World of Chemistry, 3rd edition
World of Chemistry, 3rd edition
3rd Edition
ISBN: 9781133109655
Author: Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher: Brooks / Cole / Cengage Learning
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Chapter 15, Problem 43A
Interpretation Introduction

Interpretation: The number of equivalents of hydroxide ions needed to react with 1.53 equivalents of hydrogen ions is to be given and explained.

Concept introduction: In neutralization reaction, one equivalent of hydroxide ion neutralizes one equivalent of hydrogen ion as both carry unit charges with opposite signs as follows:

  H+(aq)+OH(aq)H2(l)

Expert Solution & Answer
Check Mark

Answer to Problem 43A

The number of equivalents of hydroxide ions needed to react with 1.53 equivalents of hydrogen ions is to 1.53.

Explanation of Solution

In neutralization reaction, one equivalent of hydroxide ion neutralizes one equivalent of hydrogen ion as both carry unit charges with opposite signs as follows:

  H+(aq)+OH(aq)H2(l)

So, the number of equivalents of hydroxide ions needed to react with 1.53 equivalents of hydrogen ions is to 1.53. Even without a balanced chemical equation, this can be told as the chemistry of reaction of hydroxide ion and hydrogen ion is not affected by the coefficients in the balanced chemical equation.

Conclusion

The number of equivalents of hydroxide ions needed to react with 1.53 equivalents of hydrogen ions is to 1.53.

Chapter 15 Solutions

World of Chemistry, 3rd edition

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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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY