3.1.4 - Qualitative Analysis

Tests for Ions

The composition of a compound or molecule is knowledge that can have many applications. For example, a herbal remedy may be analysed to find out what exactly the molecule of compound is that works, which can then be manufactured and used as medicine.

Testing for Anions

The following tests are ones that can be conducted in a test-tube, and involve a reaction that results in an observable change so that the result can be clearly seen with the naked eye. If nothing is known about a solution, then the tests must be conducted in the order shown, as other orders will result in contaminated result if conducted on the same sample.

CARBONATE ION, CO$\small_3^{2-}$:

Carbonates will react with acids to produce carbon dioxide, a gas.

$$ \text{CO$_3$} \small^{2-} _\text{(aq)} + \text{2H$^+$} _\text{(aq)} \rightarrow \text{H$_2$O} _\text{(l)} + \text{CO$_2$} _\text{(g)} $$

An immediate observation would therefore be that bubbles are formed (effervescence). However, there is a test for carbon dioxide also. Aqueous calcium hydroxide reacts with carbon dioxide gas to produce calcium carbonate and water.

$$ \text{CO$_2$} _\text{(g)} + \text{Ca(OH)$_2$} _\text{(aq)} \rightarrow \text{CaCO$_3$} _\text{(s)} + \text{H$_2$O} _\text{(l)} $$

Since calcium carbonate is only slightly soluble in water, a milk-white precipitate is formed.

With this, the procedure is:

1. Add dilute hydrochloric acid to a sample of the unknown solution.
2. If any gas is produced, collect it.
3. To check, bubble the collected gas through limewater (calcium hydroxide solution).
4. If the limewater turns milk-white, then the original sample solution contains carbonate ions.

SULFATE ION, SO$\small_4 \small^{2-}$:

Sulfates react with barium compounds to produce barium sulfate.

$$ \text{SO$_4$} \small^{2-} _\text{(aq)} + \text{Ba$^{2+}$} _\text{(aq)} \rightarrow \text{BaSO$_4$} _\text{(s)} $$

Barium sulfate is an insoluble salt. As a result, it precipitates out of the solution, resulting in an observable white mixture.

The reason why this must be conducted after the test for carbonates is because the carbonate ion also produces an insoluble salt with barium, and so resulting a false positive for sulfate ions. Once it is established that a compound does not contain carbonate ions, then a positive test must be indicative of sulfate ions only.

With this, the procedure is:

1. Conduct a test for carbonate ions.
2. If the result is negative, continue to step 3. If positive, tests are complete.
3. Add a solution of barium chloride to the sample of the unknown solution.
4. If the mixture turns white, then the original solution solution contains sulfate ions.

HALIDE IONS, Cl$\small^{-}$, Br$\small^{-}$ or I$\small^{-}$:

Halide compounds react with silver compounds to produce silver halides.

$$ \text{X$^-$} _\text{(aq)} + \text{Ag$^+$} _\text{(aq)} \rightarrow \text{AgX} _\text{(s)} $$

Silver halides are insoluble salts. As a result, they precipitates out of the solution, resulting in an observable colour change from colourless to:

• White if chloride ions were present
• Cream if bromide ions were present
• Yellow of iodide ions were present

If the colour is unclear, then a further test may be done. By adding a different solvent that silver halides are soluble in, the disappearance of the colour will help to indicate the halide present. Silver chloride is soluble in dilute aqueous ammonia, whereas silvers bromide and iodide are not. Both silvers chloride and bromide are soluble in concentrated ammonia, whereas silver iodide is not. Below is a table that shows the unique observations for each halide:

The reason why this test must be conducted after the test for sulfates is because the sulfate ion also produces an white insoluble salt with silver, and so resulting a false positive for chloride ions. Once it is established that a compound does not contain sulfate ions, then a positive test must be indicative of halide ions only.

With this, the procedure is:

1. Conduct a test for carbonate and sulfate ions.
2. If both results are negative, continue to step 3. If positive, tests are complete.
3. Dissolve the sample of unknown solution in nitric acid.
4. Add silver nitrate solution dropwise.
5. If a precipitate is formed, then the original solution contains halide ions.
6. Add cold, dilute ammonia to the mixture. Record observations.
7. Then add cold, concentrated ammonia to the mixture. Record observations.
8. Confer with the above table to determine which halide is present.

Testing for Cations

The only cation of which knowledge of the qualitative test is required by this course is the one for ammonium ions, NH$\small_4 \small^+$. This test involves the production of the toxic gas, ammonia, so it must be conducted in a fume cupboard.

Ammonium compounds react with hydroxides to form ammonia gas.

$$ \text{NH$_4$} \small^+ _\text{(aq)} + \text{OH$^-$} _\text{(aq)} \rightarrow \text{NH$_3$} _\text{(g)} + \text{H$_2$O} _\text{(l)} $$

The gas released will dissolve in water to produce an alkaline solution. Therefore it will stain red litmus paper blue. Instead of catching the gas and actively dissolving it in water, the resultant ammonium hydroxide can be quickly produced by the litmus paper being damp to begin with.

With this, the procedure is:

1. In a fume cupboard, add a solution of warm sodium hydroxide to a new sample of the unknown solution.
2. If a gas is produced, hold a strip of damp, red litmus paper to the open end of the test tube.
3. If the paper turns blue, then the original solution contains ammonium ions.