5.3.2 Qualitative Analysis

Keep the plan practical: use small samples, clean apparatus and a separate dropping pipette for each reagent. Write down what you add, what you see and what that tells you. “No visible change” is an observation too.

If you test the same portion for anions, OCR’s order is carbonate → sulfate → halide. You can also use fresh portions of the unknown for separate tests, but say clearly what you are doing so the reagents from one test cannot confuse another.

Add a dilute acid to the sample. Carbonate gives effervescence because CO₂ is formed. Bubble the gas through limewater: a positive result turns the limewater cloudy or milky.

The ionic equation is worth knowing because it shows exactly what the acid is doing: CO₃²⁻ + 2H⁺ → CO₂ + H₂O.

For a chloride test, acidify with dilute nitric acid, then add aqueous silver nitrate. Chloride gives a white AgCl precipitate.

To distinguish it from the other silver halides, add dilute aqueous ammonia: AgCl dissolves. If you are learning the three together, I’d remember white chloride — dissolves dilute.

Acidify with dilute nitric acid and add aqueous silver nitrate. Bromide gives a cream AgBr precipitate.

AgBr does not dissolve in dilute ammonia, but it does dissolve in concentrated aqueous ammonia. That puts bromide neatly between chloride and iodide in the ammonia test.

With dilute nitric acid followed by aqueous silver nitrate, iodide gives a yellow AgI precipitate.

AgI stays insoluble in both dilute and concentrated aqueous ammonia. So the three silver halides run white/cream/yellow, while their solubility in ammonia decreases Cl⁻ → Br⁻ → I⁻.

Acidify the sample, then add a source of Ba²⁺. Sulfate gives a white BaSO₄ precipitate. The acid step matters because carbonate also forms an insoluble white barium salt.

If you are carrying the anion tests out sequentially in one tube, use dilute HNO₃ and Ba(NO₃)₂ here. That avoids adding chloride ions before the later silver-nitrate halide test.

Add aqueous NaOH and warm gently. If NH₄⁺ is present, ammonia gas is released. Hold damp red litmus paper near the mouth of the tube; the ammonia turns it blue.

For the equation, write NH₄⁺ + OH⁻ → NH₃ + H₂O. Keep the litmus damp: ammonia has to dissolve in water before its alkaline behaviour shows.

Cu²⁺ gives a blue Cu(OH)₂ precipitate with a small amount of either NaOH or NH₃. In excess NaOH the precipitate remains.

Excess NH₃ is the distinctive part: the blue precipitate dissolves to give a deep blue solution of [Cu(NH₃)₄(H₂O)₂]²⁺. That final deep-blue solution is a strong clue for Cu²⁺.

Fe²⁺ gives a dark-green Fe(OH)₂ precipitate with NaOH or NH₃, and the precipitate is insoluble in excess reagent.

On standing in air it turns brown as the iron(II) compound is oxidised. That change is useful evidence when you are separating Fe²⁺ from the other green transition-metal observations.

Fe³⁺ gives an orange-brown Fe(OH)₃ precipitate with aqueous NaOH and with aqueous NH₃.

The precipitate is insoluble in excess of either reagent. For this ion, the colour is doing most of the identification work: orange-brown is the observation to know.

Mn²⁺ gives a pale-brown Mn(OH)₂ precipitate with aqueous NaOH or NH₃. It stays insoluble when either reagent is added in excess.

The precipitate darkens on standing in air as it is oxidised. I’d keep “pale brown, then darker” separate from Fe²⁺, which starts green and turns brown.

Cr³⁺ first gives a dark-green or grey-green Cr(OH)₃ precipitate with NaOH or NH₃. The useful part is what happens when you add more reagent.

In excess NaOH it dissolves to give dark-green [Cr(OH)₆]³⁻. In excess NH₃ it dissolves to give purple [Cr(NH₃)₆]³⁺. Those two excess-reagent results make Cr³⁺ much easier to distinguish from Fe²⁺.