World of Chemistry
World of Chemistry
7th Edition
ISBN: 9780618562763
Author: Steven S. Zumdahl
Publisher: Houghton Mifflin College Div
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Chapter 15, Problem 47A
Interpretation Introduction

Interpretation: Volume of 0.172N H2SO4 required to neutralize 56.2ml of 0.145M

  NaOH is to be calculated.

Concept introduction: One normal solution is a measure of the number of moles of hydrogen ions furnished by one mol of an acid. One molar solution is a measure of the number of moles of molecules of acid in one mol of an acid. One molecule of H2SO4 furnishes 2 hydrogen ions. So, one mole of H2SO4 furnishes 2 mol of hydrogen ions.

2 normal H2SO4 solution = 1 molar H2SO4 solution

Balanced chemical equation for the reaction of H2SO4 and NaOH is as follows:

  2NaOH+H2SO4Na2SO4+2H2O

Also the volume of 0.172N H2SO4 required to neutralize 56.2ml of 0.145M

  NaOH can be calculated using the molarity equation written as follows:

  (NaOH)M1V1n1=M2V2n2(H2SO4)

Expert Solution & Answer
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Answer to Problem 47A

Volume of 0.172N H2SO4 required to neutralize 56.2ml of 0.145M

  NaOH is 47.37 mL .

Explanation of Solution

One normal solution is a measure of the number of moles of hydrogen ions furnished by one mol of an acid. One molar solution is a measure of the number of moles of molecules of acid in one mol of an acid. One molecule of H2SO4 furnishes 2 hydrogen ions. So, one mole of H2SO4 furnishes 2 mol of hydrogen ions.

2 normal H2SO4 solution = 1 molar H2SO4 solution

The given acid solution is 0.172N H2SO4

Thus, molarity of H2SO4 solution is half of 0.172N H2SO4

Therefore, molarity will be 0.086 M.

Balanced chemical equation for the reaction of H2SO4 and NaOH is as follows:

  2NaOH+H2SO4Na2SO4+2H2O

Also the volume of 0.172N H2SO4 required to neutralize 56.2ml of 0.145M

  NaOH can be calculated using the molarity equation written as follows:

  (NaOH)M1V1n1=M2V2n2(H2SO4)

Putting in all the values,

  0.145×56.22=0.086×V21V2=0.145×56.22×0.086V2=47.37mL

Conclusion

Volume of 0.172N H2SO4 required to neutralize 56.2ml of 0.145M

  NaOH is 47.37mL .

Chapter 15 Solutions

World of Chemistry

Ch. 15.2 - Prob. 5RQCh. 15.2 - Prob. 6RQCh. 15.2 - Prob. 7RQCh. 15.3 - Prob. 1RQCh. 15.3 - Prob. 2RQCh. 15.3 - Prob. 3RQCh. 15.3 - Prob. 4RQCh. 15.3 - Prob. 5RQCh. 15.3 - Prob. 6RQCh. 15.3 - Prob. 7RQCh. 15.3 - Prob. 8RQCh. 15 - Prob. 1ACh. 15 - Prob. 2ACh. 15 - Prob. 3ACh. 15 - Prob. 4ACh. 15 - Prob. 5ACh. 15 - Prob. 6ACh. 15 - Prob. 7ACh. 15 - Prob. 8ACh. 15 - Prob. 9ACh. 15 - Prob. 10ACh. 15 - Prob. 11ACh. 15 - Prob. 12ACh. 15 - Prob. 13ACh. 15 - Prob. 14ACh. 15 - Prob. 15ACh. 15 - Prob. 16ACh. 15 - Prob. 17ACh. 15 - Prob. 18ACh. 15 - Prob. 19ACh. 15 - Prob. 20ACh. 15 - Prob. 21ACh. 15 - Prob. 22ACh. 15 - Prob. 23ACh. 15 - Prob. 24ACh. 15 - Prob. 25ACh. 15 - Prob. 26ACh. 15 - Prob. 27ACh. 15 - Prob. 28ACh. 15 - Prob. 29ACh. 15 - Prob. 30ACh. 15 - Prob. 31ACh. 15 - Prob. 32ACh. 15 - Prob. 33ACh. 15 - Prob. 34ACh. 15 - Prob. 35ACh. 15 - Prob. 36ACh. 15 - Prob. 37ACh. 15 - Prob. 38ACh. 15 - Prob. 39ACh. 15 - Prob. 40ACh. 15 - Prob. 41ACh. 15 - Prob. 42ACh. 15 - Prob. 43ACh. 15 - Prob. 44ACh. 15 - Prob. 45ACh. 15 - Prob. 46ACh. 15 - Prob. 47ACh. 15 - Prob. 48ACh. 15 - Prob. 49ACh. 15 - Prob. 50ACh. 15 - Prob. 51ACh. 15 - Prob. 52ACh. 15 - Prob. 53ACh. 15 - Prob. 54ACh. 15 - Prob. 55ACh. 15 - Prob. 56ACh. 15 - Prob. 57ACh. 15 - Prob. 58ACh. 15 - Prob. 59ACh. 15 - Prob. 60ACh. 15 - Prob. 61ACh. 15 - Prob. 62ACh. 15 - Prob. 63ACh. 15 - Prob. 1STPCh. 15 - Prob. 2STPCh. 15 - Prob. 3STPCh. 15 - Prob. 4STPCh. 15 - Prob. 5STPCh. 15 - Prob. 6STPCh. 15 - Prob. 7STPCh. 15 - Prob. 8STPCh. 15 - Prob. 9STPCh. 15 - Prob. 10STPCh. 15 - Prob. 11STP
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