4.8.3.1,4.8.3.2 - Flame tests and metal hydroxide tests

4.8.3.1,4.8.3.2 - Flame tests and metal hydroxide tests

This lesson identifies metal ions using flame colours and precipitates formed with sodium hydroxide solution. It teaches the cation methods in Required Practical 7, explains how extra tests distinguish similar results, and builds balanced equations for insoluble metal hydroxides.

Flame test method and colours

Some metal ions produce characteristic colours when a small sample of their compound is placed in a flame.

A positively charged ion is a cation. A flame test can help identify the metal cation in a compound; it does not directly identify the negative ion. Use a clean nichrome or platinum wire loop. Clean it using the supervised dilute hydrochloric acid procedure, then hold it in a blue, non-luminous Bunsen flame until it gives no interfering colour. Allow it to cool before taking up a small sample, then place the sample on the wire in the blue flame and record the flame colour.

A blue flame gives less background colour than a luminous yellow flame. Cleaning the loop removes residues from previous samples. Use separate labelled samples and clean the loop between them. A method saying only “hold it over a Bunsen burner” does not specify that the sample must enter the flame.

Metal ion in the compoundCharacteristic flame colour
LithiumCrimson
SodiumYellow
PotassiumLilac
CalciumOrange-red
CopperGreen

Heating supplies energy to the sample and light is emitted. Flame emission spectroscopy separates that light into more detailed evidence. A green flame supports the presence of copper ions; it does not mean the original solution must have been green.

A mixture can hide a colour. For example, a strong sodium yellow may mask a weaker colour from another ion. A yellow result therefore does not prove that sodium is the only metal present. Contamination can have the same effect. Wear eye protection, tie back long hair, keep flammable materials clear and handle the hot wire safely under supervision. The written method develops Required Practical 7 knowledge; the practical itself must be performed in the appropriate laboratory setting.

Hydroxide precipitates

Sodium hydroxide solution supplies hydroxide ions that can react with dissolved metal ions to form insoluble solids.

A precipitate is an insoluble solid formed from substances in solution. The solid appears as particles or a cloudy suspension; it is not merely the original solution becoming a different colour. Put a small measured portion of the unknown solution in a labelled test tube. Add sodium hydroxide solution drop by drop, mix carefully and record the precipitate's colour. Small additions let you observe the first precipitate before testing the effect of excess reagent.

Metal ions presentPrecipitate with sodium hydroxide
Aluminium, calcium or magnesiumWhite
Copper(II)Blue
Iron(II)Green
Iron(III)Brown

The Roman numeral specifies the charge of the metal ion: iron(II) is FeX2+\ce{Fe^2+}, while iron(III) is FeX3+\ce{Fe^3+}. Their different hydroxides give different observations. A solution's starting colour must not be confused with the colour of the newly formed solid.

Building balanced hydroxide equations

Hydroxide is OHX−\ce{OH^-}, so a metal with charge 2+2+ requires two hydroxide groups in its neutral hydroxide: Mg(OH)X2\ce{Mg(OH)2}, for example. A metal with charge 3+3+ requires three: Al(OH)X3\ce{Al(OH)3}. The brackets mean that the whole hydroxide group is repeated. Write the correct formula first, then balance by changing coefficients, never by changing subscripts.

The following balanced full equations use soluble chloride salts as examples. The sodium chloride stays dissolved; the metal hydroxide is the solid.

MgClX2(aq)+2 NaOH(aq)→Mg(OH)X2(s)+2 NaCl(aq)\ce{MgCl2(aq) + 2NaOH(aq) -> Mg(OH)2(s) + 2NaCl(aq)} CaClX2(aq)+2 NaOH(aq)→Ca(OH)X2(s)+2 NaCl(aq)\ce{CaCl2(aq) + 2NaOH(aq) -> Ca(OH)2(s) + 2NaCl(aq)} CuClX2(aq)+2 NaOH(aq)→Cu(OH)X2(s)+2 NaCl(aq)\ce{CuCl2(aq) + 2NaOH(aq) -> Cu(OH)2(s) + 2NaCl(aq)} FeClX2(aq)+2 NaOH(aq)→Fe(OH)X2(s)+2 NaCl(aq)\ce{FeCl2(aq) + 2NaOH(aq) -> Fe(OH)2(s) + 2NaCl(aq)} FeClX3(aq)+3 NaOH(aq)→Fe(OH)X3(s)+3 NaCl(aq)\ce{FeCl3(aq) + 3NaOH(aq) -> Fe(OH)3(s) + 3NaCl(aq)} AlClX3(aq)+3 NaOH(aq)→Al(OH)X3(s)+3 NaCl(aq)\ce{AlCl3(aq) + 3NaOH(aq) -> Al(OH)3(s) + 3NaCl(aq)}

Balancing a magnesium hydroxide reaction

First write the supplied formulae on the correct sides. Magnesium chloride has two chlorine atoms, so two sodium chlorides are needed. Two sodium hydroxides then supply the two sodium atoms and two hydroxide groups.

MgClX2+NaOH→Mg(OH)X2+NaCl\ce{MgCl2 + NaOH -> Mg(OH)2 + NaCl} MgClX2+2 NaOH→Mg(OH)X2+2 NaCl\ce{MgCl2 + 2NaOH -> Mg(OH)2 + 2NaCl}

Counting Mg, Cl, Na, O and H gives the same total on each side.

Balancing an aluminium hydroxide reaction

Aluminium chloride has three chlorine atoms, so the sodium chloride coefficient is three. Aluminium hydroxide contains three hydroxide groups, supplied by three sodium hydroxides.

AlClX3+NaOH→Al(OH)X3+NaCl\ce{AlCl3 + NaOH -> Al(OH)3 + NaCl} AlClX3+3 NaOH→Al(OH)X3+3 NaCl\ce{AlCl3 + 3NaOH -> Al(OH)3 + 3NaCl}

The balanced coefficients are 1:3:1:31:3:1:3. Neither calculation changes a compound's formula.

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