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ChemistryIntroductory Chemistry: A Foundation9th Edition

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Chapter 15, Problem 85QAP
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Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

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Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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85. How many milliliters of 0.50 N NaOH are required to neutralize exactly 15.0 mL of 0.35 N H2SO3?

Interpretation Introduction

Interpretation:

The milliliters of 0.50 N

NaOH required to neutralize the given H2SO4 solution is to be calculated.

Concept Introduction:

A chemical reaction that occurs between acid and base is known as neutralization reaction. In the neutralization reaction, the OH− ions of the base combines with H+ ions of an acid. During neutralization reaction, water is released as a side product.

Molarity is used to determine the concentration of the substance in a solution.

Explanation

The value of N1,  N2 and V1 is given to be 0.35 N, 0.50 N and 15.0 mL respectively.

The relationship between normality and volume of both solutions is shown below.

N1V1=N2V2

Where,

  • N1 is the normality of H2SO4 solution.
  • V1 is the volume of H2SO4 solution.
  • N2 is the concentration of NaOH solution.
  • V2 is the volume of NaOH solution.

Rearrange an above expression to calculate V2 which is needed to neutralize the H2SO4 solution

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Sect-15.3 P-15.1SCSect-15.3 P-15.2SCSect-15.4 P-15.3SCSect-15.4 P-15.4SCSect-15.4 P-15.5SCSect-15.4 P-15.6SCSect-15.5 P-15.7SCSect-15.6 P-1CTSect-15.6 P-15.8SCSect-15.7 P-15.9SC
Sect-15.8 P-15.10SCSect-15.8 P-15.11SCCh-15 P-1ALQCh-15 P-2ALQCh-15 P-3ALQCh-15 P-4ALQCh-15 P-5ALQCh-15 P-6ALQCh-15 P-7ALQCh-15 P-8ALQCh-15 P-9ALQCh-15 P-10ALQCh-15 P-11ALQCh-15 P-12ALQCh-15 P-13ALQCh-15 P-14ALQCh-15 P-1QAPCh-15 P-2QAPCh-15 P-3QAPCh-15 P-4QAPCh-15 P-5QAPCh-15 P-6QAPCh-15 P-7QAPCh-15 P-8QAPCh-15 P-9QAPCh-15 P-10QAPCh-15 P-11QAPCh-15 P-12QAPCh-15 P-13QAPCh-15 P-14QAPCh-15 P-15QAPCh-15 P-16QAPCh-15 P-17QAPCh-15 P-18QAPCh-15 P-19QAPCh-15 P-20QAPCh-15 P-21QAPCh-15 P-22QAPCh-15 P-23QAPCh-15 P-24QAPCh-15 P-25QAPCh-15 P-26QAPCh-15 P-27QAPCh-15 P-28QAPCh-15 P-29QAPCh-15 P-30QAPCh-15 P-31QAPCh-15 P-32QAPCh-15 P-33QAPCh-15 P-34QAPCh-15 P-35QAPCh-15 P-36QAPCh-15 P-37QAPCh-15 P-38QAPCh-15 P-39QAPCh-15 P-40QAPCh-15 P-41QAPCh-15 P-42QAPCh-15 P-43QAPCh-15 P-44QAPCh-15 P-45QAPCh-15 P-46QAPCh-15 P-47QAPCh-15 P-48QAPCh-15 P-49QAPCh-15 P-50QAPCh-15 P-51QAPCh-15 P-52QAPCh-15 P-53QAPCh-15 P-54QAPCh-15 P-55QAPCh-15 P-56QAPCh-15 P-57QAPCh-15 P-58QAPCh-15 P-59QAPCh-15 P-60QAPCh-15 P-61QAPCh-15 P-62QAPCh-15 P-63QAPCh-15 P-64QAPCh-15 P-65QAPCh-15 P-66QAPCh-15 P-67QAPCh-15 P-68QAPCh-15 P-69QAPCh-15 P-70QAPCh-15 P-71QAPCh-15 P-72QAPCh-15 P-73QAPCh-15 P-74QAPCh-15 P-75QAPCh-15 P-76QAPCh-15 P-77QAPCh-15 P-78QAPCh-15 P-79QAPCh-15 P-80QAPCh-15 P-81QAPCh-15 P-82QAPCh-15 P-83QAPCh-15 P-84QAPCh-15 P-85QAPCh-15 P-86QAPCh-15 P-87QAPCh-15 P-88QAPCh-15 P-89APCh-15 P-90APCh-15 P-91APCh-15 P-92APCh-15 P-93APCh-15 P-94APCh-15 P-95APCh-15 P-96APCh-15 P-97APCh-15 P-98APCh-15 P-99APCh-15 P-100APCh-15 P-101APCh-15 P-102APCh-15 P-103APCh-15 P-104APCh-15 P-105APCh-15 P-106APCh-15 P-107APCh-15 P-108APCh-15 P-109APCh-15 P-110APCh-15 P-111APCh-15 P-112APCh-15 P-113APCh-15 P-114APCh-15 P-115APCh-15 P-116APCh-15 P-117APCh-15 P-118APCh-15 P-119APCh-15 P-120APCh-15 P-121APCh-15 P-122APCh-15 P-123APCh-15 P-124APCh-15 P-125APCh-15 P-126APCh-15 P-127APCh-15 P-128APCh-15 P-129APCh-15 P-130APCh-15 P-131APCh-15 P-132APCh-15 P-133APCh-15 P-134APCh-15 P-135APCh-15 P-136APCh-15 P-137CPCh-15 P-138CPCh-15 P-139CPCh-15 P-140CPCh-15 P-141CPCh-15 P-142CPCh-15 P-143CPCh-15 P-1CRCh-15 P-2CRCh-15 P-3CRCh-15 P-4CRCh-15 P-5CRCh-15 P-6CRCh-15 P-7CRCh-15 P-8CRCh-15 P-9CRCh-15 P-10CRCh-15 P-11CRCh-15 P-12CRCh-15 P-13CRCh-15 P-14CRCh-15 P-15CRCh-15 P-16CRCh-15 P-17CRCh-15 P-18CRCh-15 P-19CRCh-15 P-20CRCh-15 P-21CRCh-15 P-22CRCh-15 P-23CRCh-15 P-24CRCh-15 P-25CRCh-15 P-26CRCh-15 P-27CRCh-15 P-28CRCh-15 P-29CRCh-15 P-30CRCh-15 P-31CRCh-15 P-32CRCh-15 P-33CRCh-15 P-34CRCh-15 P-35CRCh-15 P-36CRCh-15 P-37CRCh-15 P-38CRCh-15 P-39CRCh-15 P-40CRCh-15 P-41CRCh-15 P-42CR

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