Weight out accurately 0.14-0.15g of pure dry KIO3 , dissolve it in 25ml of distilled water. Add 2 g of potassium iodate and 2ml of dil. H2SO4. Titrate liberated I2 with the thiosulphate solution with constant stirring. When the color of the liquid has become pale yellow, dilute the solution to ~200ml with distilled water, add 2ml of starch solution and continue the titration until the color changes from blue to colorless. (Volume of Na2S2O3 in trial 01, 20.30ml and trial 02 20.60ml)

Question

Weight out accurately 0.14-0.15g of pure dry KIO3 , dissolve it in 25ml of distilled water. Add 2 g of potassium iodate and 2ml of dil. H2SO4. Titrate liberated I2 with the thiosulphate solution with constant stirring. When the color of the liquid has become pale yellow, dilute the solution to ~200ml with distilled water, add 2ml of starch solution and continue the titration until the color changes from blue to colorless. (Volume of Na2S2O3 in trial 01, 20.30ml and trial 02 20.60ml)

So calculate the concentration of the Na2S2O3 solution?

Answer 1

IO3^- + 5I^- + 6H^+ ==> 3I2 + 3H2O. The titration step is

2S2O3^2- + I2 --> S4O6^2- + 2I^-
So how much KIO3 did you weight out?
mols KIO3 = grams/molar mass = ?
mols KIO3 from above x (3 mols I2/1 mol KIO3) x (2 moles S2O3^2-/1 mol I2) = moles S2O3^2-
Then M S2O3^2- = mols S2O3^2-/L S2O3^- = ?
Post your work if you get stuck.