Experiment 55

Objective

To identify the presence of iodide in a supplied sample through the formation of silver iodide precipitate.

Introduction

When iodide ions in a salt react with silver nitrate, they form a yellow precipitate of silver iodide (AgI), which is insoluble in both nitric acid and ammonium hydroxide. This property distinguishes iodide from other halides in qualitative analysis.

Chemical Reaction:

NaI (aq)+AgNO₃ (aq)→AgI↓ (yellow precipitate)+NaNO₃ (aq)

Materials

Chemicals

●      Supplied sample (containing iodide)

●      Silver nitrate (AgNO₃)

●      Ammonium hydroxide (NH₄OH)

Apparatus:

●      Test tube

●      Test tube holder

●      Bunsen burner

Reagents:

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

Procedure

1.    Sample Preparation: Place 5-6 ml of the sample solution in a test tube.

2.    Silver Nitrate Addition: Add 2-3 ml of silver nitrate to the sample.

3.    Observation of Initial Reaction: Observe any changes in the reaction, especially the color of the precipitate.

4.    Ammonium Hydroxide Test: Add ammonium hydroxide to the test tube and observe any further reaction changes.

Observations

●      Upon adding silver nitrate, a yellow precipitate forms, indicating the presence of iodide ions.

●      When ammonium hydroxide is added, the yellow precipitate remains insoluble, further confirming the presence of iodide.

Questions

1.    What color forms during the iodide test?

2.    What reagent is used in the iodide test?

3.    What happens when ammonium hydroxide is added?

4.    What is an iodide salt?