The volume of 0.115 M HClO 4 required to neutralize 50 mL of 0 .0875 M NaOH . Concept introduction: The volume of HClO 4 required for neutralization is calculated by the formula, M 1 V 1 = M 2 V 2 V 1 = M 2 V 2 M 1 Here, M 1 is the molarity of HClO 4 . M 2 is the molarity of NaOH . V 1 is the volume of HClO 4 . V 2 is the volume of NaOH .

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Answer 1

Question

Chapter 13, Problem 43E

(a)

Interpretation Introduction

Interpretation: The volume of $0.115 M HClO4$ required to neutralize $50 mL$ of $0.0875 M$

$NaOH$ .

Concept introduction:
The volume of $HClO4$ required for neutralization is calculated by the formula,

$M1V1=M2V2V1=M2V2M1$

Here,
$M1$ is the molarity of $HClO4$ .
$M2$ is the molarity of $NaOH$ .
$V1$ is the volume of $HClO4$ .
$V2$ is the volume of $NaOH$ .

(b)

Interpretation Introduction

Interpretation: The volume of $0.128 M$

$HCl$ required to neutralize $2.87 g Mg(OH)2$ .

Concept introduction:

The number of moles is defined as the ratio of given mass of the compound in grams to the molar mass of the compound.
And,
The volume is calculated by the formula,

$M=n of HClV( mL)×1000 mLV(mL)=nM×1000 mL$

(c)

Interpretation Introduction

Interpretation: The molarity of $Cl−$ solution.

Concept introduction:
The molarity is defined as the ratio of number of moles of solute to the volume of the solution in liters.
$Molarity=Moles of soluteVolume of solution$

(d)

Interpretation Introduction

Interpretation: The mass of $KOH$ present in the solution if $45.3 mL$ of $0.108 M$

$HCl$ solution is required to neutralize the $KOH$ solution.

Concept introduction:
The mass of $KOH$ is calculated by the formula,

$n=mMm=(n×M)$

Here,
$n$ is the number of moles of $KOH$ .
$m$ is the given mass of $KOH$ .
$M$ is the molar mass of $KOH$ .

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Answer 2

Question

Chapter 13, Problem 43E

(a)

Interpretation Introduction

Interpretation: The volume of $0.115 M HClO4$ required to neutralize $50 mL$ of $0.0875 M$

$NaOH$ .

Concept introduction:
The volume of $HClO4$ required for neutralization is calculated by the formula,

$M1V1=M2V2V1=M2V2M1$

Here,
$M1$ is the molarity of $HClO4$ .
$M2$ is the molarity of $NaOH$ .
$V1$ is the volume of $HClO4$ .
$V2$ is the volume of $NaOH$ .

(b)

Interpretation Introduction

Interpretation: The volume of $0.128 M$

$HCl$ required to neutralize $2.87 g Mg(OH)2$ .

Concept introduction:

The number of moles is defined as the ratio of given mass of the compound in grams to the molar mass of the compound.
And,
The volume is calculated by the formula,

$M=n of HClV( mL)×1000 mLV(mL)=nM×1000 mL$

(c)

Interpretation Introduction

Interpretation: The molarity of $Cl−$ solution.

Concept introduction:
The molarity is defined as the ratio of number of moles of solute to the volume of the solution in liters.
$Molarity=Moles of soluteVolume of solution$

(d)

Interpretation Introduction

Interpretation: The mass of $KOH$ present in the solution if $45.3 mL$ of $0.108 M$

$HCl$ solution is required to neutralize the $KOH$ solution.

Concept introduction:
The mass of $KOH$ is calculated by the formula,

$n=mMm=(n×M)$

Here,
$n$ is the number of moles of $KOH$ .
$m$ is the given mass of $KOH$ .
$M$ is the molar mass of $KOH$ .

Expert Solution & Answer

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Check out a sample textbook solution

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Answer 3

Question

Chapter 13, Problem 43E

(a)

Interpretation Introduction

Interpretation: The volume of $0.115 M HClO4$ required to neutralize $50 mL$ of $0.0875 M$

$NaOH$ .

Concept introduction:
The volume of $HClO4$ required for neutralization is calculated by the formula,

$M1V1=M2V2V1=M2V2M1$

Here,
$M1$ is the molarity of $HClO4$ .
$M2$ is the molarity of $NaOH$ .
$V1$ is the volume of $HClO4$ .
$V2$ is the volume of $NaOH$ .

(b)

Interpretation Introduction

Interpretation: The volume of $0.128 M$

$HCl$ required to neutralize $2.87 g Mg(OH)2$ .

Concept introduction:

The number of moles is defined as the ratio of given mass of the compound in grams to the molar mass of the compound.
And,
The volume is calculated by the formula,

$M=n of HClV( mL)×1000 mLV(mL)=nM×1000 mL$

(c)

Interpretation Introduction

Interpretation: The molarity of $Cl−$ solution.

Concept introduction:
The molarity is defined as the ratio of number of moles of solute to the volume of the solution in liters.
$Molarity=Moles of soluteVolume of solution$

(d)

Interpretation Introduction

Interpretation: The mass of $KOH$ present in the solution if $45.3 mL$ of $0.108 M$

$HCl$ solution is required to neutralize the $KOH$ solution.

Concept introduction:
The mass of $KOH$ is calculated by the formula,

$n=mMm=(n×M)$

Here,
$n$ is the number of moles of $KOH$ .
$m$ is the given mass of $KOH$ .
$M$ is the molar mass of $KOH$ .

Expert Solution & Answer

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Answer 4

Question

Chapter 13, Problem 43E

(a)

Interpretation Introduction

Interpretation: The volume of $0.115 M HClO4$ required to neutralize $50 mL$ of $0.0875 M$

$NaOH$ .

Concept introduction:
The volume of $HClO4$ required for neutralization is calculated by the formula,

$M1V1=M2V2V1=M2V2M1$

Here,
$M1$ is the molarity of $HClO4$ .
$M2$ is the molarity of $NaOH$ .
$V1$ is the volume of $HClO4$ .
$V2$ is the volume of $NaOH$ .

(b)

Interpretation Introduction

Interpretation: The volume of $0.128 M$

$HCl$ required to neutralize $2.87 g Mg(OH)2$ .

Concept introduction:

The number of moles is defined as the ratio of given mass of the compound in grams to the molar mass of the compound.
And,
The volume is calculated by the formula,

$M=n of HClV( mL)×1000 mLV(mL)=nM×1000 mL$

(c)

Interpretation Introduction

Interpretation: The molarity of $Cl−$ solution.

Concept introduction:
The molarity is defined as the ratio of number of moles of solute to the volume of the solution in liters.
$Molarity=Moles of soluteVolume of solution$

(d)

Interpretation Introduction

Interpretation: The mass of $KOH$ present in the solution if $45.3 mL$ of $0.108 M$

$HCl$ solution is required to neutralize the $KOH$ solution.

Concept introduction:
The mass of $KOH$ is calculated by the formula,

$n=mMm=(n×M)$

Here,
$n$ is the number of moles of $KOH$ .
$m$ is the given mass of $KOH$ .
$M$ is the molar mass of $KOH$ .

Expert Solution & Answer

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See solution

Answer 5

Chapter 13, Problem 43E

(a)

Interpretation Introduction

Interpretation: The volume of $0.115 M HClO4$ required to neutralize $50 mL$ of $0.0875 M$

$NaOH$ .

Concept introduction:
The volume of $HClO4$ required for neutralization is calculated by the formula,

$M1V1=M2V2V1=M2V2M1$

Here,
$M1$ is the molarity of $HClO4$ .
$M2$ is the molarity of $NaOH$ .
$V1$ is the volume of $HClO4$ .
$V2$ is the volume of $NaOH$ .

(b)

Interpretation Introduction

Interpretation: The volume of $0.128 M$

$HCl$ required to neutralize $2.87 g Mg(OH)2$ .

Concept introduction:

The number of moles is defined as the ratio of given mass of the compound in grams to the molar mass of the compound.
And,
The volume is calculated by the formula,

$M=n of HClV( mL)×1000 mLV(mL)=nM×1000 mL$

(c)

Interpretation Introduction

Interpretation: The molarity of $Cl−$ solution.

Concept introduction:
The molarity is defined as the ratio of number of moles of solute to the volume of the solution in liters.
$Molarity=Moles of soluteVolume of solution$

(d)

Interpretation Introduction

Interpretation: The mass of $KOH$ present in the solution if $45.3 mL$ of $0.108 M$

$HCl$ solution is required to neutralize the $KOH$ solution.

Concept introduction:
The mass of $KOH$ is calculated by the formula,

$n=mMm=(n×M)$

Here,
$n$ is the number of moles of $KOH$ .
$m$ is the given mass of $KOH$ .
$M$ is the molar mass of $KOH$ .

Answer 6

Chapter 13, Problem 43E

(a)

Interpretation Introduction

Interpretation: The volume of $0.115 M HClO4$ required to neutralize $50 mL$ of $0.0875 M$

$NaOH$ .

Concept introduction:
The volume of $HClO4$ required for neutralization is calculated by the formula,

$M1V1=M2V2V1=M2V2M1$

Here,
$M1$ is the molarity of $HClO4$ .
$M2$ is the molarity of $NaOH$ .
$V1$ is the volume of $HClO4$ .
$V2$ is the volume of $NaOH$ .

Answer 7

Chapter 13, Problem 43E

(a)

Interpretation Introduction

Interpretation: The volume of $0.115 M HClO4$ required to neutralize $50 mL$ of $0.0875 M$

$NaOH$ .

Concept introduction:
The volume of $HClO4$ required for neutralization is calculated by the formula,

$M1V1=M2V2V1=M2V2M1$

Here,
$M1$ is the molarity of $HClO4$ .
$M2$ is the molarity of $NaOH$ .
$V1$ is the volume of $HClO4$ .
$V2$ is the volume of $NaOH$ .

Answer 8

(b)

Interpretation Introduction

Interpretation: The volume of $0.128 M$

$HCl$ required to neutralize $2.87 g Mg(OH)2$ .

Concept introduction:

The number of moles is defined as the ratio of given mass of the compound in grams to the molar mass of the compound.
And,
The volume is calculated by the formula,

$M=n of HClV( mL)×1000 mLV(mL)=nM×1000 mL$

Answer 9

(b)

Interpretation Introduction

Interpretation: The volume of $0.128 M$

$HCl$ required to neutralize $2.87 g Mg(OH)2$ .

Concept introduction:

The number of moles is defined as the ratio of given mass of the compound in grams to the molar mass of the compound.
And,
The volume is calculated by the formula,

$M=n of HClV( mL)×1000 mLV(mL)=nM×1000 mL$

Answer 10

(c)

Interpretation Introduction

Interpretation: The molarity of $Cl−$ solution.

Concept introduction:
The molarity is defined as the ratio of number of moles of solute to the volume of the solution in liters.
$Molarity=Moles of soluteVolume of solution$

Answer 11

(c)

Interpretation Introduction

Interpretation: The molarity of $Cl−$ solution.

Concept introduction:
The molarity is defined as the ratio of number of moles of solute to the volume of the solution in liters.
$Molarity=Moles of soluteVolume of solution$

Answer 12

(d)

Interpretation Introduction

Interpretation: The mass of $KOH$ present in the solution if $45.3 mL$ of $0.108 M$

$HCl$ solution is required to neutralize the $KOH$ solution.

Concept introduction:
The mass of $KOH$ is calculated by the formula,

$n=mMm=(n×M)$

Here,
$n$ is the number of moles of $KOH$ .
$m$ is the given mass of $KOH$ .
$M$ is the molar mass of $KOH$ .

Answer 13

(d)

Interpretation Introduction

Interpretation: The mass of $KOH$ present in the solution if $45.3 mL$ of $0.108 M$

$HCl$ solution is required to neutralize the $KOH$ solution.

Concept introduction:
The mass of $KOH$ is calculated by the formula,

$n=mMm=(n×M)$

Here,
$n$ is the number of moles of $KOH$ .
$m$ is the given mass of $KOH$ .
$M$ is the molar mass of $KOH$ .

Answer 14

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Answer 15

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Answer 16

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Answer 17

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