Experiment 47

Objective

To identify the presence of ethanoic acid in a given sample.

Introduction

Ethanoic acid, also known as acetic acid, reacts with sodium hydroxide (NaOH) to form sodium ethanoate. When sodium ethanoate is further reacted with arsenic oxide, it produces a distinct odor of cacodyl oxide, which serves as an indicator for the presence of ethanoic acid.

Chemical Reactions:

1.    Formation of Sodium Ethanoate:

CH3COOH + NaOH → CH3COOHNa + H2O

2.    Odor-inducing reaction with arsenic oxide:

4CH3COONa + AsO3 → (Odor of cacodyl oxide)

Materials

Chemicals

●      Ethanoic acid

●      Sodium hydroxide (NaOH)

●      Arsenic oxide (As₂O₃)

Apparatus

●      Test tube

●      Test tube holder

●      Dropper

●      Bunsen burner

Preparation of Reagents

●      Place 5-6 mL of ethanoic acid in a test tube.

●      Add 5-6 mL of NaOH solution to the ethanoic acid.

●      Add arsenic oxide dropwise during the procedure.

Procedure

1.    Place the ethanoic acid sample in a test tube.

2.    Secure the test tube in a test tube holder.

3.    Add NaOH solution to the sample.

4.    Add arsenic oxide dropwise to the solution.

5.    Heat the test tube gently for 1-2 minutes.

6.    When a distinct odor is detected, indicating cacodyl oxide, stop heating.

Observation

The formation of a characteristic odor (cacodyl oxide) indicates the presence of ethanoic acid in the sample.

Questions

1.    What is ethanoic acid?

2.    How is ethanoic acid identified in this experiment?

3.    What happens when arsenic oxide is added to sodium ethanoate?