World of Chemistry
World of Chemistry
7th Edition
ISBN: 9780618562763
Author: Steven S. Zumdahl
Publisher: Houghton Mifflin College Div
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Chapter 15.3, Problem 4RQ
Interpretation Introduction

Interpretation: The critical information necessary to find the volume of the substance needs to be determined.

Concept introduction: In an acid−base reaction when number of moles of H+ is exactly equal to the number of moles of OH it is called point of neutralization and the reaction is called neutralization reaction.

Expert Solution & Answer
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Answer to Problem 4RQ

Concentration of the substance whose volume is being determined is needed.

Explanation of Solution

At neutralization point of a reaction,

  number of moles of acid=number of moles of of baseM1V1=M2V2

Where,

  M1=concentration of acid.

  V1= volume of the acid.

  M2=concentration of base.

  V2= volume of the base.

Say we need to determine the volume of acid. First step is to determine the moles of base present in solution initially.

Moles of base =M2×V2=n (say) . If we know the concentration of the acid then from M1V1=M2V2 equation we can easily calculate the volume of acid ( V1 ) needed in neutralization reaction.

Conclusion

Concentration of the substance whose volume is being determined is needed.

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