0.1 M Sodium Thiosulfate Standardization Calculation
Introduction
Sodium thiosulfate (Na2S2O3) is a strong reducing agent commonly used as a titrant in oxidation-reduction reactions. To ensure accurate results, it is essential to standardize the sodium thiosulfate solution. In this article, we will discuss the calculation involved in standardizing a 0.1 M sodium thiosulfate solution using potassium dichromate (K2Cr2O7) as the primary standard.
Principle
The standardization of sodium thiosulfate is based on the reaction between sodium thiosulfate and potassium dichromate:
Na2S2O3 + K2Cr2O7 → Na2CrO4 + K2S2O3
Materials
- 0.1 M sodium thiosulfate solution (approximate concentration)
- Potassium dichromate (K2Cr2O7) crystals
- Distilled water
- Burette and pipettes
- Conical flask
Procedure
- Preparation of Potassium Dichromate Solution: Weigh accurately about 1.2 g of potassium dichromate crystals and dissolve it in 100 mL of distilled water in a conical flask. This solution will serve as the primary standard.
- Titration: Pipette 25 mL of the potassium dichromate solution into a conical flask. Add a few drops of indicator (e.g., diphenylamine) and slowly add the sodium thiosulfate solution from the burette until the endpoint is reached. Repeat the titration until three concordant readings are obtained.
- Calculation: The amount of potassium dichromate reacted with sodium thiosulfate is calculated using the following formula:
Number of moles of K2Cr2O7 = (Volume of K2Cr2O7 solution x Molarity of K2Cr2O7) / 2
where the factor 2 is due to the reaction stoichiometry.
Let the volume of K2Cr2O7 solution = 25 mL = 0.025 L Molarity of K2Cr2O7 = 0.1 M ( exact molarity can be calculated separately)
Number of moles of K2Cr2O7 = (0.025 L x 0.1 M) / 2 = 0.00125 mol
The number of moles of sodium thiosulfate reacted is equal to the number of moles of potassium dichromate reacted, since the reaction is 1:1.
Number of moles of Na2S2O3 = 0.00125 mol
The concentration of sodium thiosulfate can be calculated as:
Molarity of Na2S2O3 = Number of moles of Na2S2O3 / Volume of Na2S2O3 solution (in liters)
Let the volume of Na2S2O3 solution used in the titration = x mL = x/1000 L
Molarity of Na2S2O3 = 0.00125 mol / (x/1000) L = 1.25/x M
Example Calculation
Suppose the volume of sodium thiosulfate solution used in the titration is 23.50 mL.
Molarity of Na2S2O3 = 1.25 / (23.50/1000) = 0.106 M
Therefore, the standardized concentration of the sodium thiosulfate solution is 0.106 M.
Conclusion
The standardization of sodium thiosulfate is a crucial step in ensuring accurate results in oxidation-reduction reactions. By following the procedure outlined above and performing the necessary calculations, the concentration of sodium thiosulfate can be accurately determined.
