Question

Data: (10 marks) Part 1: Standardization of 0.0025M sodium thiosulphate solution No. of drops of 0.0025 M sodium thiosulphate solution Solution Trial 1 Trial 2 Trial 3 Average 1 mL of potassium iodate solution + 0.2 mL 3 M HCI + 1 mL D.I. water + 3 drops of starch 2) 22 121.67 N Part 2: Dissolved oxygen in water sample No. of drops of 0.0025 M sodium thiosulphate solution Solution Trial 1 Trial 2 Trial 3 Average Tap Water 17 7 2 Polluted water Boiled water Data treatment A. Calculate the exact concentration of sodium thiosulphate (10 marks) Volume of each drop = 0.045 mL Total volume of solution added = 0.045mL x no. of drops added Conc. of potassium iodate solution (KIO3): 4.24 Xlo M (recorded from the bottle provided) (a) Volume of Na2S2O3 used to titrate 1 mL potassium iodate solution (mL): No. of drops x 0.045 mL = 21.67 X 0.045 ml = 0.97515 mL (b) Exact conc. of Na2S2O3 (M): 1 mL xconc. of KIO3 (M) ~3x2Vol. of Na2S2O3 used (mL)

doing experiment about standardization of 0.0025M sodium thiosulphate solution,
can someone tell me how this formula (calculating exact conc of Na2S2o3) (at the bottom) comes ？

Answer 1

--- The reaction of war Le03, k103 & #1 6 Nor de Op tk IO3 + 6 Hel - 3 Naz 2406+KI + GNach +3th 8 Here given [x203) = 4.2 4X10 Moles of k 10 in 10 4.24x10ken KZ03 solution = 4.24 x 10-mo x m l 1.24 x 10-7 mol Now according to stoichiometric ration of the reaction 6 linol KZ02 react with 6 moles Nag Se Oz .-. 4o2YX70* 6x 29X7017 - 2.544x18 wil Narses Here, volume of warde og used [ da - 0.97515 er – 0.97515 810 L Number of moler Now, molarity = volume of solution in Exact conc. of war 03 (1) - 2.544x16anos – 2.609 410 mol - - 0.97515X1GX 1 = 2.669 x103 Now wsing given formalar: Exact conc. of Nanse (an) - 1 x conc. of k203 (l) x 3X2 - vol. of Nor selges - 1 X 4.24x16 M X 3 X2 - 2.609 810 sm 0.97515 ml used nl)