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Monday, April 25, 2022

Identification test for sodium thiosulphate

Learn the identification test for sodium thiosulphate through a laboratory experiment or practical.


Aim:

To perform identification test for sodium thiosulphate.

Reference: Indian Pharmacopoeia (IP)

Requirements:

Glasswares and miscellaneous:
Test tube, measuring cylinder, pipette, glass rod, volumetric flask, beaker, funnel, dropper, rubber stopper, test tube stand, test tube holder, filter paper, balance, etc.

Chemicals:
Sodium thiosulphate (Na2S2O3), ferric chloride (FeCl3), iodine (I2), bromine solution (Br), hydrochloric acid (HCl), etc.

Theory:

This identification of Na2S2O3 is usually based on the reactions of ions and cations present in the sample, by which we can determine the quality of the drug. It comes in the form of crystalline powder which is odorless and colorless that dissolves well in water. It is mostly used in different industries, also used in analytical chemistry, gold mining, the pharmaceutics sector, and as an antidote to cyanide poisoning.

Procedure:

To perform the following tests, prepare a sample solution by dissolving 01.00 g in 20.00 ml of distilled water.

Identification tests for sodium-ion:

  1. Take 02 ml of prepared sample solution and 2 ml (15% w/v) of potassium carbonate solution, and heat it to boiling. No precipitate is formed, and then adds 04 ml of freshly antimonite solution and heated to a boil. Then cool in an ice bath and rub the inside wall of the test tube with a glass rod. A dense white precipitate of sodium antimonite is formed.
  2. Using acetic acid, acidify 01 ml of the above solution, and add additional magnesium uranyl acetate solution. The yellow crystalline precipitate of sodium magnesium uranyl acetate is formed.

Identification tests for thiosulphate ion:

  1. Take 0.1 g of the sample in 5 ml of water and add 02 ml of HCl. A white precipitate is formed, which quickly turns yellow because of the dissociation of sulfur, and sulfur dioxide is produced, which is identified by its odor.
  2. Take 0.1 g of the sample in 5 and add 02 ml of ferric chloride test solution, which produces a deep purple color that fades quickly. 
  3. Add a few drops of iodine solution to a 10% w/v sample solution. The color has discharged. It happens because of an oxidation-reduction reaction.

Result:

The given sample is identified as sodium thiosulphate.


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