8.4 - Tests for Cations
The flame test
- A cation is a positively charged ion.
- When a metal compound is heated in a flame, it gives off a characteristic colour that identifies the metal ion.
Carrying out the test
-
Clean the wire. Dip a nichrome or platinum wire loop in hydrochloric acid, then hold it in a blue Bunsen flame until it burns with no colour. Any residue left from an earlier test would mask the sample's colour. Hydrochloric acid is used because chloride ions give no flame colour.
-
Pick up the sample. Dip the clean wire loop into the solid sample so a little clings to it.
-
Hold the loop in the flame and record the colour. Use the roaring blue flame (air hole open), which is the hottest part and won't add any yellow colour of its own.
Flame colours
| Metal ion | Flame colour |
|---|---|
| Lithium, Li+ | crimson (red) |
| Sodium, Na+ | yellow (yellow-orange) |
| Potassium, K+ | lilac |
| Calcium, Ca2+ | orange-red (brick red) |
| Copper(II), Cu2+ | blue-green (green) |

What the test cannot do
- A flame test only works reliably on a single metal ion.
- If a sample contains a mixture of ions, the colours blend and a strong colour masks a weaker one (e.g. sodium's yellow completely masks potassium's lilac).
Precipitates with sodium hydroxide
- Adding sodium hydroxide solution to a solution of a metal compound identifies the positive ion (cation) it contains.
- Most metal hydroxides are insoluble in water. When mixed, the insoluble metal hydroxide forms as a solid suspended in the liquid - a precipitate.
- Method - Dissolve the sample in water, add sodium hydroxide solution a few drops at a time, and record the colour of the precipitate.
| Metal ion | Precipitate formed | Colour |
|---|---|---|
| Copper(II), Cu2+ | copper(II) hydroxide, Cu(OH)2 | blue |
| Iron(II), Fe2+ | iron(II) hydroxide, Fe(OH)2 | green |
| Iron(III), Fe3+ | iron(III) hydroxide, Fe(OH)3 | brown |
| Aluminium, Al3+ | aluminium hydroxide, Al(OH)3 | white |
| Calcium, Ca2+ | calcium hydroxide, Ca(OH)2 | white |
| Magnesium, Mg2+ | magnesium hydroxide, Mg(OH)2 | white |

Telling the white precipitates apart
Add sodium hydroxide solution until it is in excess.
| White precipitate | With sodium hydroxide in excess |
|---|---|
| Aluminium, Al3+ | dissolves to give a colourless solution |
| Calcium, Ca2+ | does not dissolve |
| Magnesium, Mg2+ | does not dissolve |
- Only aluminium hydroxide dissolves in excess sodium hydroxide. Calcium and magnesium precipitates remain unchanged.
- This test cannot identify Group 1 metals. Lithium, sodium, and potassium hydroxides are soluble in water, so no precipitate forms. These are identified using flame tests instead.

Equations for the hydroxide precipitates
When sodium hydroxide solution reacts with a soluble metal salt, an insoluble metal hydroxide precipitate and a soluble sodium salt form:
Metal salt + sodium hydroxide → metal hydroxide + sodium salt
Writing the balanced equation
Example using copper(II) sulfate:
- Reactants - CuSO4(aq) and NaOH(aq).
- Products - Cu(OH)2(s) and Na2SO4(aq).
- Formula of hydroxide - Cu2+ requires two OH- ions to balance the charge → Cu(OH)2.
- Balanced equation:
- CuSO4(aq) + 2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq)
Other hydroxide precipitate equations
| Metal ion | Equation |
|---|---|
| Iron(II), Fe2+ | FeCl2(aq) + 2NaOH(aq) → Fe(OH)2(s) + 2NaCl(aq) |
| Iron(III), Fe3+ | FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq) |
| Aluminium, Al3+ | AlCl3(aq) + 3NaOH(aq) → Al(OH)3(s) + 3NaCl(aq) |
| Calcium, Ca2+ | CaCl2(aq) + 2NaOH(aq) → Ca(OH)2(s) + 2NaCl(aq) |
| Magnesium, Mg2+ | MgCl2(aq) + 2NaOH(aq) → Mg(OH)2(s) + 2NaCl(aq) |
The number of hydroxide ions and NaOH units matches the magnitude of the metal ion's charge (2+ ions need 2NaOH; 3+ ions need 3NaOH).
Worked example - Equation for iron(III) hydroxide
Sodium hydroxide solution is added to a solution of iron(III) chloride, and a brown precipitate of iron(III) hydroxide forms. Write the balanced symbol equation for the reaction.
Step 1 - Write the formulae of the reactants
- Iron(III) chloride: Fe3+ and three Cl- ions → FeCl3(aq)
- Sodium hydroxide: NaOH(aq)
Step 2 - Determine the products
Iron(III) hydroxide and sodium chloride: Fe(OH)3(s) and NaCl(aq)
Step 3 - Balance the equation and add state symbols
FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq)
Worked example - Identifying a compound from tests
A white solid is thought to be a single ionic compound. Four tests are carried out. Name the compound and give its formula.
- A flame test on the solid gives no colour.
- Adding sodium hydroxide solution to a solution of the solid gives a green precipitate.
- Adding dilute hydrochloric acid to a fresh portion of the solid produces no bubbles.
- Acidifying a fresh portion of the solution with dilute hydrochloric acid, then adding barium chloride solution, gives a white precipitate.
Step 1 - Analyse each observation
- No flame colour - Rules out Li+, Na+, K+, Ca2+, and Cu2+.
- Green precipitate with NaOH - Confirms iron(II) hydroxide, so the cation is Fe2+.
- No bubbles with dilute acid - Rules out carbonate (CO32-).
- White precipitate with HCl and BaCl2 - Confirms barium sulfate precipitate, so the anion is sulfate, SO42-.
Step 2 - Combine the ions to find the formula
- Fe2+ and SO42- balance in a 1:1 ratio.
- Name - Iron(II) sulfate
- Formula - FeSO4