Experiment 42

Objective

To identify the presence of chloride ions in a sample by observing the formation of silver chloride.

Introduction

In the presence of chloride ions, silver nitrate (AgNO₃) forms a white precipitate of silver chloride (AgCl). This precipitate is insoluble in nitric acid but dissolves in ammonium hydroxide (NH₄OH), forming a soluble complex known as diamine silver chloride.

Reactions:

1.    Formation of silver chloride:

NaCl (aq) + AgNO3 (aq) → AgCl ↓ + NaNO3 (aq)

2.    Dissolution in ammonium hydroxide:

AgCl + 2NH4OH → [Ag(NH3)2]Cl + H2O

Materials

Chemicals

●      Supplied sample containing chloride

●      Silver nitrate (AgNO₃)

●      Ammonium hydroxide (NH₄OH)

Apparatus

●      Test tube

●      Test tube holder

●      Bunsen burner

Reagents

●      Dissolve 3.657 g of the sample in 250 ml of distilled water.

Procedure

1.    Place 5–6 ml of the sample solution in a test tube.

2.    Add 2–3 ml of silver nitrate solution to the sample.

3.    Observe the reaction and note any changes.

4.    Add ammonium hydroxide to the test tube and observe the reaction change.

Observation

A white precipitate of AgCl forms upon the addition of silver nitrate. This precipitate dissolves upon the addition of ammonium hydroxide, forming a clear solution.

Questions

1.    What color forms during the chloride test?

2.    What reagents are used in the chloride test?

3.    What happens when ammonium hydroxide is added?