The amount of I3 - (aq) in a solution can be determined by titration with a solution containing a known concentration of S2O3^2 - (aq) (thiosulfate ion).

The determination is based on the net ionic equation

2S2O3^2 - (aq) + I3^ - (aq) - --> S4O6^2 - (aq) + 31^ - (aq)

Given that it requires 35.8 mL of 0.350 M Na2S2O3 (aq) to titrate a 30.0-mL sample of I3^ - (aq) calculate the molarity of I3^ - (aq) in the solution

molarity of I3^ - (aq) in the solution = 0.21 M

Explanation:

We are given the balanced chemical reaction and the volume and molarity of the Na₂S₂O₃ so we can calculate the moles of the thiosulfate that were required, and then can calculate the molarity the I₃⁻ by the definition of molarity.

First lets convert the volume of Na₂S₂O₃ to liters:

35.8 mL x 1 L/1000 mL = 0.0358 L

# moles Na₂S₂O₃ = 0.350 mol/L x 0.0358 L = 0.0125 mol

From the stoichiometry of the reaction we know 2 mol Na₂S₂O₃ s react with 1 mol I₃⁻, therefore mol of I₃⁻ will be given by

1 mol I₃⁻ / 2 mol Na₂S₂O₃ x 0.0125 mol Na₂S₂O₃ = 0.0063 mol I₃⁻

and its molarity is:

0.0063 mol I₃⁻ / 0.030 L = 0.21 M