10.1.1 Principles of ion tests and flame tests

A test is only useful if its result is unique

1. A test identifies an ion by giving a result that no other ion gives.
2. A result shared by two ions leaves the answer ambiguous.
3. A white precipitate on its own identifies nothing, because several ions give one.
4. What makes a test unique is the combination of reagent and observation.
5. Where one test cannot separate two ions, a second test is used alongside it.

A flame test identifies a metal ion by colour

1. A small sample of the solid is held in a flame and the colour is observed.
2. The wire holding the sample is cleaned first, usually by dipping it in acid.
3. A clean wire is checked by holding it in the flame until it gives no colour.
4. The flame used is a blue Bunsen flame, since a yellow flame would hide the result.
5. The colour comes from the metal ion, so the same ion gives the same colour in any compound.

The five flame colours

1. Lithium, ${Li}^{+}$, gives a red flame.
2. Sodium, ${Na}^{+}$, gives a yellow flame.
3. Potassium, $K^{+}$, gives a lilac flame.
4. Calcium, ${Ca}^{2+}$, gives an orange-red flame.
5. Copper, ${Cu}^{2+}$, gives a blue-green flame.

What a flame test cannot do

1. Only one ion can be identified at a time, so a mixture gives a confused result.
2. Sodium's strong yellow colour masks the paler colours of other ions in the same sample.
3. Lithium's red and calcium's orange-red are similar, so they can be confused.
4. A flame test says nothing about the negative ion in the compound.
5. A full identification therefore needs a separate test for the negative ion.

10.1.2 Tests for cations and for ammonia

Sodium hydroxide solution precipitates metal hydroxides

1. Sodium hydroxide solution is added to a solution of the unknown salt.
2. The hydroxide ions react with the metal ion to form an insoluble hydroxide.
3. That insoluble solid appears as a precipitate.
4. The colour of the precipitate is what identifies the metal ion.
5. Adding sodium hydroxide in excess gives a second piece of evidence.

The coloured precipitates

1. Copper, ${Cu}^{2+}$, gives a blue precipitate.
2. Iron(II), ${Fe}^{2+}$, gives a green precipitate.
3. Iron(III), ${Fe}^{3+}$, gives a brown precipitate.
4. The reaction for copper is: $${Cu}^{2+}(aq)+2{OH}^{-}(aq)\to {Cu(OH)}_2(s)$$
5. None of these three dissolves when more sodium hydroxide is added.

The two white precipitates are told apart by excess

1. Aluminium, ${Al}^{3+}$, gives a white precipitate.
2. Calcium, ${Ca}^{2+}$, also gives a white precipitate.
3. Adding excess sodium hydroxide separates them.
4. The aluminium precipitate dissolves in excess, giving a colourless solution.
5. The calcium precipitate does not dissolve, however much is added.

The ammonium ion gives off ammonia

1. Sodium hydroxide solution is added to the sample and the mixture is warmed gently.
2. The ammonium ion, ${NH}_4^{+}$, releases ammonia gas.
3. The reaction is: $${NH}_4^{+}+{OH}^{-}\to {NH}_3+H_{2}O$$
4. No precipitate forms, since the ammonium ion is not a metal ion.
5. The gas is then identified by the test for ammonia.

The chemical test for ammonia

1. Damp red litmus paper is held at the mouth of the tube.
2. Ammonia is alkaline in solution, so it turns the paper blue.
3. The paper must be damp, because the gas has to dissolve before it can act.
4. Ammonia also has a sharp, choking smell, though smell alone is not the test.
5. A blue result confirms the ammonium ion was present in the original sample.

10.1.3 Tests for anions

Carbonates give off carbon dioxide with dilute acid

1. Dilute acid is added to the solid or to its solution.
2. A carbonate fizzes, because a gas is being released.
3. The reaction releases carbon dioxide: $$carbonate+acid\to salt+water+carbon\; dioxide$$
4. The gas is passed through limewater, which turns milky.
5. Fizzing alone is not enough, since the gas has to be identified as well.

Sulfates give a white precipitate with barium chloride

1. Dilute hydrochloric acid is added to the solution first.
2. Barium chloride solution is then added.
3. A white precipitate of barium sulfate shows a sulfate ion is present.
4. The precipitate forms as: $${Ba}^{2+}(aq)+{SO}_4^{2-}(aq)\to {BaSO}_4(s)$$
5. The acid is added first to remove any carbonate, which would also precipitate.

Halides give coloured precipitates with silver nitrate

1. Dilute nitric acid is added to the solution first.
2. Silver nitrate solution is then added.
3. A chloride gives a white precipitate.
4. A bromide gives a cream precipitate.
5. An iodide gives a yellow precipitate.

Why the acid comes first

1. A carbonate in the sample would give a precipitate with silver or barium ions.
2. That precipitate would be mistaken for a positive halide or sulfate result.
3. Adding acid first destroys any carbonate, releasing it as carbon dioxide.
4. Nitric acid is used before silver nitrate, so that no chloride is introduced.
5. Hydrochloric acid is used before barium chloride, for the same kind of reason.

10.1.4 Identification of ions in unknown salts

An unknown salt needs both of its ions identified

1. Every salt contains a positive ion and a negative ion.
2. Naming the salt means identifying both, not just one.
3. The tests are chosen so that each gives a unique result.
4. Results are recorded as observations first, then interpreted.
5. The name of the salt is the conclusion, put together at the end.

A workable order for the tests

1. Start with a flame test on the solid, which points to the metal ion.
2. Dissolve a fresh sample and add sodium hydroxide solution, then add it in excess.
3. Those two together usually settle the positive ion.
4. For the negative ion, add dilute acid first to check for a carbonate.
5. Then test a fresh portion with barium chloride or with silver nitrate as appropriate.

Combining the results

1. Each observation narrows the possibilities rather than giving the answer outright.
2. A white precipitate with sodium hydroxide leaves two candidates until excess is added.
3. A result that rules an ion out is as useful as one that rules it in.
4. The two ions are then written together with the correct charges.
5. The formula follows once the charges are balanced.

Where an identification goes wrong

1. A contaminated wire or tube carries an ion from the previous test.
2. Skipping the excess step leaves aluminium and calcium unseparated.
3. Skipping the acid step lets a carbonate give a false precipitate.
4. Reading a colour in poor light, or against a coloured background, misleads.
5. Stating a conclusion without the observation behind it leaves the answer unsupported.

10.1.5 Instrumental methods of analysis

Instrumental methods do the same job with a machine

1. Chemical tests rely on an observation made by a person.
2. An instrumental method takes the same measurement with a machine.
3. Machines improve the sensitivity, so a far smaller amount can be detected.
4. They improve the accuracy, so the result is closer to the true value.
5. They improve the speed, so many samples can be analysed in the time one test takes.

Why those three advantages matter

1. A trace of an ion too small to colour a flame can still be detected by an instrument.
2. A judgement between lilac and red depends on the observer, while a machine reads a number.
3. A laboratory testing hundreds of samples needs the throughput a machine gives.
4. An instrument also gives a numerical result, which a colour never does.
5. Against that, instruments are expensive and need trained operators.

A flame photometer measures the light from a flame

1. The sample is put into a flame, as in a flame test.
2. The instrument measures the light the flame gives out.
3. The pattern of light is compared with reference data to identify the metal ion.
4. The intensity of the light depends on how much of the ion is present.
5. No knowledge of how the instrument works is needed to use its output.

Using a calibration curve

1. Solutions of known concentration are measured first.
2. Their readings are plotted against concentration to give a curve.
3. The unknown solution is then measured in the same instrument.
4. Its reading is found on the vertical axis and traced across to the curve, then down to the concentration.
5. A reading beyond the range of the curve cannot be read off, so the sample is diluted first.