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ChemistryIntroductory Chemistry: A Foundation9th Edition

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Chapter 15, Problem 74QAP
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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
61 views

74. What Volume of 0.101 M HNO3 is required to neutralize each of the following solutions?

a. 12.7 mL of 0.501 M NaOH

b. 24.9 mL of 0.00491 M Ba(OH)2

c. 49.1 mL of 0.103 MNH3

d. 1.21 L of 0.102 MKOH

Interpretation Introduction

(a)

Interpretation:

The volume of 0.101 M

HNO3 required to neutralize the given solution is to be calculated.

12.7 mL of 0.501 M

NaOH.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

Moles=MassgMolar mass

The molarity is calculated by the formula,

Molarity=Number of moles of soluteVolume of solution.

Explanation

The value of M1,  V1 and M2 is given to be 0.501 M, 12.7 mL and 0.101 M respectively. V2 is the volume which is required to neutralize the given reaction.

The balanced equation when HNO3 reacts with NaOH is shown below.

NaOH+HNO3→H2O+NaNO3

The above reaction indicates that one equivalent of HNO3 required to neutralize one equivalent of NaOH.

The relationship between concentration and volume of NaOH and HNO3 solutions is shown below.

M1V1=M2V2

Where,

  • M1 is the molarity of NaOH solution.
  • V1 is the volume of NaOH solution
Interpretation Introduction

(b)

Interpretation:

The volume of 0.101 M

HNO3 required to neutralize the given solution is to be calculated.

24.9 mL of 0.00491 M

BaOH2.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

Moles=MassgMolar mass

The molarity is calculated by the formula,

Molarity=Number of moles of soluteVolume of solution.

Interpretation Introduction

(c)

Interpretation:

The volume of 0.101 M

HNO3 required to neutralize the given solution is to be calculated.

49.1 mL of 0.103 M

NH3.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

Moles=MassgMolar mass

The molarity is calculated by the formula,

Molarity=Number of moles of soluteVolume of solution.

Interpretation Introduction

(d)

Interpretation:

The volume of 0.101 M

HNO3 required to neutralize the given solution is to be calculated.

1.21 L of 0.102 M

KOH.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

Moles=MassgMolar mass

The molarity is calculated by the formula,

Molarity=Number of moles of soluteVolume of 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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