Introduction To Chemistry
Introduction To Chemistry
5th Edition
ISBN: 9781259911149
Author: BAUER, Richard C., Birk, James P., Marks, Pamela
Publisher: Mcgraw-hill Education,
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Chapter 11, Problem 80QP

(a)

Interpretation Introduction

Interpretation:

The volume of 0.5000 MHCl required to neutralize 35.00mL of 0.0500 MNaOH is to be calculated.

(a)

Expert Solution
Check Mark

Explanation of Solution

The balanced chemical equation is,

HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l)

The moles of NaOH present in 35.00mL of 0.0500 MNaOH after converting millilitre to litre unit is calculated as follows,

Moles of  NaOH=35.00mL×1L1000mL×0.0500mol/L=0.00175mol

From the balanced chemical reaction, it is clear that 1mol of HCl reaction with 1mol of NaOH . The mol of HCl is calculated using the mol ratio as shown below.

Moles of HCl=0.00175molNaOH×1mol HCl1mol NaOH=0.00175mol HCl

The relationship between molarity, moles of solute, and volume of the solution is,

Molarity=Moles of soluteVolume of solution in litre    ......1

Equation 1 can be rearranged to the following equation.

Volume in litre=Moles of soluteMolarity

The volume of HCl required is,

Volume of HCl in litre=0.00175mol0.5000mol L1=0.0035L

The required volume in liter unit is very small. Thus, it must be converted to milliliter unit.

Volumeof NaOH in millilitre=0.0035L×1000mL1L=3.50mL

Hence, 3.50mL of 0.0500 MNaOH is required to neutralize 35.0mL of 0.5000 MHCl .

(b)

Interpretation Introduction

Interpretation:

The volume of 0.5000 MHCl required to neutralize 10.00mL of 0.2000 MBa(OH)2 is to be calculated.

(b)

Expert Solution
Check Mark

Explanation of Solution

The balanced chemical equation is,

Ba(OH)2(aq)+2HCl(aq)BaCl2(aq)+2H2O(l)

The moles of Ba(OH)2 present in 15.00mL of 0.2000 MBa(OH)2 after converting milliliter into liter unit is calculated as follows,

Moles of Ba(OH)2=15.00mL×1L1000mL×0.2000mol L1=0.003mol

From the balanced chemical reaction, it is clear that 2moles of HCl react with 1mol of Ba(OH)2 . The moles of HCl is calculated using the mol ratio as shown below.

Moles of HCl=0.003mol Ba(OH)2×2mol HCl1mol Ba(OH)2=0.006mol HCl

The relationship between molarity, moles of solute, and volume of the solution is,

Molarity=Moles of soluteVolume of solution in litre    ......1

Equation 1 can be rearranged to the following equation.

Volume in litre=Moles of soluteMolarity

The volume of HCl required is as shown below.

Volumeof HCl in litre=0.006mol0.5000mol L1=0.012L

The required volume in liter unit is very small. Thus, it must be converted to milliliter.

Volumeof HCl in millilitre=0.012L×1000mL1L=12.0mL

Hence, 12.0mL of 0.5000 MHCl required to neutralize 15.00mL of 0.2000 MBa(OH)2 .

(c)

Interpretation Introduction

Interpretation:

The volume of 0.5000 MHCl required to neutralize 15.00mL of 0.2500MNH3 is to be calculated.

(c)

Expert Solution
Check Mark

Explanation of Solution

The balanced chemical equation is,

HCl(aq)+NH3(aq)NH4Cl(aq)

The moles of NH3 present in 15.00mL of 0.2500 MNH3 after converting milliliter to liter unit is calculated as follows,

Moles NH3=15.00mL×1L1000mL×0.2500mol L1=0.00375mol

From the balanced chemical reaction, it is clear that 1mol of HCl reacts with 1mol of NH3 . The moles of HCl is calculated using the mol ratio as shown below.

Moles HCl=0.00375mol NH3×1mol HCl1mol NH3=0.00375mol HCl

The relationship between molarity, moles of solute, and volume of the solution is,

Molarity=Moles of soluteVolume of solution in litre    ......1

Equation 1 can be rearranged to the following equation.

Volume in litre=Moles of soluteMolarity

The volume of HCl required is as given below.

Volumeof HCl in litre=0.00375mol0.5000mol L1=0.0075L

The required volume in liter unit is very small. Thus, it must be converted into milliliters.

Volumeof HCl in mollilitre=0.0075L×1000mL1L=7.5mL

Hence, 7.5mL of 0.5000 MHCl required to neutralize 15.00mL of 0.2500MNH3 .

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Chapter 11 Solutions

Introduction To Chemistry

Ch. 11 - Prob. 5PPCh. 11 - Prob. 6PPCh. 11 - Prob. 7PPCh. 11 - Prob. 8PPCh. 11 - Prob. 9PPCh. 11 - Prob. 10PPCh. 11 - Prob. 11PPCh. 11 - Prob. 1QPCh. 11 - Prob. 2QPCh. 11 - Prob. 3QPCh. 11 - Prob. 4QPCh. 11 - Prob. 5QPCh. 11 - Prob. 6QPCh. 11 - Prob. 7QPCh. 11 - Prob. 8QPCh. 11 - Prob. 9QPCh. 11 - Prob. 10QPCh. 11 - Prob. 11QPCh. 11 - Prob. 12QPCh. 11 - Prob. 13QPCh. 11 - Prob. 14QPCh. 11 - Prob. 15QPCh. 11 - Prob. 16QPCh. 11 - Prob. 17QPCh. 11 - Prob. 18QPCh. 11 - Prob. 19QPCh. 11 - When NaOH dissolves in water, the solution feels...Ch. 11 - Prob. 21QPCh. 11 - Prob. 22QPCh. 11 - Prob. 23QPCh. 11 - Prob. 24QPCh. 11 - Prob. 25QPCh. 11 - Prob. 26QPCh. 11 - Use the rue “like dissolves like� to predict...Ch. 11 - Prob. 28QPCh. 11 - Prob. 29QPCh. 11 - Prob. 30QPCh. 11 - Use intermolecular forces to explain why NaCl is...Ch. 11 - Prob. 32QPCh. 11 - Prob. 33QPCh. 11 - Prob. 34QPCh. 11 - Prob. 35QPCh. 11 - Prob. 36QPCh. 11 - Prob. 37QPCh. 11 - Prob. 38QPCh. 11 - Prob. 39QPCh. 11 - Prob. 40QPCh. 11 - Prob. 41QPCh. 11 - Prob. 42QPCh. 11 - Prob. 43QPCh. 11 - Prob. 44QPCh. 11 - How might you prepare a saturated solution of a...Ch. 11 - Prob. 46QPCh. 11 - Prob. 47QPCh. 11 - Prob. 48QPCh. 11 - Prob. 49QPCh. 11 - Prob. 50QPCh. 11 - Prob. 51QPCh. 11 - Prob. 52QPCh. 11 - Prob. 53QPCh. 11 - Prob. 54QPCh. 11 - Prob. 55QPCh. 11 - Prob. 56QPCh. 11 - Prob. 57QPCh. 11 - Prob. 58QPCh. 11 - The chemical trichloroethylene (TCE) is a...Ch. 11 - Prob. 60QPCh. 11 - Prob. 61QPCh. 11 - Prob. 62QPCh. 11 - Prob. 63QPCh. 11 - Prob. 64QPCh. 11 - Prob. 65QPCh. 11 - Prob. 66QPCh. 11 - Prob. 67QPCh. 11 - Prob. 68QPCh. 11 - Drinking water may contain a low concentration of...Ch. 11 - Prob. 70QPCh. 11 - Prob. 71QPCh. 11 - Prob. 72QPCh. 11 - Prob. 73QPCh. 11 - Prob. 74QPCh. 11 - Prob. 75QPCh. 11 - Prob. 76QPCh. 11 - Prob. 77QPCh. 11 - Prob. 78QPCh. 11 - Prob. 79QPCh. 11 - Prob. 80QPCh. 11 - Prob. 81QPCh. 11 - Prob. 82QPCh. 11 - Prob. 83QPCh. 11 - Prob. 84QPCh. 11 - Prob. 85QPCh. 11 - Prob. 86QPCh. 11 - Prob. 87QPCh. 11 - Prob. 88QPCh. 11 - Prob. 89QPCh. 11 - Prob. 90QPCh. 11 - Prob. 91QPCh. 11 - Prob. 92QPCh. 11 - Prob. 93QPCh. 11 - Prob. 94QPCh. 11 - Prob. 95QPCh. 11 - Prob. 96QPCh. 11 - Prob. 97QPCh. 11 - Prob. 98QPCh. 11 - Prob. 99QPCh. 11 - Prob. 100QPCh. 11 - Prob. 101QPCh. 11 - Prob. 102QPCh. 11 - Prob. 103QPCh. 11 - Prob. 104QPCh. 11 - Prob. 105QPCh. 11 - Prob. 106QPCh. 11 - Prob. 107QPCh. 11 - The solubility of KNO3 increases as the...Ch. 11 - Prob. 109QPCh. 11 - Prob. 110QPCh. 11 - Prob. 111QPCh. 11 - Prob. 112QPCh. 11 - Prob. 113QPCh. 11 - Prob. 114QPCh. 11 - Prob. 115QPCh. 11 - Prob. 116QPCh. 11 - Prob. 117QPCh. 11 - Prob. 118QPCh. 11 - Prob. 119QPCh. 11 - Prob. 120QPCh. 11 - A salad dressing can be made by shaking together...Ch. 11 - Prob. 122QPCh. 11 - Prob. 123QPCh. 11 - Prob. 124QPCh. 11 - Prob. 125QPCh. 11 - Prob. 126QPCh. 11 - Prob. 127QPCh. 11 - Prob. 128QPCh. 11 - Prob. 129QPCh. 11 - Prob. 130QPCh. 11 - Prob. 131QPCh. 11 - Prob. 132QPCh. 11 - Prob. 133QPCh. 11 - Lead(II) iodide, PbI2, is a yellow solid with a...Ch. 11 - Prob. 135QPCh. 11 - Prob. 136QPCh. 11 - Prob. 137QPCh. 11 - Prob. 138QPCh. 11 - Prob. 139QP
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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