Question

Suppose a student needs to standardize a sodium thiosulfate, Na, S,Og, solution for a titration experiment. To do so, he or she will react it with a solution of iodine. The student adds a 1.00 mL aliquot of 0.0200 M KIO2 solution to a flask, followed by 3 mL of distilled water, 0.2 g of solid KI, and 1 mL H,SO,. The student then titrates the solution with sodium thiosulfate solution in order to determine the exact concentration of Na, S,O2. The end point of the titration is reached after 0.82 mL of Na, S,03 is dispensed from a microburet. What is the concentration of the standard sodium thiosulfate solution? concentration =

Answer 1

Reaction between KIO₃ and KI in acidic medium

net ionic equation

IO₃^- + 5I^- + 6H⁺ ---------> 3 IO₂ + 3H₂O

reaction between I₂ with thiosulphate

Na₂S₂O₃ -------> 2Na⁺ +    S₂O₃^-2

3I₂ + 6 S₂O₃^-2 -------> 6I^- + S₄O₆^-2

net reaction IO₃^- + 6H⁺ + 6 S₂O₃^-2 --------> I^- + 3S₄O₆^-2 + 3H₂O

moles of KIO₃ used = molarity X volume (L)

= 0.02 mol/L X 0.001 L

= 0.00002 mol

1 mol of KIO₃ reacts with 6 moles of S₂O₃^-2

0.00002 mol of  KIO₃ reacts with 0.00012 moles of S₂O₃^-2

moles of S₂O₃^-2 used = 0.00012 mol

volume = 0.82 mL

1 mL = 0.001 L

0.82 mL = 0.00082 L

molarity of S₂O₃^-2 = molarity / volume (L)

= 0.00012 mol / 0.00082 L

= 0.146 M

molarity of Na₂S₂O₃ = 0.146 M

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