Question

A 0.357 g of sample of an unknown oxidant is added to an excess solution of KI resulting in the production of I₂ (assume a 1:1 ratio for oxidant:I₂). The resulting solution was then titrated with a 0.103 M solution of Na₂S₂O₃. The endpoint was reached after adding 16.82 ml of the titrant. Determine the Formula weight of the oxidant to nearest whole number.

Answer 1

Given,

Mass of sample of unknown oxidant = 0.357 g

Concentration of Na₂S₂O₃ solution = 0.103 M

The volume of Na₂S₂O₃ solution required to reach the endpoint = 16.82 mL x( 1L /1000 mL) = 0.01682 L

Also given, the mol ratio for oxidant :I₂ = 1:1

Now, The balanced reaction between thiosulfate and Iodine is,

2S₂O₃^2-(aq) + I₂(aq)   S₄O₆^2-(aq) + 2I^-(aq)

Now, calculating the number of moles of thiosulfate ions,

= 0.103 M x 0.01682 L

= 0.001732 mol of Na₂S₂O₃

= 0.001732 mol of S₂O₃^2-

Now, calculating the number of moles of iodine required to react completely with 0.001732 mol of S₂O₃^2-,

= 0.001732 mol of S₂O₃^2- x ( 1 mol of I₂ / 2 mol of S₂O₃^2- )

= 0.0008662 mol of I₂

Since, the mole ratio between I₂ and oxidant is 1:1,

Moles of unknown oxidant = 0.0008662 mol

Now,

Molar mass of unknown oxidant = 0.357 g /0.0008662 mol

Molar mass of unknown oxidant = 412 g/mol