   Chapter 4, Problem 70PS

Chapter
Section
Textbook Problem

What volume of 0.955 M HCl, in milliliters, is required to titrate 2.152 g of Na2CO3 to the equivalence point?Na2CO3(aq) + 2 HCl(aq) → H2O(l) + CO2(g) + NaCl(aq)

Interpretation Introduction

Interpretation:

The volume of 0.955MHCl required to titrate 2.152g of Na2CO3 to the equivalence point has to be calculated in milliliters.

Concept introduction:

Numberofmole=GivenmassofthesubstanceMolarmass

• 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
• Volumeofthesubstance=Amountof substanceConcentrationofsubstance
Explanation

Balanced chemical equation for the given reaction is,

Na2CO3(aq)+2HCl(aq)H2O(l)+CO2(g)+2NaCl(aq)

The amount (moles) of Na2CO3 available can be calculated as follows,

2.152gNa2CO3×1molNa2CO3106g=0.02030molNa2CO3

The balanced equation for the reaction shows that 1mol of Na2CO3 requires 2mol of HCl.  This is the required stoichiometric factor to obtain the amount of HCl present.

Thus the amount of HCl can be calculated as follows,

0

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