Testing for Metal Cations with Sodium Hydroxide

Identification of Metal Cations using Sodium Hydroxide

A common method for identifying metal ions (cations) in an unknown aqueous solution is to add aqueous sodium hydroxide, $NaOH(aq)$, dropwise. Because most metal hydroxides are insoluble in water, they will form a solid precipitate when the hydroxide ions from the sodium hydroxide react with the metal cations in the solution.

The general ionic equation for this reaction is: $M^{n+}(aq) + nOH^-(aq) \rightarrow M(OH)_n(s)$

Where $M$ represents the metal and $n+$ represents its ionic charge. The colour and behaviour of the resulting precipitate allow us to deduce which metal ion is present.

Cations that form White Precipitates

Three specific cations in the ESAT specification produce white precipitates when sodium hydroxide is added. These are aluminium ($Al^{3+}$), calcium ($Ca^{2+}$), and magnesium ($Mg^{2+}$).

1. Aluminium ions, $Al^{3+}$: These react to form aluminium hydroxide. The ionic equation is $Al^{3+}(aq) + 3OH^-(aq) \rightarrow Al(OH)_3(s)$. The precipitate is white.
2. Calcium ions, $Ca^{2+}$: These react to form calcium hydroxide. The ionic equation is $Ca^{2+}(aq) + 2OH^-(aq) \rightarrow Ca(OH)_2(s)$. The precipitate is white.
3. Magnesium ions, $Mg^{2+}$: These react to form magnesium hydroxide. The ionic equation is $Mg^{2+}(aq) + 2OH^-(aq) \rightarrow Mg(OH)_2(s)$. The precipitate is white.

Because all three ions produce a white precipitate, further testing is often required to distinguish between them. For example, if excess sodium hydroxide is added, the white precipitate of aluminium hydroxide will redissolve to form a colourless solution, whereas the precipitates of calcium hydroxide and magnesium hydroxide will not redissolve.

Cations that form Coloured Precipitates

Transition metal ions often form coloured precipitates, making them much easier to identify through simple observation.

Copper(II) Ions

When sodium hydroxide is added to a solution containing $Cu^{2+}$ ions, a distinct blue precipitate of copper(II) hydroxide is formed. This is a standard test for the presence of copper salts.

Equation: $Cu^{2+}(aq) + 2OH^-(aq) \rightarrow Cu(OH)_2(s)$

Iron(II) and Iron(III) Ions

Iron can exist in different oxidation states, and the sodium hydroxide test is a reliable way to distinguish between them.

1. Iron(II) ions, $Fe^{2+}$: These produce a green precipitate of iron(II) hydroxide. Over time, the surface of this precipitate may turn brown as it reacts with oxygen in the air to be oxidised to iron(III) hydroxide. Equation: $Fe^{2+}(aq) + 2OH^-(aq) \rightarrow Fe(OH)_2(s)$

2. Iron(III) ions, $Fe^{3+}$: These produce a characteristic brown precipitate (often described as foxy red or rust coloured) of iron(III) hydroxide. Equation: $Fe^{3+}(aq) + 3OH^-(aq) \rightarrow Fe(OH)_3(s)$

Worked Examples

Example 1: Identifying an Unknown Solution A student is given a solution that may contain either $Cu^{2+}$ or $Fe^{2+}$. Upon adding a few drops of sodium hydroxide, a pale green precipitate forms. Identify the ion.

Reasoning: Since the precipitate is green, the ion must be iron(II), $Fe^{2+}$. A blue precipitate would have indicated $Cu^{2+}$.

Example 2: Writing Ionic Equations Write the ionic equation for the test that produces a rust brown precipitate.

Step 1: Identify the precipitate. A brown precipitate with $NaOH$ indicates $Fe^{3+}$ ions reacting to form $Fe(OH)_3$. Step 2: Balance the charges. The $Fe^{3+}$ ion requires three $OH^-$ ions to form a neutral solid. Step 3: Include state symbols. $Fe^{3+}(aq) + 3OH^-(aq) \rightarrow Fe(OH)_3(s)$.