The pH of 0.12 M solution of KNO 2 needs to be determined. Concept Introduction: A base is the substance that gives OH - ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium. BOH (aq) ⇌ OH - (aq) +B - (aq) An acid is the substance that gives H + ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium. HA (aq) + H 2 O (l) ⇌ H 3 O + (aq) + A - (aq)

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Answer 1

Question

Chapter 7, Problem 109E

(a)

Interpretation Introduction

Interpretation: The pH of 0.12 M solution of $KNO2$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(a)

Expert Solution

Answer to Problem 109E

$pH = 8.24$

Explanation of Solution

[ $KNO2$ ] = 0.12 M

$KNO2$ is a basic salt. The ionization of $KNO2$ forms respective ions:

$KNO2→K++NO2-$

Here $NO2-$ can act as string conjugate base of weak acid $HNO2$ therefore the salt is slightly basic in nature as $K+$ is neither acidic nor basic.

The reaction of $NO2-$ ions with water can be written as:

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)$

Initial concentration of $NO2-$ = 0.12 M

$Kb$ for $NO2-$ = $2.5× 10-11$

Make ICE table:

Concentration	$NO2-$	$OH-$	$HNO2$
Initial	0.12	0	0
Change	-x	x	x
Equilibrium	0.12−x	x	x

Calculate $H3O+$ :

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)Kb= [OH- ][HNO2] [NO2-]2.5× 10-11 = [x] [x][0.12-x](0.12>>x)2.5× 10-11 = [x] [x][0.12][x]2 =2.5× 10-11 ×0.12[x]= 1.73×10-6 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[1.73×10-6 ]pOH = 5.76pH = 14 - pOH = 14- 5.76 = 8.24$

(b)

Interpretation Introduction

Interpretation: The pH of 0.45 M solution of $NaOCl$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(b)

Expert Solution

Answer to Problem 109E

$pH = 10.5$

Explanation of Solution

[ $NaOCl$ ] = 0.45 M

$NaOCl$ is a basic salt. The ionization of $NaOCl$ forms respective ions:

$NaOCl→Na++OCl-$

Here $OCl-$ can act as strong conjugate base of weak acid $HClO$ therefore the salt is slightly basic in nature as $Na+$ is neither acidic nor basic.

The reaction of $OCl-$ ions with water can be written as:

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)$

Initial concentration of $OCl-$ = 0.45 M

$Kb$ for $OCl-$ = $2.8× 10-7$

Make ICE table:

Concentration	$OCl-$	$OH-$	$HClO$
Initial	0.45	0	0
Change	-x	x	x
Equilibrium	0.45−x	x	x

Calculate $H3O+$ :

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)Kb= [OH-][HClO] [OCl-]2.8× 10-7 = [x] [x][0.45-x](0.45>>x)2.8× 10-7 = [x] [x][0.45][x]2 =2.8× 10-7 ×0.45[x]= 3.55×10-4 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[3.55×10-4 ]pOH = 3.45pH = 14 - pOH = 14- 3.45 = 10.5$

(c)

Interpretation Introduction

Interpretation: The pH of 0.40 M solution of $NH4ClO4$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(c)

Expert Solution

Answer to Problem 109E

$pH = 4.8$

Explanation of Solution

[ $NH4ClO4$ ] = 0.40 M

$NH4ClO4$ is aacidic salt. The ionization of $NH4ClO4$ forms respective ions:

$NH4ClO4→NH4++ClO4-$

Here $ClO4-$ can act as weak conjugate base of strong acid $HClO4$ therefore the salt is slightly acidic in nature as $NH4+$ is strong conjugate acid of weak base NH₃.

The reaction of $NH4+$ ions with water can be written as:

$NH4+(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)$

Initial concentration of $OCl-$ = 0.40 M

$Ka$ for $NH4+$ = $5.6× 10-10$

Make ICE table:

Concentration	$NH4+$	$NH3$	$H3O+$
Initial	0.40	0	0
Change	-x	x	x
Equilibrium	0.40−x	x	x

Calculate $H3O+$ :

$NH4 +(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)Ka= [NH3 ][H3O+] [NH4 +]5.6× 10-10 = [x] [x][0.40-x](0.40>>x)5.6× 10-10 = [x] [x][0.40][x]2 =5.6× 10-10×0.40[x]= 1.49×10-5 =[H3O+]$

Calculate pH:

$pH = -log[H3O+] pH = -log[1.49×10-5]pH = 4.8$

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Answer 2

Question

Chapter 7, Problem 109E

(a)

Interpretation Introduction

Interpretation: The pH of 0.12 M solution of $KNO2$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(a)

Expert Solution

Answer to Problem 109E

$pH = 8.24$

Explanation of Solution

[ $KNO2$ ] = 0.12 M

$KNO2$ is a basic salt. The ionization of $KNO2$ forms respective ions:

$KNO2→K++NO2-$

Here $NO2-$ can act as string conjugate base of weak acid $HNO2$ therefore the salt is slightly basic in nature as $K+$ is neither acidic nor basic.

The reaction of $NO2-$ ions with water can be written as:

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)$

Initial concentration of $NO2-$ = 0.12 M

$Kb$ for $NO2-$ = $2.5× 10-11$

Make ICE table:

Concentration	$NO2-$	$OH-$	$HNO2$
Initial	0.12	0	0
Change	-x	x	x
Equilibrium	0.12−x	x	x

Calculate $H3O+$ :

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)Kb= [OH- ][HNO2] [NO2-]2.5× 10-11 = [x] [x][0.12-x](0.12>>x)2.5× 10-11 = [x] [x][0.12][x]2 =2.5× 10-11 ×0.12[x]= 1.73×10-6 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[1.73×10-6 ]pOH = 5.76pH = 14 - pOH = 14- 5.76 = 8.24$

(b)

Interpretation Introduction

Interpretation: The pH of 0.45 M solution of $NaOCl$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(b)

Expert Solution

Answer to Problem 109E

$pH = 10.5$

Explanation of Solution

[ $NaOCl$ ] = 0.45 M

$NaOCl$ is a basic salt. The ionization of $NaOCl$ forms respective ions:

$NaOCl→Na++OCl-$

Here $OCl-$ can act as strong conjugate base of weak acid $HClO$ therefore the salt is slightly basic in nature as $Na+$ is neither acidic nor basic.

The reaction of $OCl-$ ions with water can be written as:

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)$

Initial concentration of $OCl-$ = 0.45 M

$Kb$ for $OCl-$ = $2.8× 10-7$

Make ICE table:

Concentration	$OCl-$	$OH-$	$HClO$
Initial	0.45	0	0
Change	-x	x	x
Equilibrium	0.45−x	x	x

Calculate $H3O+$ :

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)Kb= [OH-][HClO] [OCl-]2.8× 10-7 = [x] [x][0.45-x](0.45>>x)2.8× 10-7 = [x] [x][0.45][x]2 =2.8× 10-7 ×0.45[x]= 3.55×10-4 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[3.55×10-4 ]pOH = 3.45pH = 14 - pOH = 14- 3.45 = 10.5$

(c)

Interpretation Introduction

Interpretation: The pH of 0.40 M solution of $NH4ClO4$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(c)

Expert Solution

Answer to Problem 109E

$pH = 4.8$

Explanation of Solution

[ $NH4ClO4$ ] = 0.40 M

$NH4ClO4$ is aacidic salt. The ionization of $NH4ClO4$ forms respective ions:

$NH4ClO4→NH4++ClO4-$

Here $ClO4-$ can act as weak conjugate base of strong acid $HClO4$ therefore the salt is slightly acidic in nature as $NH4+$ is strong conjugate acid of weak base NH₃.

The reaction of $NH4+$ ions with water can be written as:

$NH4+(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)$

Initial concentration of $OCl-$ = 0.40 M

$Ka$ for $NH4+$ = $5.6× 10-10$

Make ICE table:

Concentration	$NH4+$	$NH3$	$H3O+$
Initial	0.40	0	0
Change	-x	x	x
Equilibrium	0.40−x	x	x

Calculate $H3O+$ :

$NH4 +(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)Ka= [NH3 ][H3O+] [NH4 +]5.6× 10-10 = [x] [x][0.40-x](0.40>>x)5.6× 10-10 = [x] [x][0.40][x]2 =5.6× 10-10×0.40[x]= 1.49×10-5 =[H3O+]$

Calculate pH:

$pH = -log[H3O+] pH = -log[1.49×10-5]pH = 4.8$

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Answer 3

Question

Chapter 7, Problem 109E

(a)

Interpretation Introduction

Interpretation: The pH of 0.12 M solution of $KNO2$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(a)

Expert Solution

Answer to Problem 109E

$pH = 8.24$

Explanation of Solution

[ $KNO2$ ] = 0.12 M

$KNO2$ is a basic salt. The ionization of $KNO2$ forms respective ions:

$KNO2→K++NO2-$

Here $NO2-$ can act as string conjugate base of weak acid $HNO2$ therefore the salt is slightly basic in nature as $K+$ is neither acidic nor basic.

The reaction of $NO2-$ ions with water can be written as:

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)$

Initial concentration of $NO2-$ = 0.12 M

$Kb$ for $NO2-$ = $2.5× 10-11$

Make ICE table:

Concentration	$NO2-$	$OH-$	$HNO2$
Initial	0.12	0	0
Change	-x	x	x
Equilibrium	0.12−x	x	x

Calculate $H3O+$ :

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)Kb= [OH- ][HNO2] [NO2-]2.5× 10-11 = [x] [x][0.12-x](0.12>>x)2.5× 10-11 = [x] [x][0.12][x]2 =2.5× 10-11 ×0.12[x]= 1.73×10-6 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[1.73×10-6 ]pOH = 5.76pH = 14 - pOH = 14- 5.76 = 8.24$

(b)

Interpretation Introduction

Interpretation: The pH of 0.45 M solution of $NaOCl$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(b)

Expert Solution

Answer to Problem 109E

$pH = 10.5$

Explanation of Solution

[ $NaOCl$ ] = 0.45 M

$NaOCl$ is a basic salt. The ionization of $NaOCl$ forms respective ions:

$NaOCl→Na++OCl-$

Here $OCl-$ can act as strong conjugate base of weak acid $HClO$ therefore the salt is slightly basic in nature as $Na+$ is neither acidic nor basic.

The reaction of $OCl-$ ions with water can be written as:

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)$

Initial concentration of $OCl-$ = 0.45 M

$Kb$ for $OCl-$ = $2.8× 10-7$

Make ICE table:

Concentration	$OCl-$	$OH-$	$HClO$
Initial	0.45	0	0
Change	-x	x	x
Equilibrium	0.45−x	x	x

Calculate $H3O+$ :

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)Kb= [OH-][HClO] [OCl-]2.8× 10-7 = [x] [x][0.45-x](0.45>>x)2.8× 10-7 = [x] [x][0.45][x]2 =2.8× 10-7 ×0.45[x]= 3.55×10-4 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[3.55×10-4 ]pOH = 3.45pH = 14 - pOH = 14- 3.45 = 10.5$

(c)

Interpretation Introduction

Interpretation: The pH of 0.40 M solution of $NH4ClO4$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(c)

Expert Solution

Answer to Problem 109E

$pH = 4.8$

Explanation of Solution

[ $NH4ClO4$ ] = 0.40 M

$NH4ClO4$ is aacidic salt. The ionization of $NH4ClO4$ forms respective ions:

$NH4ClO4→NH4++ClO4-$

Here $ClO4-$ can act as weak conjugate base of strong acid $HClO4$ therefore the salt is slightly acidic in nature as $NH4+$ is strong conjugate acid of weak base NH₃.

The reaction of $NH4+$ ions with water can be written as:

$NH4+(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)$

Initial concentration of $OCl-$ = 0.40 M

$Ka$ for $NH4+$ = $5.6× 10-10$

Make ICE table:

Concentration	$NH4+$	$NH3$	$H3O+$
Initial	0.40	0	0
Change	-x	x	x
Equilibrium	0.40−x	x	x

Calculate $H3O+$ :

$NH4 +(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)Ka= [NH3 ][H3O+] [NH4 +]5.6× 10-10 = [x] [x][0.40-x](0.40>>x)5.6× 10-10 = [x] [x][0.40][x]2 =5.6× 10-10×0.40[x]= 1.49×10-5 =[H3O+]$

Calculate pH:

$pH = -log[H3O+] pH = -log[1.49×10-5]pH = 4.8$

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Answer 4

Question

Chapter 7, Problem 109E

(a)

Interpretation Introduction

Interpretation: The pH of 0.12 M solution of $KNO2$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(a)

Expert Solution

Answer to Problem 109E

$pH = 8.24$

Explanation of Solution

[ $KNO2$ ] = 0.12 M

$KNO2$ is a basic salt. The ionization of $KNO2$ forms respective ions:

$KNO2→K++NO2-$

Here $NO2-$ can act as string conjugate base of weak acid $HNO2$ therefore the salt is slightly basic in nature as $K+$ is neither acidic nor basic.

The reaction of $NO2-$ ions with water can be written as:

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)$

Initial concentration of $NO2-$ = 0.12 M

$Kb$ for $NO2-$ = $2.5× 10-11$

Make ICE table:

Concentration	$NO2-$	$OH-$	$HNO2$
Initial	0.12	0	0
Change	-x	x	x
Equilibrium	0.12−x	x	x

Calculate $H3O+$ :

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)Kb= [OH- ][HNO2] [NO2-]2.5× 10-11 = [x] [x][0.12-x](0.12>>x)2.5× 10-11 = [x] [x][0.12][x]2 =2.5× 10-11 ×0.12[x]= 1.73×10-6 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[1.73×10-6 ]pOH = 5.76pH = 14 - pOH = 14- 5.76 = 8.24$

(b)

Interpretation Introduction

Interpretation: The pH of 0.45 M solution of $NaOCl$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(b)

Expert Solution

Answer to Problem 109E

$pH = 10.5$

Explanation of Solution

[ $NaOCl$ ] = 0.45 M

$NaOCl$ is a basic salt. The ionization of $NaOCl$ forms respective ions:

$NaOCl→Na++OCl-$

Here $OCl-$ can act as strong conjugate base of weak acid $HClO$ therefore the salt is slightly basic in nature as $Na+$ is neither acidic nor basic.

The reaction of $OCl-$ ions with water can be written as:

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)$

Initial concentration of $OCl-$ = 0.45 M

$Kb$ for $OCl-$ = $2.8× 10-7$

Make ICE table:

Concentration	$OCl-$	$OH-$	$HClO$
Initial	0.45	0	0
Change	-x	x	x
Equilibrium	0.45−x	x	x

Calculate $H3O+$ :

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)Kb= [OH-][HClO] [OCl-]2.8× 10-7 = [x] [x][0.45-x](0.45>>x)2.8× 10-7 = [x] [x][0.45][x]2 =2.8× 10-7 ×0.45[x]= 3.55×10-4 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[3.55×10-4 ]pOH = 3.45pH = 14 - pOH = 14- 3.45 = 10.5$

(c)

Interpretation Introduction

Interpretation: The pH of 0.40 M solution of $NH4ClO4$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(c)

Expert Solution

Answer to Problem 109E

$pH = 4.8$

Explanation of Solution

[ $NH4ClO4$ ] = 0.40 M

$NH4ClO4$ is aacidic salt. The ionization of $NH4ClO4$ forms respective ions:

$NH4ClO4→NH4++ClO4-$

Here $ClO4-$ can act as weak conjugate base of strong acid $HClO4$ therefore the salt is slightly acidic in nature as $NH4+$ is strong conjugate acid of weak base NH₃.

The reaction of $NH4+$ ions with water can be written as:

$NH4+(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)$

Initial concentration of $OCl-$ = 0.40 M

$Ka$ for $NH4+$ = $5.6× 10-10$

Make ICE table:

Concentration	$NH4+$	$NH3$	$H3O+$
Initial	0.40	0	0
Change	-x	x	x
Equilibrium	0.40−x	x	x

Calculate $H3O+$ :

$NH4 +(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)Ka= [NH3 ][H3O+] [NH4 +]5.6× 10-10 = [x] [x][0.40-x](0.40>>x)5.6× 10-10 = [x] [x][0.40][x]2 =5.6× 10-10×0.40[x]= 1.49×10-5 =[H3O+]$

Calculate pH:

$pH = -log[H3O+] pH = -log[1.49×10-5]pH = 4.8$

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Answer 5

Chapter 7, Problem 109E

(a)

Interpretation Introduction

Interpretation: The pH of 0.12 M solution of $KNO2$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(a)

Expert Solution

Answer to Problem 109E

$pH = 8.24$

Explanation of Solution

[ $KNO2$ ] = 0.12 M

$KNO2$ is a basic salt. The ionization of $KNO2$ forms respective ions:

$KNO2→K++NO2-$

Here $NO2-$ can act as string conjugate base of weak acid $HNO2$ therefore the salt is slightly basic in nature as $K+$ is neither acidic nor basic.

The reaction of $NO2-$ ions with water can be written as:

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)$

Initial concentration of $NO2-$ = 0.12 M

$Kb$ for $NO2-$ = $2.5× 10-11$

Make ICE table:

Concentration	$NO2-$	$OH-$	$HNO2$
Initial	0.12	0	0
Change	-x	x	x
Equilibrium	0.12−x	x	x

Calculate $H3O+$ :

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)Kb= [OH- ][HNO2] [NO2-]2.5× 10-11 = [x] [x][0.12-x](0.12>>x)2.5× 10-11 = [x] [x][0.12][x]2 =2.5× 10-11 ×0.12[x]= 1.73×10-6 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[1.73×10-6 ]pOH = 5.76pH = 14 - pOH = 14- 5.76 = 8.24$

(b)

Interpretation Introduction

Interpretation: The pH of 0.45 M solution of $NaOCl$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(b)

Expert Solution

Answer to Problem 109E

$pH = 10.5$

Explanation of Solution

[ $NaOCl$ ] = 0.45 M

$NaOCl$ is a basic salt. The ionization of $NaOCl$ forms respective ions:

$NaOCl→Na++OCl-$

Here $OCl-$ can act as strong conjugate base of weak acid $HClO$ therefore the salt is slightly basic in nature as $Na+$ is neither acidic nor basic.

The reaction of $OCl-$ ions with water can be written as:

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)$

Initial concentration of $OCl-$ = 0.45 M

$Kb$ for $OCl-$ = $2.8× 10-7$

Make ICE table:

Concentration	$OCl-$	$OH-$	$HClO$
Initial	0.45	0	0
Change	-x	x	x
Equilibrium	0.45−x	x	x

Calculate $H3O+$ :

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)Kb= [OH-][HClO] [OCl-]2.8× 10-7 = [x] [x][0.45-x](0.45>>x)2.8× 10-7 = [x] [x][0.45][x]2 =2.8× 10-7 ×0.45[x]= 3.55×10-4 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[3.55×10-4 ]pOH = 3.45pH = 14 - pOH = 14- 3.45 = 10.5$

(c)

Interpretation Introduction

Interpretation: The pH of 0.40 M solution of $NH4ClO4$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(c)

Expert Solution

Answer to Problem 109E

$pH = 4.8$

Explanation of Solution

[ $NH4ClO4$ ] = 0.40 M

$NH4ClO4$ is aacidic salt. The ionization of $NH4ClO4$ forms respective ions:

$NH4ClO4→NH4++ClO4-$

Here $ClO4-$ can act as weak conjugate base of strong acid $HClO4$ therefore the salt is slightly acidic in nature as $NH4+$ is strong conjugate acid of weak base NH₃.

The reaction of $NH4+$ ions with water can be written as:

$NH4+(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)$

Initial concentration of $OCl-$ = 0.40 M

$Ka$ for $NH4+$ = $5.6× 10-10$

Make ICE table:

Concentration	$NH4+$	$NH3$	$H3O+$
Initial	0.40	0	0
Change	-x	x	x
Equilibrium	0.40−x	x	x

Calculate $H3O+$ :

$NH4 +(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)Ka= [NH3 ][H3O+] [NH4 +]5.6× 10-10 = [x] [x][0.40-x](0.40>>x)5.6× 10-10 = [x] [x][0.40][x]2 =5.6× 10-10×0.40[x]= 1.49×10-5 =[H3O+]$

Calculate pH:

$pH = -log[H3O+] pH = -log[1.49×10-5]pH = 4.8$

Answer 6

Chapter 7, Problem 109E

(a)

Interpretation Introduction

Interpretation: The pH of 0.12 M solution of $KNO2$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(a)

Expert Solution

Answer to Problem 109E

$pH = 8.24$

Explanation of Solution

[ $KNO2$ ] = 0.12 M

$KNO2$ is a basic salt. The ionization of $KNO2$ forms respective ions:

$KNO2→K++NO2-$

Here $NO2-$ can act as string conjugate base of weak acid $HNO2$ therefore the salt is slightly basic in nature as $K+$ is neither acidic nor basic.

The reaction of $NO2-$ ions with water can be written as:

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)$

Initial concentration of $NO2-$ = 0.12 M

$Kb$ for $NO2-$ = $2.5× 10-11$

Make ICE table:

Concentration	$NO2-$	$OH-$	$HNO2$
Initial	0.12	0	0
Change	-x	x	x
Equilibrium	0.12−x	x	x

Calculate $H3O+$ :

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)Kb= [OH- ][HNO2] [NO2-]2.5× 10-11 = [x] [x][0.12-x](0.12>>x)2.5× 10-11 = [x] [x][0.12][x]2 =2.5× 10-11 ×0.12[x]= 1.73×10-6 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[1.73×10-6 ]pOH = 5.76pH = 14 - pOH = 14- 5.76 = 8.24$

Answer 7

Chapter 7, Problem 109E

(a)

Interpretation Introduction

Interpretation: The pH of 0.12 M solution of $KNO2$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

Answer 8

(a)

Expert Solution

Answer to Problem 109E

$pH = 8.24$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

[ $KNO2$ ] = 0.12 M

$KNO2$ is a basic salt. The ionization of $KNO2$ forms respective ions:

$KNO2→K++NO2-$

Here $NO2-$ can act as string conjugate base of weak acid $HNO2$ therefore the salt is slightly basic in nature as $K+$ is neither acidic nor basic.

The reaction of $NO2-$ ions with water can be written as:

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)$

Initial concentration of $NO2-$ = 0.12 M

$Kb$ for $NO2-$ = $2.5× 10-11$

Make ICE table:

Concentration	$NO2-$	$OH-$	$HNO2$
Initial	0.12	0	0
Change	-x	x	x
Equilibrium	0.12−x	x	x

Calculate $H3O+$ :

$NO2-(aq)+ H2O(l)⇌HNO2(aq)+ OH-(aq)Kb= [OH- ][HNO2] [NO2-]2.5× 10-11 = [x] [x][0.12-x](0.12>>x)2.5× 10-11 = [x] [x][0.12][x]2 =2.5× 10-11 ×0.12[x]= 1.73×10-6 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[1.73×10-6 ]pOH = 5.76pH = 14 - pOH = 14- 5.76 = 8.24$

Answer 13

(b)

Interpretation Introduction

Interpretation: The pH of 0.45 M solution of $NaOCl$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(b)

Expert Solution

Answer to Problem 109E

$pH = 10.5$

Explanation of Solution

[ $NaOCl$ ] = 0.45 M

$NaOCl$ is a basic salt. The ionization of $NaOCl$ forms respective ions:

$NaOCl→Na++OCl-$

Here $OCl-$ can act as strong conjugate base of weak acid $HClO$ therefore the salt is slightly basic in nature as $Na+$ is neither acidic nor basic.

The reaction of $OCl-$ ions with water can be written as:

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)$

Initial concentration of $OCl-$ = 0.45 M

$Kb$ for $OCl-$ = $2.8× 10-7$

Make ICE table:

Concentration	$OCl-$	$OH-$	$HClO$
Initial	0.45	0	0
Change	-x	x	x
Equilibrium	0.45−x	x	x

Calculate $H3O+$ :

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)Kb= [OH-][HClO] [OCl-]2.8× 10-7 = [x] [x][0.45-x](0.45>>x)2.8× 10-7 = [x] [x][0.45][x]2 =2.8× 10-7 ×0.45[x]= 3.55×10-4 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[3.55×10-4 ]pOH = 3.45pH = 14 - pOH = 14- 3.45 = 10.5$

Answer 14

(b)

Interpretation Introduction

Interpretation: The pH of 0.45 M solution of $NaOCl$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

Answer 15

(b)

Expert Solution

Answer to Problem 109E

$pH = 10.5$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

[ $NaOCl$ ] = 0.45 M

$NaOCl$ is a basic salt. The ionization of $NaOCl$ forms respective ions:

$NaOCl→Na++OCl-$

Here $OCl-$ can act as strong conjugate base of weak acid $HClO$ therefore the salt is slightly basic in nature as $Na+$ is neither acidic nor basic.

The reaction of $OCl-$ ions with water can be written as:

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)$

Initial concentration of $OCl-$ = 0.45 M

$Kb$ for $OCl-$ = $2.8× 10-7$

Make ICE table:

Concentration	$OCl-$	$OH-$	$HClO$
Initial	0.45	0	0
Change	-x	x	x
Equilibrium	0.45−x	x	x

Calculate $H3O+$ :

$OCl-(aq)+ H2O(l)⇌HClO(aq)+ OH-(aq)Kb= [OH-][HClO] [OCl-]2.8× 10-7 = [x] [x][0.45-x](0.45>>x)2.8× 10-7 = [x] [x][0.45][x]2 =2.8× 10-7 ×0.45[x]= 3.55×10-4 =[OH−]$

Calculate pOH and pH:

$pOH = -log[OH−] pOH = -log[3.55×10-4 ]pOH = 3.45pH = 14 - pOH = 14- 3.45 = 10.5$

Answer 20

(c)

Interpretation Introduction

Interpretation: The pH of 0.40 M solution of $NH4ClO4$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

(c)

Expert Solution

Answer to Problem 109E

$pH = 4.8$

Explanation of Solution

[ $NH4ClO4$ ] = 0.40 M

$NH4ClO4$ is aacidic salt. The ionization of $NH4ClO4$ forms respective ions:

$NH4ClO4→NH4++ClO4-$

Here $ClO4-$ can act as weak conjugate base of strong acid $HClO4$ therefore the salt is slightly acidic in nature as $NH4+$ is strong conjugate acid of weak base NH₃.

The reaction of $NH4+$ ions with water can be written as:

$NH4+(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)$

Initial concentration of $OCl-$ = 0.40 M

$Ka$ for $NH4+$ = $5.6× 10-10$

Make ICE table:

Concentration	$NH4+$	$NH3$	$H3O+$
Initial	0.40	0	0
Change	-x	x	x
Equilibrium	0.40−x	x	x

Calculate $H3O+$ :

$NH4 +(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)Ka= [NH3 ][H3O+] [NH4 +]5.6× 10-10 = [x] [x][0.40-x](0.40>>x)5.6× 10-10 = [x] [x][0.40][x]2 =5.6× 10-10×0.40[x]= 1.49×10-5 =[H3O+]$

Calculate pH:

$pH = -log[H3O+] pH = -log[1.49×10-5]pH = 4.8$

Answer 21

(c)

Interpretation Introduction

Interpretation: The pH of 0.40 M solution of $NH4ClO4$ needs to be determined.

Concept Introduction: A base is the substance that gives $OH-$ ions in its aqueous solution. On the basis of base dissociation, they can be classified as strong and weak base. On the contrary, a weak base ionized partially and reaches to equilibrium.

$BOH(aq)⇌OH-(aq)+B-(aq)$

An acid is the substance that gives $H+$ ions in its aqueous solution. On the basis of acid dissociation, acids can be classified as strong and weak acid. On the contrary, a weak acid ionized partially and reaches to equilibrium.

$HA(aq)+ H2O(l)⇌H3O+(aq)+ A-(aq)$

Answer 22

(c)

Expert Solution

Answer to Problem 109E

$pH = 4.8$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

[ $NH4ClO4$ ] = 0.40 M

$NH4ClO4$ is aacidic salt. The ionization of $NH4ClO4$ forms respective ions:

$NH4ClO4→NH4++ClO4-$

Here $ClO4-$ can act as weak conjugate base of strong acid $HClO4$ therefore the salt is slightly acidic in nature as $NH4+$ is strong conjugate acid of weak base NH₃.

The reaction of $NH4+$ ions with water can be written as:

$NH4+(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)$

Initial concentration of $OCl-$ = 0.40 M

$Ka$ for $NH4+$ = $5.6× 10-10$

Make ICE table:

Concentration	$NH4+$	$NH3$	$H3O+$
Initial	0.40	0	0
Change	-x	x	x
Equilibrium	0.40−x	x	x

Calculate $H3O+$ :

$NH4 +(aq)+ H2O(l)⇌NH3(aq)+ H3O+(aq)Ka= [NH3 ][H3O+] [NH4 +]5.6× 10-10 = [x] [x][0.40-x](0.40>>x)5.6× 10-10 = [x] [x][0.40][x]2 =5.6× 10-10×0.40[x]= 1.49×10-5 =[H3O+]$