Can you provide me calculation for standardization of 0.1 M Iodine solution against 0.1 M Sodium thiosulfate. I can’t get an answer as per my calculation. (0.1 M sodium thio sulfate volumetric solutionis equivalent to 0.01269 g of iodine.
The principle of standardization of sodium thiosulphate solution is based on Redox iodometric titration
with potassium iodate as the primary standard. Potassium iodate a strong oxidizing agent is treated with excess potassium iodide in acidic media which liberates iodine which is back titrated with sodium
thiosulphate. Uniformity of reactions between iodine and sodium thiosulphate forms the basis for utilizing the standard solution of iodine in the analysis of sodium thiosulphate and use of sodium thiosulphate in the analysis of iodine. A starch indicator is used as an indicator showing the disappearance of blue color at the end point.
The chemical reaction is represented below,
KIO3 + 5KI + 3H2SO4 → 3K2SO4 + 3H2O + 3I2
I2 + 2Na2S2O3 → 2NaI + Na2S4O6
Therefore, 2 moles of Sodium thiosulphate = 1 mole of Iodine
(Considering molecular weights of Sodium thiosulphate = 158.11 g/mol and
Iodine I2 = 253.81 g/mol)
Hence, 1 g/mol of Sodium thiosulphate = 1/2 g/mole of Iodine
Hence, 1000 mL of 1 molar Sodium thiosulphate solution = 253.81/2 of Iodine = 126.90 g of Iodine
Hence, 1 mL of 0.1 molar Sodium thiosulphate solution =
[1 X 0.1 X 126.90] / 1000 = 0.01269 g of Iodine
I hope you got the correct calculation to get Iodine amount equivalent to 0.1 M Sodium thiosulphate solution.