Introductory Chemistry: An Active Learning Approach
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN: 9781305079250
Author: Mark S. Cracolice, Ed Peters
Publisher: Cengage Learning
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Chapter 16, Problem 105E
Interpretation Introduction

Interpretation:

The normality of sodium carbonate solution of volume 51.89 mL titrated with 5.038 g of HC7H5O2 is to be calculated.

Concept introduction:

An acid base reaction is one of the types of reaction which is known as neutralization reaction. Normality is used to express concentration of any solution. The n factor of any species is defined in different way depending upon the type of reaction. In an acid base reaction n-factor is defined as the number of replaceable H+ of an acid and for a base it is the total charge present on a species after removing all metal ions. Equation which equates the equivalents of an acid and a base is given below.

n1M1V1equivalents of acid =n2M2V2equivalents of base 

Expert Solution & Answer
Check Mark

Answer to Problem 105E

The normality of sodium carbonate of volume 51.89 mL which is titrated with 5.038 g of HC7H5O2 is 0.7952 N.

Explanation of Solution

The equation which is used to calculate the normality of sodium carbonate is given below.

n1M1V1equivalents of acid =n2M2V2equivalents of base  …(1)

The molarity is given by the formula shown below.

M1=w(g)Molar mass of acid(g/ mol)×Volume of acid(V in mL)M1V1=( g)Molar mass of acid ( g / mol) …(2)

Substitute relation (2) in (1) as shown below.

n1M1V1equivalents of acid =n2M2V2equivalents of base n1× w( g ) M o ( g / mol )equivalents of acid HC7H5O2=n2M2V2equivalents of Sodium Carbonate …(3)

Where,

n1 is the n-factor of benzoic acid(HC7H5O2).

n2 is the n-factor of sodium carbonate base.

w is the mass used of benzoic acid.

M2 is the molarity of sodium carbonate base.

Mo is the molar mass of acid used.

V2 is the volume of sodium carbonate base used.

Benzoic acid has 1H+. So, n-factor of benzoic acid is 1 and sodium carbonate has n-factor 2.

The given value n1 is 1.

The given value of w is 5.038 g.

The given value of M is 122.12 g/mol.

The given value of n2 is 2.

The given value of V2 is 51.89 mL.

Substitute all the values in the equation (2) as shown below.

1×5.038g122.12g/mol=2×M2×51.89mLM2=1×5.038g122.12g/mol×2×51.89 mLM2=0.0003976mol/mL×(1000mL1L)=0.3976 mol/mL

Relation between molarity and normality is written below.

N=n×M

Where,

N is the normality of the solution.

The calculation of normality of sodium carbonate is shown below.

N=n×MN2=2×0.3976 M=0.7952 N

Therefore, normality of sodium carbonate of volume 51.89 mL which is titrated with 5.038 g of HC7H5O2 is 0.7952 N.

Conclusion

The normality of sodium carbonate of volume 51.89 mL titrated with 5.038 g of HC7H5O2 is calculated as 0.7952 N.

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Chapter 16 Solutions

Introductory Chemistry: An Active Learning Approach

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