Question

Suppose you need to standardize a sodium thiosulfate solution for a titration experiment. To do so, you will react it with a solution of iodine. You add a 1.00 mL aliquot of 0.0200 M KIO3 solution to a flask, followed by 3 mL of distilled water, 0.2 g of solid KI, and 1 mL H2SO4. You then titrate the solution with sodium thiosulfate solution in order to determine the exact concentration of Na2S2O3. The end point of the titration is reached after 0.86 mL of Na2S2O3 is dispensed from a microburet. What is the concentration of the standard sodium thiosulfate solution?

____M

Answer 1

KIO₃ + 5 KI + 3 H₂SO₄     3 K₂SO₄ + 3 H₂O + 3I₂

Moles of KIO₃ = (Molarity x Volume) / 1000 = 0.02 x 1 /1000 = 2 x 10^-5 moles

Moles of KI = Mass/Molar mass = 0.2/166 = 1.20 x 10^-3 moles  

One mole of KIO₃ react with 5 moles of KI

2 x 10^-5 moles of KIO₃ react with 1.0 x 10^-4 moles of KI

Therefore, KIO₃ is the limiting reagent

Moles of I₂ produced = 2 x 10^-5 x 3 = 6 x 10^-5 moles

2 Na₂S₂O₃ + I₂          Na₂S₄O₆ + 2 NaI

One mole of I₂ react with 2 moles of Na₂S₂O₃

Therefore, 6.0 x 10^-5 moles of I₂ react with (2 x 6.0 x 10^-5) = 1.2 x 10^-4 moles of Na₂S₂O₃

Molarity of standard sodium thiosulfate solution = [(1.2 x 10^-4) x1000]/0.86 = 0.139 M