Test 1 and 2: Testing for group 2 metal cation

Test 1: Dilute Sodium Hydroxide

• Place 10 drops of 0.1M Barium Chloride in a clean dry test tube

• Add 10 drops of 0.6M Sodium Hydroxide and mix well

• Add the Sodium Hydroxide dropwise until in excess, and note any changes in a suitable results table

Repeat the test for:

• Calcium bromide

• Magnesium chloride

• Strontium bromide

BaCl₂ + 2NaOH → Ba(OH)₂ + 2NaCl

Group 2 hydroxides are mostly insoluble; however, the solubility of the hydroxide increases as you move down the group due to the increasing radius of the cation.

Ions of similar size experience stronger attractions to one another. As the hydroxide ion is quite small, the smaller metal cation will create stronger ionic compounds, therefore will be less soluble.

Test 2: Dilute Sulfuric Acid

• Place 10 drops of 0.1M Barium Chloride in a clean, dry test tube

• Add about 10 drops of 0.1M Sulfuric Acid and mix well

• Continue to add the dilute sulfuric acid, dropwise, gently shaking until in excess

Repeat the test with:

• Calcium bromide

• Magnesium chloride

• Strontium bromide

Group 2 metals form metal sulfates when reacted with sulfuric acid

BaCl₂ + H₂SO₄ → BaSO₄ + 2HCL

The solubility of metal sulfates is opposite the hydroxides.

The sulfate ion is large so the larger cations will form stronger ionic bonds, and therefore will be less soluble.

Test 3: Testing for ammonia ions in aqueous solution

• Place 10 drops of 0.1M Ammonium Chloride into a clean, dry test tube

• Add 10 drops of 0.4M Sodium Hydroxide solution and shake the mixture

• Warm the mixture gently using a water bath

• Test the fumes released by using forceps to hold a piece of damp red litmus paper over the mouth of the test tube

• Record your observations in a suitable table

NH₄Cl + HaOH → NH₃ + NaCl + H₂O

The reaction when warmed produces ammonia, the fumes will turn the red litmus paper blue.

Test 4: Testing for Hydroxide Ions

• Place 1ml of 0.4M Sodium Hydroxide solution in a test tube, and test with red litmus paper

• Record your observations in a suitable table, then dispose of the test tube contents.

This method can be used to test for ammonia gas. Which forms hydroxide ions when it comes into contact with water:

NH₃ + H₂O → NH₄⁺ + HO^-

• Place 5 drops of 1m ammonia solution and place on an filter paper in a petri dish with a lid

• Dampen a piece of red litmus paper with distilled water and place on the other side of the petri dish, and place the lid on the dish

• Record your observations in a suitable table

Litmus paper should turn blue because of the hydroxide ions produced by the ammonia and water reaction.

Test 5: Test for Carbonate ions in aqueous solution

• Place 2mls of Calcium Hydroxide (limewater) into one test tube

• To a second test tube, add 3ml of the 0.5M Sodium Carbonate solution, then add an equal volume of 1.0M dilute Hydrochloric acid

• Immediately place a delivery tube into the tube containing the limewater. Ensure the end of the delivery tube is completely submerged in the limewater

• Record your observations in a suitable table

The reaction:

Na₂CO₃ + 2HCL → 2NaCl + H₂O + CO₂

The release of carbon dioxide gas will turn the limewater cloudy. As the carbon dioxide reacts with the calcium hydroxide to form insoluble precipitate of calcium carbonate

CO₂ + Ca(OH)₂ → CaCO₃ + H₂O

Test 6: Test for Sulfate ions in aqueous solution

• Add 1ml of 0.1M Magnesium Sulfate solution to a test tube, add an equal volume of dilute Hydrochloric Acid, followed by an equal volume of Barium Chloride

• Record your results in a suitable table

Barium ions will react with the sulphate ions to form an insoluble precipitate of barium sulfate

The solution is acidified first using HCL. Carbonate ions also produce a white precipitate with barium chloride the acid reacts with any carbonate present, and this prevents a false positive result

The reaction:

MgSO₄ + BaCl₂ → BaSO₄ + MgCl

Test 7: Test for Halide ions

• Place 10 drops of 0.1M Potassium Chloride into a clean, dry test tube

• Add 5 drops of dilute Nitric Acid, and shake well

• To the solution add 10 drops of 0.05M Silver Nitrate solution

• Add an excess of 2m ammonia solution and mix thoroughly. Observe your results and record in a suitable table.

Note: This step needs to be carried out in a fume cupboard.

The halide ions react with silver nitrate and form an insoluble silver halide

The solution is acidified first to remove any carbonate ions present as carbonate ions will react to form insoluble silver carbonate precipitate

Silver chloride – white ppt

Silver bromide – cream ppt

Silver iodide – yellow ppt

Balanced equation -

AgNO₃ + NaCl → NaNO₃ + AgBr

Ammonia can be used to be sure of the identity of the halide ions as different halides have different solubility in ammonia solution

Test 8: Test for halide ions in solids salts using concentrated sulfuric acid.

Note: Gloves should be worn for all the experiments below

All experiments should be done in a fume cupboard.

• Place a small spatula of solid Potassium Chloride into a clean, dry test tube

• Carefully and slowly add a few drops of Concentrated Sulfuric Acid

• Observe what happens and record your results in a suitable table

• Test the gas evolved with moist blue litmus paper. Take care to ensure the paper doesn’t touch the side of the test tube.

• Repeat the experiment with solid Potassium Bromide, but this time test the gas using a narrow strip of filter paper that has been dipped in acidified Potassium Dichromate solution

• Repeat using solid Potassium Iodide, but this time test the gas using a narrow strip of filter paper dipped in Lead Nitrate solution

Explanation:

Potassium Chloride gives of acidic fumes of HCL:

KLC + H₂SO₄ → HCL + KHSO₄

So will turn the damp blue litmus paper red

Potassium Bromide will react to form acidic fumes of HBr + brown bromide vapour

An acid base reaction occurs making:

KBr + H₂SO₄ → KHSO₄ + HBr

As Bromine is a more powerful reducing agent a redox reaction also takes place. The Bromide ion reduces the Sulfuric acid so we get the evolution of Sulfur Dioxide gas and Bromine which is the orange/brown vapour

2H⁺ + 2Br → SO₂ + 2H₂O + Br₂

Potassium Iodide acid base reaction

KI + H₂SO₄ → KHSO₄ + HI

Iodide is also a powerful reducing agent and reduces the Sulfur in the Sulfuric Acid to several different oxidation