One way of identifying metal ions in an aqueous solution is by observing the colour of the precipitate they form when we add a few drops of sodium hydroxide (NaOH). These are metal hydroxide precipitates and they form when a small amount of NaOH is added.
However, some metal ions such as magnesium, aluminium, and calcium all form a white precipitate. In this case, we need to add sodium hydroxide until it is in excess to distinguish between the ions. This is because some of the precipitates may dissolve in excess sodium hydroxide.
Let’s look at the method for the sodium hydroxide test:
1. Dissolve a small amount of the substance in water.
2. Place about 5cm³ of the sample solution into a test tube.
3. Slowly add a few drops of sodium hydroxide solution.
4. Note down the colour of any precipitate that forms.
5. Add sodium hydroxide in excess and note down the result.
The table below shows the expected results:
|Metal ion||Result of adding NaOH||Effect of adding excess NaOH|
|Copper(II) – Cu²⁺||Blue precipitate||Blue precipitate remains|
|Iron(II) – Fe²⁺||Green precipitate||Green precipitate remains|
|Iron(III) – Fe³⁺||Red-brown precipitate||Red-brown precipitate remains|
|Magnesium – Mg²⁺||White precipitate||White precipitate remains|
|Aluminium – Al³⁺||White precipitate||White precipitate dissolves to form a colourless solution|
|Calcium – Ca²⁺||White precipitate||White precipitate remains|
After adding excess sodium hydroxide, a positive result is observed for aluminium ions. However, we cannot distinguish between the magnesium and calcium ions as they both remain unchanged. Therefore, we need to perform another test such as a flame test to differentiate between the two.
The precipitate formed from the calcium ion produces an orange-red colour in a flame test, while the precipitate from the magnesium ion gives no colour.
These equations show the reactions between the metal ions and sodium hydroxide to form the precipitate: