Experiment 30

Objective

To identify the presence of aluminum ions in a sample by observing characteristic precipitation reactions with sodium hydroxide and ammonium chloride.

Introduction

Aluminum salts react with sodium hydroxide (NaOH) to form a white, jelly-like precipitate of aluminum hydroxide (Al(OH)₃). On adding excess NaOH, this precipitate dissolves, forming sodium aluminate (NaAlO₂). When ammonium chloride (NH₄Cl) is added to this solution, aluminum hydroxide precipitates again, indicating the presence of aluminum ions.

Reactions:

1.    Formation of aluminum hydroxide:

AlCl3 + NaOH → Al(OH)3 ↓ + HCl

2.    Formation of sodium aluminate:

Al(OH)3 + NaOH → NaAlO2 + 2H2 O

3.    Re-precipitation of aluminum hydroxide:

NaAlO2 + NH4Cl + H2O → Al(OH)3 ↓ + NaCl + NH3

Materials

Chemicals

·       Aluminum chloride (AlCl₃)

·       Sodium hydroxide (NaOH)

·       Ammonium chloride (NH₄Cl)

·       Distilled water

Apparatus

·       Test tube

·       Test tube holder

·       Bunsen burner

Reagents

·       Dissolve 3–4 ml of the sample in a test tube.

·       Prepare a 5% NaOH solution in a 250 ml beaker.

Procedure

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

2.    Add 3–4 ml of NaOH to the sample and observe the formation of a white precipitate.

3.    Add an additional 7–8 ml of NaOH and observe that the precipitate dissolves.

4.    Add ammonium chloride to the solution and observe the reappearance of the white precipitate.

Observation

Upon adding NaOH, a white precipitate of Al(OH)₃ forms. With excess NaOH, the precipitate dissolves, forming a clear solution. The addition of ammonium chloride causes the white precipitate to reappear, confirming the presence of aluminum ions.

Questions

1.    What is an aluminum salt?

2.    How is aluminum salt identified in this test?

3.    What happens when aluminum salt reacts with NaOH?

4.    What is sodium aluminate?