(a) Interpretation: The molarity of a solution of HClO 4 of amount 25.00mL is to be determined by the student. The student is also using 0.731M KOH. He forgets to add the indicator in solution after the addition of KOH of volume 42.35 mL into the solution. At this condition, he takes the pH of the solution which is 12.39. It has to be determined if the student go beyond equivalence point or not. Concept introduction: The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration. It can also be known as volumetric titration.

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Answer 1

Question

Chapter 14, Problem 80QAP

Interpretation Introduction

(a)

Interpretation:

The molarity of a solution of HClO₄ of amount 25.00mL is to be determined by the student. The student is also using 0.731M KOH. He forgets to add the indicator in solution after the addition of KOH of volume 42.35 mL into the solution. At this condition, he takes the pH of the solution which is 12.39. It has to be determined if the student go beyond equivalence point or not.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration. It can also be known as volumetric titration.

Interpretation Introduction

(b)

Interpretation:

The molarity of strong acid is to be determined.

Concept introduction:

The molarity can be calculated as follows:

molarity=number of moles of solute volume of solution(L)

The relation between pH and pOH is as follows:

pH+pOH=14

The pOH can be calculated as follows:

pOH=−log10[OH−]

Interpretation Introduction

(c)

Interpretation:

The volume of KOH is to be determined which is added in excess if he added. Or the volume of KOH is to be determined which is required to add if he didn’t add.

Concept introduction:

The formula which will be used is −

M1V1=M2V2

Here, M is molarity and V is volume.

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Students have asked these similar questions

(a) A 50.0 mL solution is prepared to be 1.29 M acetylsalicylic acid. In the first step, 8.55 mL of NaOH is titrated into the solution until the pH is exactly 5.0. What is the concentration of the titrant (NaOH)? (b) In the second step, enough 5.85 M nitric acid is added to the solution after the titration in part (a) is complete until the pH is one unit lower than the pKa of acetylsalicylic acid. What volume (mL) of nitric acid was added?

The major component of vinegar is acetic acid, CH3COOH. Its Ka is 1.8 × 10-5 . One student used 1.000 M NaOH to titrate 25.00 mL vinegar. At the end point, 21.82 mL NaOH was used. (a) What is the concentration of CH3COOH in vinegar? (b) What is the pH of the solution at the end point? (c) What indicator(s) the student should use in this titration? Explain

A 0.018 M solution of salicylic acid, HOC6H4CO2H, has the same pH as 0.0038 M HNO3solution. (a) Write an equation for the ionization of salicylic acid in aqueous solution. (Assume only the –CO2H portion of the molecule ionizes.) (b) What is the pH of solution containing 0.018 M salicylic acid? (c) Calculate the Ka of salicylic acid.

Answer 2

Question

Chapter 14, Problem 80QAP

Interpretation Introduction

(a)

Interpretation:

The molarity of a solution of HClO₄ of amount 25.00mL is to be determined by the student. The student is also using 0.731M KOH. He forgets to add the indicator in solution after the addition of KOH of volume 42.35 mL into the solution. At this condition, he takes the pH of the solution which is 12.39. It has to be determined if the student go beyond equivalence point or not.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration. It can also be known as volumetric titration.

Interpretation Introduction

(b)

Interpretation:

The molarity of strong acid is to be determined.

Concept introduction:

The molarity can be calculated as follows:

molarity=number of moles of solute volume of solution(L)

The relation between pH and pOH is as follows:

pH+pOH=14

The pOH can be calculated as follows:

pOH=−log10[OH−]

Interpretation Introduction

(c)

Interpretation:

The volume of KOH is to be determined which is added in excess if he added. Or the volume of KOH is to be determined which is required to add if he didn’t add.

Concept introduction:

The formula which will be used is −

M1V1=M2V2

Here, M is molarity and V is volume.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 14, Problem 80QAP

Interpretation Introduction

(a)

Interpretation:

The molarity of a solution of HClO₄ of amount 25.00mL is to be determined by the student. The student is also using 0.731M KOH. He forgets to add the indicator in solution after the addition of KOH of volume 42.35 mL into the solution. At this condition, he takes the pH of the solution which is 12.39. It has to be determined if the student go beyond equivalence point or not.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration. It can also be known as volumetric titration.

Interpretation Introduction

(b)

Interpretation:

The molarity of strong acid is to be determined.

Concept introduction:

The molarity can be calculated as follows:

molarity=number of moles of solute volume of solution(L)

The relation between pH and pOH is as follows:

pH+pOH=14

The pOH can be calculated as follows:

pOH=−log10[OH−]

Interpretation Introduction

(c)

Interpretation:

The volume of KOH is to be determined which is added in excess if he added. Or the volume of KOH is to be determined which is required to add if he didn’t add.

Concept introduction:

The formula which will be used is −

M1V1=M2V2

Here, M is molarity and V is volume.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 14, Problem 80QAP

Interpretation Introduction

(a)

Interpretation:

The molarity of a solution of HClO₄ of amount 25.00mL is to be determined by the student. The student is also using 0.731M KOH. He forgets to add the indicator in solution after the addition of KOH of volume 42.35 mL into the solution. At this condition, he takes the pH of the solution which is 12.39. It has to be determined if the student go beyond equivalence point or not.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration. It can also be known as volumetric titration.

Interpretation Introduction

(b)

Interpretation:

The molarity of strong acid is to be determined.

Concept introduction:

The molarity can be calculated as follows:

molarity=number of moles of solute volume of solution(L)

The relation between pH and pOH is as follows:

pH+pOH=14

The pOH can be calculated as follows:

pOH=−log10[OH−]

Interpretation Introduction

(c)

Interpretation:

The volume of KOH is to be determined which is added in excess if he added. Or the volume of KOH is to be determined which is required to add if he didn’t add.

Concept introduction:

The formula which will be used is −

M1V1=M2V2

Here, M is molarity and V is volume.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 14, Problem 80QAP

Interpretation Introduction

(a)

Interpretation:

The molarity of a solution of HClO₄ of amount 25.00mL is to be determined by the student. The student is also using 0.731M KOH. He forgets to add the indicator in solution after the addition of KOH of volume 42.35 mL into the solution. At this condition, he takes the pH of the solution which is 12.39. It has to be determined if the student go beyond equivalence point or not.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration. It can also be known as volumetric titration.

Interpretation Introduction

(b)

Interpretation:

The molarity of strong acid is to be determined.

Concept introduction:

The molarity can be calculated as follows:

molarity=number of moles of solute volume of solution(L)

The relation between pH and pOH is as follows:

pH+pOH=14

The pOH can be calculated as follows:

pOH=−log10[OH−]

Interpretation Introduction

(c)

Interpretation:

The volume of KOH is to be determined which is added in excess if he added. Or the volume of KOH is to be determined which is required to add if he didn’t add.

Concept introduction:

The formula which will be used is −

M1V1=M2V2

Here, M is molarity and V is volume.

Answer 6

Chapter 14, Problem 80QAP

Interpretation Introduction

(a)

Interpretation:

The molarity of a solution of HClO₄ of amount 25.00mL is to be determined by the student. The student is also using 0.731M KOH. He forgets to add the indicator in solution after the addition of KOH of volume 42.35 mL into the solution. At this condition, he takes the pH of the solution which is 12.39. It has to be determined if the student go beyond equivalence point or not.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration. It can also be known as volumetric titration.

Answer 7

Chapter 14, Problem 80QAP

Interpretation Introduction

(a)

Interpretation:

The molarity of a solution of HClO₄ of amount 25.00mL is to be determined by the student. The student is also using 0.731M KOH. He forgets to add the indicator in solution after the addition of KOH of volume 42.35 mL into the solution. At this condition, he takes the pH of the solution which is 12.39. It has to be determined if the student go beyond equivalence point or not.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration. It can also be known as volumetric titration.

Answer 8

Interpretation Introduction

(b)

Interpretation:

The molarity of strong acid is to be determined.

Concept introduction:

The molarity can be calculated as follows:

molarity=number of moles of solute volume of solution(L)

The relation between pH and pOH is as follows:

pH+pOH=14

The pOH can be calculated as follows:

pOH=−log10[OH−]

Answer 9

Interpretation Introduction

(b)

Interpretation:

The molarity of strong acid is to be determined.

Concept introduction:

The molarity can be calculated as follows:

molarity=number of moles of solute volume of solution(L)

The relation between pH and pOH is as follows:

pH+pOH=14

The pOH can be calculated as follows:

pOH=−log10[OH−]

Answer 10

Interpretation Introduction

(c)

Interpretation:

The volume of KOH is to be determined which is added in excess if he added. Or the volume of KOH is to be determined which is required to add if he didn’t add.

Concept introduction:

The formula which will be used is −

M1V1=M2V2

Here, M is molarity and V is volume.

Answer 11

Interpretation Introduction

(c)

Interpretation:

The volume of KOH is to be determined which is added in excess if he added. Or the volume of KOH is to be determined which is required to add if he didn’t add.

Concept introduction:

The formula which will be used is −

M1V1=M2V2

Here, M is molarity and V is volume.

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

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Answer 15

Ch. 14 - Write a net ionic equation for the reaction...Ch. 14 - Write a net ionic equation for the reaction...Ch. 14 - Write a balanced net ionic equation for the...Ch. 14 - Write a balanced net ionic equation for the...Ch. 14 - Calculate K for the reactions in Question 1.Ch. 14 - Calculate K for the reactions in Question 2.Ch. 14 - Prob. 7QAP Ch. 14 - Calculate K for the reactions in Question 4.Ch. 14 - Calculate [H+] and pH in a solution in which...Ch. 14 - Calculate [OH-] and pH in a solution in which the...

Solution Summary: The author explains how the molarity of HClO 4 is to be determined by the student, who forgets to add the indicator in the solution.

Concept explainers

Videos

(c)

Interpretation:

The volume of KOH is to be determined which is added in excess if he added. Or the volume of KOH is to be determined which is required to add if he didn’t add.

Concept introduction:

The formula which will be used is −

M1V1=M2V2

Here, M is molarity and V is volume.

Want to see the full answer?

Chapter 14 Solutions

Solution Summary: The author explains how the molarity of HClO 4 is to be determined by the student, who forgets to add the indicator in the solution.

Concept explainers

Videos

(c) Interpretation: The volume of KOH is to be determined which is added in excess if he added. Or the volume of KOH is to be determined which is required to add if he didn’t add. Concept introduction: The formula which will be used is − M1V1=M2V2 Here, M is molarity and V is volume.

Want to see the full answer?

Chapter 14 Solutions

(c)

Interpretation:

The volume of KOH is to be determined which is added in excess if he added. Or the volume of KOH is to be determined which is required to add if he didn’t add.

Concept introduction:

The formula which will be used is −

M1V1=M2V2

Here, M is molarity and V is volume.