Experiment 73

Objective

To identify the presence of sulfate ions in an unknown sample by observing the formation of barium sulphate.

Introduction

The sulphate radical, often present in sulfate salts like sulphuric acid, dissolves easily in water, with exceptions like barium sulphate (BaSO₄) and lead sulphate, which are insoluble, and calcium sulphate, which is only slightly soluble. In this test, barium nitrate (Ba(NO₃)₂) reacts with sodium sulphate (Na₂SO₄) to form a white precipitate of barium sulphate, which confirms the presence of sulphate ions.

Reaction:

Ba (NO3)2 ​+ Na2​SO4  ​→ 2NaNO3 ​+ BaSO4 ​↓ (ppt)

Materials

Chemicals

●      Barium nitrate (Ba(NO₃)₂)

●      Distilled water

●      Dilute hydrochloric acid (HCl)

●      Experimental salt as sodium sulphate (Na₂SO₄)

Apparatus

●      Two test tubes

●      Test tube holder

Reagent Preparation:

●      Dissolve 10 g of Ba(NO₃)₂ in 250 mL of distilled water.

●      Dissolve 10 g of Na₂SO₄ in 250 mL of distilled water.

Procedure

1.    Prepare a solution of sodium sulphate in one test tube.

2.    In a separate test tube, prepare a barium nitrate solution.

3.    Mix the barium nitrate solution with the sodium sulphate solution.

4.    Shake the mixture well and add dilute hydrochloric acid.

Observation

A white precipitate of BaSO₄ forms, which is insoluble in HCl, confirming the presence of sulphate ions (SO₄²⁻) in the sample.

Questions

1.    How is the solution prepared?

2.    Why is Ba(NO₃)₂ used instead of BaCl₂?

3.    What color change is observed during the sulphate identification test?