Qualitative Analysis and Identification of Ions

Flame Tests for Metal Cations

The vivid colours bursting in a fireworks display are created by heating specific metal ions. In the laboratory, a flame test identifies specific metal cations (positive ions) based on the unique colour of light they emit when heated.

To perform a flame test accurately:

1. Take a loop of unreactive wire, such as nichrome or platinum.
2. Clean the loop by dipping it into concentrated hydrochloric acid ($HCl$) and heating it in a roaring blue (non-luminous) Bunsen flame. This prevents masking, where intense colours from impurities (like sodium's bright yellow) hide the true colour of the ion being tested.
3. Dip the clean, moist loop into the solid sample or solution.
4. Place the loop at the very edge of the hot blue flame and record the first characteristic colour.

The specific OCR Gateway flame test colours are:

• Lithium ($Li^+$): Crimson
• Sodium ($Na^+$): Yellow
• Potassium ($K^+$): Lilac
• Calcium ($Ca^{2+}$): Orange-red
• Copper ($Cu^{2+}$): Blue-green

Cation Precipitation Reactions

You can often identify a dissolved metal simply by forcing it to form an insoluble solid. When sodium hydroxide ($NaOH$) solution is added to solutions containing certain metal ions, they react to form insoluble metal hydroxides.

To test an unknown cation, add a few drops of $NaOH$ and record the colour of the precipitate. Then, add $NaOH$ in excess to see if the solid dissolves.

• Copper(II) ($Cu^{2+}$): Forms a blue precipitate that is insoluble in excess $NaOH$.

• Iron(II) ($Fe^{2+}$): Forms a green precipitate that is insoluble in excess. Over time, the surface slowly turns brown as it oxidises to Iron(III).
• Iron(III) ($Fe^{3+}$): Forms a brown precipitate that is insoluble in excess.
• Aluminium ($Al^{3+}$) & Zinc ($Zn^{2+}$): Both form a white precipitate that dissolves in excess $NaOH$ to leave a colourless solution. To distinguish them, add aqueous ammonia instead: zinc's precipitate dissolves in excess ammonia, but aluminium's does not.
• Calcium ($Ca^{2+}$) & Magnesium ($Mg^{2+}$): Both form a white precipitate that remains insoluble in excess $NaOH$. A flame test will easily distinguish them (calcium gives an orange-red flame, magnesium produces no colour).

The ammonium ion ($NH_4^+$) is a unique cation because it does not form a precipitate. Instead, warming the sample with $NaOH$ releases pungent ammonia gas, which can be identified using damp red litmus paper.

Testing for Anions (Negative Ions)

Identifying a complete ionic compound requires discovering its negative half as well as its positive half. We use specific reagents to test for anions.

• Carbonates ($CO_3^{2-}$): Add dilute hydrochloric acid ($HCl$). If carbonate ions are present, you will observe effervescence (fizzing) as carbon dioxide gas escapes. You can confirm this gas is $CO_2$ using limewater.
• Sulfates ($SO_4^{2-}$): Add dilute $HCl$ followed by barium chloride ($BaCl_2$) solution. A positive result is a white precipitate of barium sulfate.

• Halides ($Cl^-$, $Br^-$, $I^-$): Add dilute nitric acid ($HNO_3$) followed by silver nitrate ($AgNO_3$) solution. The colour of the resulting silver halide precipitate identifies the specific ion:
  • Chloride ($Cl^-$): White precipitate.
  • Bromide ($Br^-$): Cream precipitate.
  • Iodide ($I^-$): Yellow precipitate.

Crucially, dilute acid must be added before the sulfate and halide tests to remove any hidden carbonate ions. If carbonates were left in the solution, they would react with the barium or silver ions to form false-positive white precipitates. We specifically use nitric acid for the halide test because hydrochloric acid contains chloride ions, which would ruin the test by creating a massive false positive.

The Systematic Identification Sequence

Like a detective following a strict protocol, chemists must test for anions in a very specific order to avoid being tricked by overlapping chemical reactions.

When analysing a single unknown sample for anions, you must follow this exact sequence: Carbonate $\rightarrow$ Sulfate $\rightarrow$ Halide.

The logical mechanism behind this is simple: barium ions react with both sulfates and carbonates, while silver ions react with halides, sulfates, and carbonates. Therefore, you must test for (and completely react away) carbonates first, then test for sulfates, and only test for halides last.

Worked Example

A student is given an unknown solid, Compound X, and dissolves it in water. They perform the following tests in order to deduce its identity.
Test 1: Add dilute $HCl$. Result: No effervescence.
Test 2: Add $BaCl_2(aq)$. Result: No precipitate.
Test 3: Add dilute $HNO_3$ and $AgNO_3(aq)$. Result: A cream precipitate forms.
Test 4: Add aqueous $NaOH$. Result: A green precipitate forms that does not dissolve in excess.

Step 1: Interpret Test 1 and 2 (Anions).

• No fizzing with acid means there is no carbonate.
• No solid with barium chloride means there is no sulfate.

Step 2: Interpret Test 3 (Anions).

• Silver nitrate produces a cream precipitate. This confirms that bromide ions ($Br^-$) are present.

Step 3: Interpret Test 4 (Cations).

• Sodium hydroxide produces an insoluble green solid. This confirms that Iron(II) ions ($Fe^{2+}$) are present.

Conclusion:

• Compound X is Iron(II) bromide ($FeBr_2$).