g) → NH₄Cl_(s)

On a reasonably warm day and with a solution containing concentrated Ammonium ions you wouldn't even need the first few steps

Transition Metal ions

As previously discussed Transition Metal solutions exist as hexaqua complexes

Here the overall charge is the charge of the central ion because the ligands are neutral water molecules.

For the purposes of qualitative testing, however, the examiners are usually be prepared to accept Fe²⁺_(aq) or Fe²⁺_(aq).

Again, the previous chapters explain that the OH- ions remove H+ ions from water ligands in an acid-base reaction.

And that they do this until the complex becomes neutral, at which point it will precipitate.

[Fe(H₂O)₆]²⁺ _(aq) + 2 OH^-_(aq) --> [Fe(H₂O)₄(OH)₂] _(s) + 2 H₂O _(l)

_and

[Fe(H₂O)₆]³⁺ _(aq) + 3 OH^-_(aq) --> [Fe(H₂O)₃(OH)₃] _(s) + 3 H₂O _(l)

But, again, the examiner will accept;

Fe²⁺_(aq) + 2 OH^-_(aq) --> Fe(OH)_{2 (s)} Fe³⁺_(aq) + 3 OH^-_(aq) --> Fe(OH)_{3 (s)}

So all we really need to do is substitute the symbol for the metal ion under consideration and try to recall the likely colours of the solution we start with and the precipitates we will make.

(*) Colours of solutions are dependent on concentration.

Note: details of behaviour with excess OH^- /NH₃ may be required and are covered in the previous chapter.

Halide ions (Cl^-,Br^-,I^-)

• • Add dilute Nitric Acid

  • Add Silver Nitrate solution

  • Note colour of precipitate:Chloride = White, Bromide = Cream, Iodide = Yellow

Ag⁺ _(aq) + X^- _(aq)--> AgX_(s)

• • Add dilute Ammonia solution, Chloride precipitate will dissolve

  • Add concentrated Ammonia solution, Bromide precipitate will dissolve

Note: The Nitric acid removes and CO₂ that has dissolved to form Carbonate ions which would also precipitate

If you were sure you had a halide solution and it didn't precipitate then you can assume that it is a Fluoride solution.

The ammonia re-dissolving is a chance to confirm your suspicions if the colour of precipitates are not absolutely clear.