The balanced net ionic equations for the neutralization of equal molar amounts of the given acids and bases has to be determined. After neutralization, it is to be determine whether the pH in each combination of the given acid and base is greater than, or equal to or, less than 7. Concept Introduction: Any strong electrolyte can dissociate completely into its ions but for weak electrolyte only certain number of ions is dissociated. Hence, pH depends on the dissociated ions.

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Answer 1

Question

Chapter 16, Problem 16.1P

Interpretation Introduction

Interpretation:

The balanced net ionic equations for the neutralization of equal molar amounts of the given acids and bases has to be determined. After neutralization, it is to be determine whether the pH in each combination of the given acid and base is greater than, or equal to or, less than 7.

Concept Introduction:

Any strong electrolyte can dissociate completely into its ions but for weak electrolyte only certain number of ions is dissociated. Hence, pH depends on the dissociated ions.

Expert Solution & Answer

Answer to Problem 16.1P

Solution:

(a) $HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$ , pH >7

(b) $H+(aq)+NH3(aq)⇄NH4+(aq)$ , pH<7

(c) $OH−(aq)+H+(aq)⇄H2O(l)$ , pH=7

Explanation of Solution

(a) $HNO2$ and $KOH$

Balanced Molecular equation:

$HNO2(aq)+KOH(aq)⇄KNO2(aq)+H2O(l)$

Total ionic equation:

$HNO2(aq)+K+(aq)+OH−(aq)⇄K+(aq)+NO2−(aq)+H2O(l)$

Net ionic equation:

$HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$

$HNO2$ is a weak acid and $KOH$ is a strong base. So at equivalence point pH is greater than 7.

(b) $HBr$ and $NH3$

Balanced Molecular equation:

$HBr(aq)+NH3(aq)⇄NH4Br(aq)$

Total ionic equation:

$H+(aq)+Br−(aq)+NH3(aq)⇄NH4+(aq)+Br−(aq)$

Net ionic equation:

$H+(aq)+NH3(aq)⇄NH4+(aq)$

$HBr$ is a strong acid and $NH3$ is a weak base. So, at equivalence point pH is less than 7.

(c) $KOH$ and $HClO4$

Balanced Molecular equation:

$HClO4(aq)+KOH(aq)⇄KClO4(aq)+H2O(l)$

Total ionic equation:

$K+(aq)+OH−(aq)+H+(aq)+ClO4−(aq)⇄K+(aq)+ClO4−(aq)+H2O(l)$

Net ionic equation:

$OH−(aq)+H+(aq)⇄H2O(l)$

$HClO4$ is a strong acid and $KOH$ is a strong base. So, at equivalence point pH is equal to 7.

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Answer 2

Question

Chapter 16, Problem 16.1P

Interpretation Introduction

Interpretation:

The balanced net ionic equations for the neutralization of equal molar amounts of the given acids and bases has to be determined. After neutralization, it is to be determine whether the pH in each combination of the given acid and base is greater than, or equal to or, less than 7.

Concept Introduction:

Any strong electrolyte can dissociate completely into its ions but for weak electrolyte only certain number of ions is dissociated. Hence, pH depends on the dissociated ions.

Expert Solution & Answer

Answer to Problem 16.1P

Solution:

(a) $HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$ , pH >7

(b) $H+(aq)+NH3(aq)⇄NH4+(aq)$ , pH<7

(c) $OH−(aq)+H+(aq)⇄H2O(l)$ , pH=7

Explanation of Solution

(a) $HNO2$ and $KOH$

Balanced Molecular equation:

$HNO2(aq)+KOH(aq)⇄KNO2(aq)+H2O(l)$

Total ionic equation:

$HNO2(aq)+K+(aq)+OH−(aq)⇄K+(aq)+NO2−(aq)+H2O(l)$

Net ionic equation:

$HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$

$HNO2$ is a weak acid and $KOH$ is a strong base. So at equivalence point pH is greater than 7.

(b) $HBr$ and $NH3$

Balanced Molecular equation:

$HBr(aq)+NH3(aq)⇄NH4Br(aq)$

Total ionic equation:

$H+(aq)+Br−(aq)+NH3(aq)⇄NH4+(aq)+Br−(aq)$

Net ionic equation:

$H+(aq)+NH3(aq)⇄NH4+(aq)$

$HBr$ is a strong acid and $NH3$ is a weak base. So, at equivalence point pH is less than 7.

(c) $KOH$ and $HClO4$

Balanced Molecular equation:

$HClO4(aq)+KOH(aq)⇄KClO4(aq)+H2O(l)$

Total ionic equation:

$K+(aq)+OH−(aq)+H+(aq)+ClO4−(aq)⇄K+(aq)+ClO4−(aq)+H2O(l)$

Net ionic equation:

$OH−(aq)+H+(aq)⇄H2O(l)$

$HClO4$ is a strong acid and $KOH$ is a strong base. So, at equivalence point pH is equal to 7.

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Answer 3

Question

Chapter 16, Problem 16.1P

Interpretation Introduction

Interpretation:

The balanced net ionic equations for the neutralization of equal molar amounts of the given acids and bases has to be determined. After neutralization, it is to be determine whether the pH in each combination of the given acid and base is greater than, or equal to or, less than 7.

Concept Introduction:

Any strong electrolyte can dissociate completely into its ions but for weak electrolyte only certain number of ions is dissociated. Hence, pH depends on the dissociated ions.

Expert Solution & Answer

Answer to Problem 16.1P

Solution:

(a) $HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$ , pH >7

(b) $H+(aq)+NH3(aq)⇄NH4+(aq)$ , pH<7

(c) $OH−(aq)+H+(aq)⇄H2O(l)$ , pH=7

Explanation of Solution

(a) $HNO2$ and $KOH$

Balanced Molecular equation:

$HNO2(aq)+KOH(aq)⇄KNO2(aq)+H2O(l)$

Total ionic equation:

$HNO2(aq)+K+(aq)+OH−(aq)⇄K+(aq)+NO2−(aq)+H2O(l)$

Net ionic equation:

$HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$

$HNO2$ is a weak acid and $KOH$ is a strong base. So at equivalence point pH is greater than 7.

(b) $HBr$ and $NH3$

Balanced Molecular equation:

$HBr(aq)+NH3(aq)⇄NH4Br(aq)$

Total ionic equation:

$H+(aq)+Br−(aq)+NH3(aq)⇄NH4+(aq)+Br−(aq)$

Net ionic equation:

$H+(aq)+NH3(aq)⇄NH4+(aq)$

$HBr$ is a strong acid and $NH3$ is a weak base. So, at equivalence point pH is less than 7.

(c) $KOH$ and $HClO4$

Balanced Molecular equation:

$HClO4(aq)+KOH(aq)⇄KClO4(aq)+H2O(l)$

Total ionic equation:

$K+(aq)+OH−(aq)+H+(aq)+ClO4−(aq)⇄K+(aq)+ClO4−(aq)+H2O(l)$

Net ionic equation:

$OH−(aq)+H+(aq)⇄H2O(l)$

$HClO4$ is a strong acid and $KOH$ is a strong base. So, at equivalence point pH is equal to 7.

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Answer 4

Question

Chapter 16, Problem 16.1P

Interpretation Introduction

Interpretation:

The balanced net ionic equations for the neutralization of equal molar amounts of the given acids and bases has to be determined. After neutralization, it is to be determine whether the pH in each combination of the given acid and base is greater than, or equal to or, less than 7.

Concept Introduction:

Any strong electrolyte can dissociate completely into its ions but for weak electrolyte only certain number of ions is dissociated. Hence, pH depends on the dissociated ions.

Expert Solution & Answer

Answer to Problem 16.1P

Solution:

(a) $HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$ , pH >7

(b) $H+(aq)+NH3(aq)⇄NH4+(aq)$ , pH<7

(c) $OH−(aq)+H+(aq)⇄H2O(l)$ , pH=7

Explanation of Solution

(a) $HNO2$ and $KOH$

Balanced Molecular equation:

$HNO2(aq)+KOH(aq)⇄KNO2(aq)+H2O(l)$

Total ionic equation:

$HNO2(aq)+K+(aq)+OH−(aq)⇄K+(aq)+NO2−(aq)+H2O(l)$

Net ionic equation:

$HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$

$HNO2$ is a weak acid and $KOH$ is a strong base. So at equivalence point pH is greater than 7.

(b) $HBr$ and $NH3$

Balanced Molecular equation:

$HBr(aq)+NH3(aq)⇄NH4Br(aq)$

Total ionic equation:

$H+(aq)+Br−(aq)+NH3(aq)⇄NH4+(aq)+Br−(aq)$

Net ionic equation:

$H+(aq)+NH3(aq)⇄NH4+(aq)$

$HBr$ is a strong acid and $NH3$ is a weak base. So, at equivalence point pH is less than 7.

(c) $KOH$ and $HClO4$

Balanced Molecular equation:

$HClO4(aq)+KOH(aq)⇄KClO4(aq)+H2O(l)$

Total ionic equation:

$K+(aq)+OH−(aq)+H+(aq)+ClO4−(aq)⇄K+(aq)+ClO4−(aq)+H2O(l)$

Net ionic equation:

$OH−(aq)+H+(aq)⇄H2O(l)$

$HClO4$ is a strong acid and $KOH$ is a strong base. So, at equivalence point pH is equal to 7.

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Answer 5

Chapter 16, Problem 16.1P

Interpretation Introduction

Interpretation:

The balanced net ionic equations for the neutralization of equal molar amounts of the given acids and bases has to be determined. After neutralization, it is to be determine whether the pH in each combination of the given acid and base is greater than, or equal to or, less than 7.

Concept Introduction:

Any strong electrolyte can dissociate completely into its ions but for weak electrolyte only certain number of ions is dissociated. Hence, pH depends on the dissociated ions.

Expert Solution & Answer

Answer to Problem 16.1P

Solution:

(a) $HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$ , pH >7

(b) $H+(aq)+NH3(aq)⇄NH4+(aq)$ , pH<7

(c) $OH−(aq)+H+(aq)⇄H2O(l)$ , pH=7

Explanation of Solution

(a) $HNO2$ and $KOH$

Balanced Molecular equation:

$HNO2(aq)+KOH(aq)⇄KNO2(aq)+H2O(l)$

Total ionic equation:

$HNO2(aq)+K+(aq)+OH−(aq)⇄K+(aq)+NO2−(aq)+H2O(l)$

Net ionic equation:

$HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$

$HNO2$ is a weak acid and $KOH$ is a strong base. So at equivalence point pH is greater than 7.

(b) $HBr$ and $NH3$

Balanced Molecular equation:

$HBr(aq)+NH3(aq)⇄NH4Br(aq)$

Total ionic equation:

$H+(aq)+Br−(aq)+NH3(aq)⇄NH4+(aq)+Br−(aq)$

Net ionic equation:

$H+(aq)+NH3(aq)⇄NH4+(aq)$

$HBr$ is a strong acid and $NH3$ is a weak base. So, at equivalence point pH is less than 7.

(c) $KOH$ and $HClO4$

Balanced Molecular equation:

$HClO4(aq)+KOH(aq)⇄KClO4(aq)+H2O(l)$

Total ionic equation:

$K+(aq)+OH−(aq)+H+(aq)+ClO4−(aq)⇄K+(aq)+ClO4−(aq)+H2O(l)$

Net ionic equation:

$OH−(aq)+H+(aq)⇄H2O(l)$

$HClO4$ is a strong acid and $KOH$ is a strong base. So, at equivalence point pH is equal to 7.

Answer 6

Chapter 16, Problem 16.1P

Interpretation Introduction

Interpretation:

The balanced net ionic equations for the neutralization of equal molar amounts of the given acids and bases has to be determined. After neutralization, it is to be determine whether the pH in each combination of the given acid and base is greater than, or equal to or, less than 7.

Concept Introduction:

Any strong electrolyte can dissociate completely into its ions but for weak electrolyte only certain number of ions is dissociated. Hence, pH depends on the dissociated ions.

Expert Solution & Answer

Answer to Problem 16.1P

Solution:

(a) $HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$ , pH >7

(b) $H+(aq)+NH3(aq)⇄NH4+(aq)$ , pH<7

(c) $OH−(aq)+H+(aq)⇄H2O(l)$ , pH=7

Explanation of Solution

(a) $HNO2$ and $KOH$

Balanced Molecular equation:

$HNO2(aq)+KOH(aq)⇄KNO2(aq)+H2O(l)$

Total ionic equation:

$HNO2(aq)+K+(aq)+OH−(aq)⇄K+(aq)+NO2−(aq)+H2O(l)$

Net ionic equation:

$HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$

$HNO2$ is a weak acid and $KOH$ is a strong base. So at equivalence point pH is greater than 7.

(b) $HBr$ and $NH3$

Balanced Molecular equation:

$HBr(aq)+NH3(aq)⇄NH4Br(aq)$

Total ionic equation:

$H+(aq)+Br−(aq)+NH3(aq)⇄NH4+(aq)+Br−(aq)$

Net ionic equation:

$H+(aq)+NH3(aq)⇄NH4+(aq)$

$HBr$ is a strong acid and $NH3$ is a weak base. So, at equivalence point pH is less than 7.

(c) $KOH$ and $HClO4$

Balanced Molecular equation:

$HClO4(aq)+KOH(aq)⇄KClO4(aq)+H2O(l)$

Total ionic equation:

$K+(aq)+OH−(aq)+H+(aq)+ClO4−(aq)⇄K+(aq)+ClO4−(aq)+H2O(l)$

Net ionic equation:

$OH−(aq)+H+(aq)⇄H2O(l)$

$HClO4$ is a strong acid and $KOH$ is a strong base. So, at equivalence point pH is equal to 7.

Answer 7

Chapter 16, Problem 16.1P

Interpretation Introduction

Interpretation:

The balanced net ionic equations for the neutralization of equal molar amounts of the given acids and bases has to be determined. After neutralization, it is to be determine whether the pH in each combination of the given acid and base is greater than, or equal to or, less than 7.

Concept Introduction:

Any strong electrolyte can dissociate completely into its ions but for weak electrolyte only certain number of ions is dissociated. Hence, pH depends on the dissociated ions.

Answer 8

Expert Solution & Answer

Answer to Problem 16.1P

Solution:

(a) $HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$ , pH >7

(b) $H+(aq)+NH3(aq)⇄NH4+(aq)$ , pH<7

(c) $OH−(aq)+H+(aq)⇄H2O(l)$ , pH=7

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

(a) $HNO2$ and $KOH$

Balanced Molecular equation:

$HNO2(aq)+KOH(aq)⇄KNO2(aq)+H2O(l)$

Total ionic equation:

$HNO2(aq)+K+(aq)+OH−(aq)⇄K+(aq)+NO2−(aq)+H2O(l)$

Net ionic equation:

$HNO2(aq)+OH−(aq)⇄NO2−(aq)+H2O(l)$

$HNO2$ is a weak acid and $KOH$ is a strong base. So at equivalence point pH is greater than 7.

(b) $HBr$ and $NH3$

Balanced Molecular equation:

$HBr(aq)+NH3(aq)⇄NH4Br(aq)$

Total ionic equation:

$H+(aq)+Br−(aq)+NH3(aq)⇄NH4+(aq)+Br−(aq)$

Net ionic equation:

$H+(aq)+NH3(aq)⇄NH4+(aq)$

$HBr$ is a strong acid and $NH3$ is a weak base. So, at equivalence point pH is less than 7.

(c) $KOH$ and $HClO4$

Balanced Molecular equation:

$HClO4(aq)+KOH(aq)⇄KClO4(aq)+H2O(l)$

Total ionic equation:

$K+(aq)+OH−(aq)+H+(aq)+ClO4−(aq)⇄K+(aq)+ClO4−(aq)+H2O(l)$

Net ionic equation:

$OH−(aq)+H+(aq)⇄H2O(l)$

$HClO4$ is a strong acid and $KOH$ is a strong base. So, at equivalence point pH is equal to 7.

Answer 12

Ch. 16 - Prob. 16.1P Ch. 16 - APPLY 16.2 Write balanced net ionic equations for...Ch. 16 - PRACTICE 16.3 Calculate the concentrations of all...Ch. 16 - APPLY 16.4 Calculate the pH of a solution prepared...Ch. 16 - Conceptual PRACTICE 16.5 The following pictures...Ch. 16 - Conceptual APPLY 16.6 The following pictures...Ch. 16 - Prob. 16.7P Ch. 16 - Prob. 16.8A Ch. 16 - Prob. 16.9P Ch. 16 - PRACTICE 16.10 Use the Henderson-Hasselbalch...

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