   Chapter 4.9, Problem 2.1ACP

Chapter
Section
Textbook Problem

Excess KI is added to a 100.0-mL sample of a soft drink that had been contaminated with bleach, NaClO. The iodine (l2) generated in the solution is then titrated with 0.0425 M Na2S2O3 and requires 25.3 mL to reach the equivalence point. What mass of NaClO was contained in the 100.0-mL sample of adulterated soft drink?

Interpretation Introduction

Interpretation:

The mass of NaClO  in the given sample has to be determined.

Concept introduction:

• The relation between the number of moles and mass of the substance is ,

Numberofmole=MassingramMolarmass

Massingramofthesubstance=Numberofmole×Molarmass

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• Concentrationofsubstance=Amountof substancevolumeofthesubstance
• Amountof substance=Concentrationofsubstance×volumeofthesubstance
• Titration is a method to determine the concentration of a substance in solution by adding a solution of known volume and concentration until the reaction is completed.
Explanation

The balanced net ionic equation for the reaction occurring in the titration is,

I2(aq)+2S2O3(aq)22I(aq)+S4O6(aq)2

Amount of Na2S2O3 can be determined as follows,

Amount ofNa2S2O3=CNa2S2O3×VNa2S2O3

=0.0425molNa2S2O31L×0.0253L

=1.075×1003molNa2S2O3

From the amount of Na2S2O3 number of moles of I2 can be calculated. Stoichiometric ratio between Na2S2O3 and I2 is 2:1

1.075×1003molNa2S2O3×1molI22molNa2S2O3=5

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