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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

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Answer:

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

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