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 73E
Interpretation Introduction

Interpretation:

The mass of NaHSO4 that must be dissolved in 7.50×102mL of solution to produce 0.200NNaHSO4 is to be calculated.

Concept introduction:

Solution is a homogeneous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. The normality of a solution is defined as the number of equivalents per liter of the solution. One equivalent of an acid is the quantity that gives 1 mole of hydrogen ions in a chemical reaction. One equivalent of a base is the quantity that reacts with one mole of hydrogen ions.

Expert Solution & Answer
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Answer to Problem 73E

The mass of NaHSO4 that must be dissolved in 7.50×102mL of solution to produce 0.200NNaHSO4 is 18.0g.

Explanation of Solution

The formula to calculate the normality is given below.

Normality=EquivalentsofsoluteVolumeofsolutioninliters …(1)

The normality of the solution is 0.200N.

The relation between N and eq/L as shown below.

1N=1eq/L

The probable conversion factors are given below.

1N1eq/Land1eq/L1N

The conversion factor to determine eq/L from N is given below.

1eq/L1N

Therefore, 0.200N can be written as shown below.

Normality=0.200NNaHSO4×1eq/L1N=0.200eqNaHSO4L

The volume of the solution is 7.50×102mL.

The relation between L and mL is given below.

1L=1000mL

The probable conversion factors are given below.

1L1000mLand1000mL1L

The conversion factor to determine L from mL is given below.

1L1000mL

Therefore, the volume in liters is calculated below.

Volume=7.50×102mL×1L1000mL=0.750L

Substitute the values of normality and volume of solution in equation (1).

0.200eqNaHSO4L=Equivalentsofsolute0.750L

Rearrange the above expression for the value of equivalents of solute in the above expression.

Equivalentsofsolute=0.200eqNaHSO4L×0.750L=0.15eqNaHSO4

Therefore, the equivalents of solute, NaHSO4 is 0.15eq.

The ionization reaction of NaHSO4 is given below.

HSO4H++SO42

In this reaction, 1 mole of NaHSO4 gives 1 mole of H+ ions during the reaction. Therefore, the number of equivalents per mole is 1.

The formula to calculate the equivalent mass is given below.

Equivalentmass=MolarmassofNaHSO4NumberofequivalentsofNaHSO4 …(2)

The molar mass of oxygen is 16.00gmol1.

The molar mass of sodium is 22.99gmol1.

The molar mass of sulfur is 32.06gmol1.

The molar mass of hydrogen is 1.008gmol1.

Therefore, the molar mass of NaHSO4 is calculated below.

Totalmolarmass=22.99gmol1+1.008gmol1+32.06gmol1+(4×16.00gmol1)=22.99gmol1+1.008gmol1+32.06gmol1+64.00gmol1=120.06gmol1

Substitute the molar mass and number of equivalents of NaHSO4 in equation (2).

EquivalentmassofNaHSO4=120.06gNaHSO41eqNaHSO4

The mass of NaHSO4 can be determined by the formula given below.

MassofNaHSO4=EquivalentsofNaHSO4×EquivalentmassofNaHSO4 …(3)

The equivalents of solute, NaHSO4 is 0.15eq.

Substitute the equivalents of NaHSO4 and the equivalent mass of NaHSO4 in equation (3).

MassofNaHSO4=0.15eqNaHSO4×120.06gNaHSO41eqNaHSO4=18.0gNaHSO4

Therefore, the mass of NaHSO4 that must be dissolved in 7.50×102mL of solution to produce 0.200NNaHSO4 is 18.0g.

Conclusion

The mass of NaHSO4 that must be dissolved in 7.50×102mL of solution to produce 0.200NNaHSO4 is 18.0g.

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Chapter 16 Solutions

Introductory Chemistry: An Active Learning Approach

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