The p H of 0.10 M aqueous solutions of H N O 2 given below has to be calculated.

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Answer 1

Question

Chapter 14, Problem 54QRT

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of HNO2 given below has to be calculated.

(a)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

HNO2(aq)+H2O(l)⇌NO2−(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NO2−][H3O+][HNO2]

A table can be set up as shown below.

HNO2(aq)H3O+(aq)NO2−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=7.4×10−4

Then,

Ka=(x)(x)(0.15−x)=7.4×10−4⇒x2=(7.4×10−4)×(0.10−x)⇒x2+7.4×10−4x−7.4×10−5=0⇒x=8.2×10−3=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(8.2×10−3)=2.09.

Therefore, the pH of 0.10M aqueous solution of HNO2 is 2.09.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NH4Cl given below has to be calculated.

(b)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NH3][H3O+][NH4+]

A table can be set up as shown below.

NH4+(aq)H3O+(aq)NH3(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=5.6×10−10

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Ka=(x)(x)(0.15−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.10)⇒x=5.6×10−11=7.5×10−6M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(7.5×10−6)=5.13.

Therefore, the pH of 0.10M aqueous solution of NH4Cl is 5.13.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaF given below has to be calculated.

(c)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

F−(aq)+H2O(l)⇌HF(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[HF][OH−][F−]

A table can be set up as shown below.

F−(aq)HF(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=1.5×10−11

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Kb=(x)(x)(0.15−x)=1.5×10−11⇒x2=(1.5×10−11)×(0.10)⇒x=1.5×10−12=1.2×10−6M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.2×10−6)=5.91.pH=14.00−pOH=14.00−5.91=8.09.

Therefore, the pH of 0.10M aqueous solution of HF is 8.09.

(d)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of Mg(CH3COO)2 given below has to be calculated.

(d)

Expert Solution

Explanation of Solution

The concentration of CH3COO− can be calculated as shown below.

[CH3COO−]=0.10 MMg(CH3COO)2×2 mol CH3COO−1 mol Mg(CH3COO)2=0.20 M.

The equilibrium reaction is given below.

CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[CH3COOH][OH−][CH3COO−]

A table can be set up as shown below.

CH3COO−(aq)CH3COOH(aq)OH−(aq)Initial conc.(M)0.2001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.20−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.20−x)=5.6×10−10

Assuming x is very small, it can be written 0.20−x≅0.20.

Then,

Kb=(x)(x)(0.20−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.20)⇒x=(5.6×10−10)×(0.20)=1.1×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.1×10−5)=4.98.pH=14.00−pOH=14.00−4.98=9.02.

Therefore, the pH of 0.10M aqueous solution of Mg(CH3COO)2 is 9.02.

(e)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of BaO given below has to be calculated.

(e)

Expert Solution

Explanation of Solution

The reaction of BaO with water is given below.

O2−(aq)+H2O(l)→2OH−(aq)

The concentration of OH− can be calculated as shown below.

[OH−]=0.10 M O2−×2 mol OH−1 mol O2−=0.20 M.

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(0.20)=0.70.pH=14.00−pOH=14.00−0.70=13.30.

Therefore, the pH of 0.10M aqueous solution of BaO is 13.30.

(f)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of KHSO4 given below has to be calculated.

(f)

Expert Solution

Explanation of Solution

HSO4− acts as both acid and base.

HSO4−(aq)+H2O(aq)⇌H3O+(aq)+SO42−(aq)Ka=[H3O+][SO42−][HSO4−]=1.1×10−2HSO4−(aq)+H2O(aq)⇌H3O+(aq)+H2SO4(aq)Kb=[H2SO4][OH−][HSO4−]=very small

HSO4− is more acidic than basic.

A table can be set up as shown below.

HSO4−(aq)H3O+(aq)SO42−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=1.1×10−2

Then,

x2=(1.1×10−2)×(0.10−x)⇒x2+1.1×10−2x−1.1×10−3=0⇒x=2.8×10−2M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(2.8×10−2)=1.55.

Therefore, the pH of 0.10M aqueous solution of KHSO4 is 1.55.

(g)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaHCO3 given below has to be calculated.

(g)

Expert Solution

Explanation of Solution

HCO3− acts as both acid and base.

HCO3−(aq)+H2O(aq)⇌H3O+(aq)+CO32−(aq)Ka=[H3O+][CO32−][HCO3−]=4.7×10−11HCO3−(aq)+H2O(aq)⇌OH−(aq)+H2CO3(aq)Kb=[H2CO3][OH−][HCO3−]=2.3×10−8

HCO3− is more basic than acidic.

A table can be set up as shown below.

HCO3−(aq)H2CO3(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=2.3×10−8

Then,

Assuming x is very small, it can be written 0.20−x≅0.20.

Kb=(x)(x)(0.10−x)=2.3×10−8⇒x2=(2.3×10−8)×(0.10)⇒x=(2.3×10−8)×(0.10)=4.8×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(4.8×10−5)=4.32.pH=14.00−pOH=14.00−4.32=9.68.

Therefore, the pH of 0.10M aqueous solution of NaHCO3 is 9.68.

(h)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions BaCl2 given below has to be calculated.

(h)

Expert Solution

Explanation of Solution

BaCl2→Ba2++2Cl−

Neither of these ions affect the pH of a water solution, because Ba2+ is the cation of Ba(OH)2 and Cl– is a weak base. So pH is 7.00 and no calculation is needed.

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Answer 2

Question

Chapter 14, Problem 54QRT

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of HNO2 given below has to be calculated.

(a)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

HNO2(aq)+H2O(l)⇌NO2−(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NO2−][H3O+][HNO2]

A table can be set up as shown below.

HNO2(aq)H3O+(aq)NO2−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=7.4×10−4

Then,

Ka=(x)(x)(0.15−x)=7.4×10−4⇒x2=(7.4×10−4)×(0.10−x)⇒x2+7.4×10−4x−7.4×10−5=0⇒x=8.2×10−3=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(8.2×10−3)=2.09.

Therefore, the pH of 0.10M aqueous solution of HNO2 is 2.09.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NH4Cl given below has to be calculated.

(b)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NH3][H3O+][NH4+]

A table can be set up as shown below.

NH4+(aq)H3O+(aq)NH3(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=5.6×10−10

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Ka=(x)(x)(0.15−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.10)⇒x=5.6×10−11=7.5×10−6M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(7.5×10−6)=5.13.

Therefore, the pH of 0.10M aqueous solution of NH4Cl is 5.13.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaF given below has to be calculated.

(c)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

F−(aq)+H2O(l)⇌HF(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[HF][OH−][F−]

A table can be set up as shown below.

F−(aq)HF(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=1.5×10−11

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Kb=(x)(x)(0.15−x)=1.5×10−11⇒x2=(1.5×10−11)×(0.10)⇒x=1.5×10−12=1.2×10−6M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.2×10−6)=5.91.pH=14.00−pOH=14.00−5.91=8.09.

Therefore, the pH of 0.10M aqueous solution of HF is 8.09.

(d)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of Mg(CH3COO)2 given below has to be calculated.

(d)

Expert Solution

Explanation of Solution

The concentration of CH3COO− can be calculated as shown below.

[CH3COO−]=0.10 MMg(CH3COO)2×2 mol CH3COO−1 mol Mg(CH3COO)2=0.20 M.

The equilibrium reaction is given below.

CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[CH3COOH][OH−][CH3COO−]

A table can be set up as shown below.

CH3COO−(aq)CH3COOH(aq)OH−(aq)Initial conc.(M)0.2001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.20−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.20−x)=5.6×10−10

Assuming x is very small, it can be written 0.20−x≅0.20.

Then,

Kb=(x)(x)(0.20−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.20)⇒x=(5.6×10−10)×(0.20)=1.1×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.1×10−5)=4.98.pH=14.00−pOH=14.00−4.98=9.02.

Therefore, the pH of 0.10M aqueous solution of Mg(CH3COO)2 is 9.02.

(e)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of BaO given below has to be calculated.

(e)

Expert Solution

Explanation of Solution

The reaction of BaO with water is given below.

O2−(aq)+H2O(l)→2OH−(aq)

The concentration of OH− can be calculated as shown below.

[OH−]=0.10 M O2−×2 mol OH−1 mol O2−=0.20 M.

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(0.20)=0.70.pH=14.00−pOH=14.00−0.70=13.30.

Therefore, the pH of 0.10M aqueous solution of BaO is 13.30.

(f)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of KHSO4 given below has to be calculated.

(f)

Expert Solution

Explanation of Solution

HSO4− acts as both acid and base.

HSO4−(aq)+H2O(aq)⇌H3O+(aq)+SO42−(aq)Ka=[H3O+][SO42−][HSO4−]=1.1×10−2HSO4−(aq)+H2O(aq)⇌H3O+(aq)+H2SO4(aq)Kb=[H2SO4][OH−][HSO4−]=very small

HSO4− is more acidic than basic.

A table can be set up as shown below.

HSO4−(aq)H3O+(aq)SO42−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=1.1×10−2

Then,

x2=(1.1×10−2)×(0.10−x)⇒x2+1.1×10−2x−1.1×10−3=0⇒x=2.8×10−2M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(2.8×10−2)=1.55.

Therefore, the pH of 0.10M aqueous solution of KHSO4 is 1.55.

(g)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaHCO3 given below has to be calculated.

(g)

Expert Solution

Explanation of Solution

HCO3− acts as both acid and base.

HCO3−(aq)+H2O(aq)⇌H3O+(aq)+CO32−(aq)Ka=[H3O+][CO32−][HCO3−]=4.7×10−11HCO3−(aq)+H2O(aq)⇌OH−(aq)+H2CO3(aq)Kb=[H2CO3][OH−][HCO3−]=2.3×10−8

HCO3− is more basic than acidic.

A table can be set up as shown below.

HCO3−(aq)H2CO3(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=2.3×10−8

Then,

Assuming x is very small, it can be written 0.20−x≅0.20.

Kb=(x)(x)(0.10−x)=2.3×10−8⇒x2=(2.3×10−8)×(0.10)⇒x=(2.3×10−8)×(0.10)=4.8×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(4.8×10−5)=4.32.pH=14.00−pOH=14.00−4.32=9.68.

Therefore, the pH of 0.10M aqueous solution of NaHCO3 is 9.68.

(h)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions BaCl2 given below has to be calculated.

(h)

Expert Solution

Explanation of Solution

BaCl2→Ba2++2Cl−

Neither of these ions affect the pH of a water solution, because Ba2+ is the cation of Ba(OH)2 and Cl– is a weak base. So pH is 7.00 and no calculation is needed.

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Answer 3

Question

Chapter 14, Problem 54QRT

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of HNO2 given below has to be calculated.

(a)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

HNO2(aq)+H2O(l)⇌NO2−(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NO2−][H3O+][HNO2]

A table can be set up as shown below.

HNO2(aq)H3O+(aq)NO2−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=7.4×10−4

Then,

Ka=(x)(x)(0.15−x)=7.4×10−4⇒x2=(7.4×10−4)×(0.10−x)⇒x2+7.4×10−4x−7.4×10−5=0⇒x=8.2×10−3=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(8.2×10−3)=2.09.

Therefore, the pH of 0.10M aqueous solution of HNO2 is 2.09.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NH4Cl given below has to be calculated.

(b)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NH3][H3O+][NH4+]

A table can be set up as shown below.

NH4+(aq)H3O+(aq)NH3(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=5.6×10−10

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Ka=(x)(x)(0.15−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.10)⇒x=5.6×10−11=7.5×10−6M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(7.5×10−6)=5.13.

Therefore, the pH of 0.10M aqueous solution of NH4Cl is 5.13.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaF given below has to be calculated.

(c)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

F−(aq)+H2O(l)⇌HF(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[HF][OH−][F−]

A table can be set up as shown below.

F−(aq)HF(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=1.5×10−11

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Kb=(x)(x)(0.15−x)=1.5×10−11⇒x2=(1.5×10−11)×(0.10)⇒x=1.5×10−12=1.2×10−6M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.2×10−6)=5.91.pH=14.00−pOH=14.00−5.91=8.09.

Therefore, the pH of 0.10M aqueous solution of HF is 8.09.

(d)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of Mg(CH3COO)2 given below has to be calculated.

(d)

Expert Solution

Explanation of Solution

The concentration of CH3COO− can be calculated as shown below.

[CH3COO−]=0.10 MMg(CH3COO)2×2 mol CH3COO−1 mol Mg(CH3COO)2=0.20 M.

The equilibrium reaction is given below.

CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[CH3COOH][OH−][CH3COO−]

A table can be set up as shown below.

CH3COO−(aq)CH3COOH(aq)OH−(aq)Initial conc.(M)0.2001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.20−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.20−x)=5.6×10−10

Assuming x is very small, it can be written 0.20−x≅0.20.

Then,

Kb=(x)(x)(0.20−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.20)⇒x=(5.6×10−10)×(0.20)=1.1×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.1×10−5)=4.98.pH=14.00−pOH=14.00−4.98=9.02.

Therefore, the pH of 0.10M aqueous solution of Mg(CH3COO)2 is 9.02.

(e)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of BaO given below has to be calculated.

(e)

Expert Solution

Explanation of Solution

The reaction of BaO with water is given below.

O2−(aq)+H2O(l)→2OH−(aq)

The concentration of OH− can be calculated as shown below.

[OH−]=0.10 M O2−×2 mol OH−1 mol O2−=0.20 M.

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(0.20)=0.70.pH=14.00−pOH=14.00−0.70=13.30.

Therefore, the pH of 0.10M aqueous solution of BaO is 13.30.

(f)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of KHSO4 given below has to be calculated.

(f)

Expert Solution

Explanation of Solution

HSO4− acts as both acid and base.

HSO4−(aq)+H2O(aq)⇌H3O+(aq)+SO42−(aq)Ka=[H3O+][SO42−][HSO4−]=1.1×10−2HSO4−(aq)+H2O(aq)⇌H3O+(aq)+H2SO4(aq)Kb=[H2SO4][OH−][HSO4−]=very small

HSO4− is more acidic than basic.

A table can be set up as shown below.

HSO4−(aq)H3O+(aq)SO42−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=1.1×10−2

Then,

x2=(1.1×10−2)×(0.10−x)⇒x2+1.1×10−2x−1.1×10−3=0⇒x=2.8×10−2M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(2.8×10−2)=1.55.

Therefore, the pH of 0.10M aqueous solution of KHSO4 is 1.55.

(g)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaHCO3 given below has to be calculated.

(g)

Expert Solution

Explanation of Solution

HCO3− acts as both acid and base.

HCO3−(aq)+H2O(aq)⇌H3O+(aq)+CO32−(aq)Ka=[H3O+][CO32−][HCO3−]=4.7×10−11HCO3−(aq)+H2O(aq)⇌OH−(aq)+H2CO3(aq)Kb=[H2CO3][OH−][HCO3−]=2.3×10−8

HCO3− is more basic than acidic.

A table can be set up as shown below.

HCO3−(aq)H2CO3(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=2.3×10−8

Then,

Assuming x is very small, it can be written 0.20−x≅0.20.

Kb=(x)(x)(0.10−x)=2.3×10−8⇒x2=(2.3×10−8)×(0.10)⇒x=(2.3×10−8)×(0.10)=4.8×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(4.8×10−5)=4.32.pH=14.00−pOH=14.00−4.32=9.68.

Therefore, the pH of 0.10M aqueous solution of NaHCO3 is 9.68.

(h)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions BaCl2 given below has to be calculated.

(h)

Expert Solution

Explanation of Solution

BaCl2→Ba2++2Cl−

Neither of these ions affect the pH of a water solution, because Ba2+ is the cation of Ba(OH)2 and Cl– is a weak base. So pH is 7.00 and no calculation is needed.

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Answer 4

Question

Chapter 14, Problem 54QRT

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of HNO2 given below has to be calculated.

(a)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

HNO2(aq)+H2O(l)⇌NO2−(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NO2−][H3O+][HNO2]

A table can be set up as shown below.

HNO2(aq)H3O+(aq)NO2−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=7.4×10−4

Then,

Ka=(x)(x)(0.15−x)=7.4×10−4⇒x2=(7.4×10−4)×(0.10−x)⇒x2+7.4×10−4x−7.4×10−5=0⇒x=8.2×10−3=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(8.2×10−3)=2.09.

Therefore, the pH of 0.10M aqueous solution of HNO2 is 2.09.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NH4Cl given below has to be calculated.

(b)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NH3][H3O+][NH4+]

A table can be set up as shown below.

NH4+(aq)H3O+(aq)NH3(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=5.6×10−10

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Ka=(x)(x)(0.15−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.10)⇒x=5.6×10−11=7.5×10−6M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(7.5×10−6)=5.13.

Therefore, the pH of 0.10M aqueous solution of NH4Cl is 5.13.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaF given below has to be calculated.

(c)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

F−(aq)+H2O(l)⇌HF(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[HF][OH−][F−]

A table can be set up as shown below.

F−(aq)HF(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=1.5×10−11

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Kb=(x)(x)(0.15−x)=1.5×10−11⇒x2=(1.5×10−11)×(0.10)⇒x=1.5×10−12=1.2×10−6M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.2×10−6)=5.91.pH=14.00−pOH=14.00−5.91=8.09.

Therefore, the pH of 0.10M aqueous solution of HF is 8.09.

(d)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of Mg(CH3COO)2 given below has to be calculated.

(d)

Expert Solution

Explanation of Solution

The concentration of CH3COO− can be calculated as shown below.

[CH3COO−]=0.10 MMg(CH3COO)2×2 mol CH3COO−1 mol Mg(CH3COO)2=0.20 M.

The equilibrium reaction is given below.

CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[CH3COOH][OH−][CH3COO−]

A table can be set up as shown below.

CH3COO−(aq)CH3COOH(aq)OH−(aq)Initial conc.(M)0.2001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.20−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.20−x)=5.6×10−10

Assuming x is very small, it can be written 0.20−x≅0.20.

Then,

Kb=(x)(x)(0.20−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.20)⇒x=(5.6×10−10)×(0.20)=1.1×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.1×10−5)=4.98.pH=14.00−pOH=14.00−4.98=9.02.

Therefore, the pH of 0.10M aqueous solution of Mg(CH3COO)2 is 9.02.

(e)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of BaO given below has to be calculated.

(e)

Expert Solution

Explanation of Solution

The reaction of BaO with water is given below.

O2−(aq)+H2O(l)→2OH−(aq)

The concentration of OH− can be calculated as shown below.

[OH−]=0.10 M O2−×2 mol OH−1 mol O2−=0.20 M.

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(0.20)=0.70.pH=14.00−pOH=14.00−0.70=13.30.

Therefore, the pH of 0.10M aqueous solution of BaO is 13.30.

(f)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of KHSO4 given below has to be calculated.

(f)

Expert Solution

Explanation of Solution

HSO4− acts as both acid and base.

HSO4−(aq)+H2O(aq)⇌H3O+(aq)+SO42−(aq)Ka=[H3O+][SO42−][HSO4−]=1.1×10−2HSO4−(aq)+H2O(aq)⇌H3O+(aq)+H2SO4(aq)Kb=[H2SO4][OH−][HSO4−]=very small

HSO4− is more acidic than basic.

A table can be set up as shown below.

HSO4−(aq)H3O+(aq)SO42−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=1.1×10−2

Then,

x2=(1.1×10−2)×(0.10−x)⇒x2+1.1×10−2x−1.1×10−3=0⇒x=2.8×10−2M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(2.8×10−2)=1.55.

Therefore, the pH of 0.10M aqueous solution of KHSO4 is 1.55.

(g)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaHCO3 given below has to be calculated.

(g)

Expert Solution

Explanation of Solution

HCO3− acts as both acid and base.

HCO3−(aq)+H2O(aq)⇌H3O+(aq)+CO32−(aq)Ka=[H3O+][CO32−][HCO3−]=4.7×10−11HCO3−(aq)+H2O(aq)⇌OH−(aq)+H2CO3(aq)Kb=[H2CO3][OH−][HCO3−]=2.3×10−8

HCO3− is more basic than acidic.

A table can be set up as shown below.

HCO3−(aq)H2CO3(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=2.3×10−8

Then,

Assuming x is very small, it can be written 0.20−x≅0.20.

Kb=(x)(x)(0.10−x)=2.3×10−8⇒x2=(2.3×10−8)×(0.10)⇒x=(2.3×10−8)×(0.10)=4.8×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(4.8×10−5)=4.32.pH=14.00−pOH=14.00−4.32=9.68.

Therefore, the pH of 0.10M aqueous solution of NaHCO3 is 9.68.

(h)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions BaCl2 given below has to be calculated.

(h)

Expert Solution

Explanation of Solution

BaCl2→Ba2++2Cl−

Neither of these ions affect the pH of a water solution, because Ba2+ is the cation of Ba(OH)2 and Cl– is a weak base. So pH is 7.00 and no calculation is needed.

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Answer 5

Chapter 14, Problem 54QRT

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of HNO2 given below has to be calculated.

(a)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

HNO2(aq)+H2O(l)⇌NO2−(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NO2−][H3O+][HNO2]

A table can be set up as shown below.

HNO2(aq)H3O+(aq)NO2−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=7.4×10−4

Then,

Ka=(x)(x)(0.15−x)=7.4×10−4⇒x2=(7.4×10−4)×(0.10−x)⇒x2+7.4×10−4x−7.4×10−5=0⇒x=8.2×10−3=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(8.2×10−3)=2.09.

Therefore, the pH of 0.10M aqueous solution of HNO2 is 2.09.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NH4Cl given below has to be calculated.

(b)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NH3][H3O+][NH4+]

A table can be set up as shown below.

NH4+(aq)H3O+(aq)NH3(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=5.6×10−10

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Ka=(x)(x)(0.15−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.10)⇒x=5.6×10−11=7.5×10−6M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(7.5×10−6)=5.13.

Therefore, the pH of 0.10M aqueous solution of NH4Cl is 5.13.

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaF given below has to be calculated.

(c)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

F−(aq)+H2O(l)⇌HF(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[HF][OH−][F−]

A table can be set up as shown below.

F−(aq)HF(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=1.5×10−11

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Kb=(x)(x)(0.15−x)=1.5×10−11⇒x2=(1.5×10−11)×(0.10)⇒x=1.5×10−12=1.2×10−6M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.2×10−6)=5.91.pH=14.00−pOH=14.00−5.91=8.09.

Therefore, the pH of 0.10M aqueous solution of HF is 8.09.

(d)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of Mg(CH3COO)2 given below has to be calculated.

(d)

Expert Solution

Explanation of Solution

The concentration of CH3COO− can be calculated as shown below.

[CH3COO−]=0.10 MMg(CH3COO)2×2 mol CH3COO−1 mol Mg(CH3COO)2=0.20 M.

The equilibrium reaction is given below.

CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[CH3COOH][OH−][CH3COO−]

A table can be set up as shown below.

CH3COO−(aq)CH3COOH(aq)OH−(aq)Initial conc.(M)0.2001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.20−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.20−x)=5.6×10−10

Assuming x is very small, it can be written 0.20−x≅0.20.

Then,

Kb=(x)(x)(0.20−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.20)⇒x=(5.6×10−10)×(0.20)=1.1×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.1×10−5)=4.98.pH=14.00−pOH=14.00−4.98=9.02.

Therefore, the pH of 0.10M aqueous solution of Mg(CH3COO)2 is 9.02.

(e)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of BaO given below has to be calculated.

(e)

Expert Solution

Explanation of Solution

The reaction of BaO with water is given below.

O2−(aq)+H2O(l)→2OH−(aq)

The concentration of OH− can be calculated as shown below.

[OH−]=0.10 M O2−×2 mol OH−1 mol O2−=0.20 M.

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(0.20)=0.70.pH=14.00−pOH=14.00−0.70=13.30.

Therefore, the pH of 0.10M aqueous solution of BaO is 13.30.

(f)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of KHSO4 given below has to be calculated.

(f)

Expert Solution

Explanation of Solution

HSO4− acts as both acid and base.

HSO4−(aq)+H2O(aq)⇌H3O+(aq)+SO42−(aq)Ka=[H3O+][SO42−][HSO4−]=1.1×10−2HSO4−(aq)+H2O(aq)⇌H3O+(aq)+H2SO4(aq)Kb=[H2SO4][OH−][HSO4−]=very small

HSO4− is more acidic than basic.

A table can be set up as shown below.

HSO4−(aq)H3O+(aq)SO42−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=1.1×10−2

Then,

x2=(1.1×10−2)×(0.10−x)⇒x2+1.1×10−2x−1.1×10−3=0⇒x=2.8×10−2M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(2.8×10−2)=1.55.

Therefore, the pH of 0.10M aqueous solution of KHSO4 is 1.55.

(g)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaHCO3 given below has to be calculated.

(g)

Expert Solution

Explanation of Solution

HCO3− acts as both acid and base.

HCO3−(aq)+H2O(aq)⇌H3O+(aq)+CO32−(aq)Ka=[H3O+][CO32−][HCO3−]=4.7×10−11HCO3−(aq)+H2O(aq)⇌OH−(aq)+H2CO3(aq)Kb=[H2CO3][OH−][HCO3−]=2.3×10−8

HCO3− is more basic than acidic.

A table can be set up as shown below.

HCO3−(aq)H2CO3(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=2.3×10−8

Then,

Assuming x is very small, it can be written 0.20−x≅0.20.

Kb=(x)(x)(0.10−x)=2.3×10−8⇒x2=(2.3×10−8)×(0.10)⇒x=(2.3×10−8)×(0.10)=4.8×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(4.8×10−5)=4.32.pH=14.00−pOH=14.00−4.32=9.68.

Therefore, the pH of 0.10M aqueous solution of NaHCO3 is 9.68.

(h)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions BaCl2 given below has to be calculated.

(h)

Expert Solution

Explanation of Solution

BaCl2→Ba2++2Cl−

Neither of these ions affect the pH of a water solution, because Ba2+ is the cation of Ba(OH)2 and Cl– is a weak base. So pH is 7.00 and no calculation is needed.

Answer 6

Chapter 14, Problem 54QRT

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of HNO2 given below has to be calculated.

(a)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

HNO2(aq)+H2O(l)⇌NO2−(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NO2−][H3O+][HNO2]

A table can be set up as shown below.

HNO2(aq)H3O+(aq)NO2−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=7.4×10−4

Then,

Ka=(x)(x)(0.15−x)=7.4×10−4⇒x2=(7.4×10−4)×(0.10−x)⇒x2+7.4×10−4x−7.4×10−5=0⇒x=8.2×10−3=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(8.2×10−3)=2.09.

Therefore, the pH of 0.10M aqueous solution of HNO2 is 2.09.

Answer 7

Chapter 14, Problem 54QRT

(a)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of HNO2 given below has to be calculated.

Answer 8

(a)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

HNO2(aq)+H2O(l)⇌NO2−(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NO2−][H3O+][HNO2]

A table can be set up as shown below.

HNO2(aq)H3O+(aq)NO2−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=7.4×10−4

Then,

Ka=(x)(x)(0.15−x)=7.4×10−4⇒x2=(7.4×10−4)×(0.10−x)⇒x2+7.4×10−4x−7.4×10−5=0⇒x=8.2×10−3=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(8.2×10−3)=2.09.

Therefore, the pH of 0.10M aqueous solution of HNO2 is 2.09.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NH4Cl given below has to be calculated.

(b)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NH3][H3O+][NH4+]

A table can be set up as shown below.

NH4+(aq)H3O+(aq)NH3(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=5.6×10−10

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Ka=(x)(x)(0.15−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.10)⇒x=5.6×10−11=7.5×10−6M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(7.5×10−6)=5.13.

Therefore, the pH of 0.10M aqueous solution of NH4Cl is 5.13.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NH4Cl given below has to be calculated.

Answer 14

(b)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

NH4+(aq)+H2O(l)⇌NH3(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

Ka=[NH3][H3O+][NH4+]

A table can be set up as shown below.

NH4+(aq)H3O+(aq)NH3(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=5.6×10−10

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Ka=(x)(x)(0.15−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.10)⇒x=5.6×10−11=7.5×10−6M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(7.5×10−6)=5.13.

Therefore, the pH of 0.10M aqueous solution of NH4Cl is 5.13.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaF given below has to be calculated.

(c)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

F−(aq)+H2O(l)⇌HF(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[HF][OH−][F−]

A table can be set up as shown below.

F−(aq)HF(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=1.5×10−11

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Kb=(x)(x)(0.15−x)=1.5×10−11⇒x2=(1.5×10−11)×(0.10)⇒x=1.5×10−12=1.2×10−6M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.2×10−6)=5.91.pH=14.00−pOH=14.00−5.91=8.09.

Therefore, the pH of 0.10M aqueous solution of HF is 8.09.

Answer 19

(c)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaF given below has to be calculated.

Answer 20

(c)

Expert Solution

Explanation of Solution

The equilibrium reaction is given below.

F−(aq)+H2O(l)⇌HF(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[HF][OH−][F−]

A table can be set up as shown below.

F−(aq)HF(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=1.5×10−11

Assuming x is very small, it can be written 0.10−x≅0.10.

Then,

Kb=(x)(x)(0.15−x)=1.5×10−11⇒x2=(1.5×10−11)×(0.10)⇒x=1.5×10−12=1.2×10−6M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.2×10−6)=5.91.pH=14.00−pOH=14.00−5.91=8.09.

Therefore, the pH of 0.10M aqueous solution of HF is 8.09.

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of Mg(CH3COO)2 given below has to be calculated.

(d)

Expert Solution

Explanation of Solution

The concentration of CH3COO− can be calculated as shown below.

[CH3COO−]=0.10 MMg(CH3COO)2×2 mol CH3COO−1 mol Mg(CH3COO)2=0.20 M.

The equilibrium reaction is given below.

CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[CH3COOH][OH−][CH3COO−]

A table can be set up as shown below.

CH3COO−(aq)CH3COOH(aq)OH−(aq)Initial conc.(M)0.2001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.20−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.20−x)=5.6×10−10

Assuming x is very small, it can be written 0.20−x≅0.20.

Then,

Kb=(x)(x)(0.20−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.20)⇒x=(5.6×10−10)×(0.20)=1.1×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.1×10−5)=4.98.pH=14.00−pOH=14.00−4.98=9.02.

Therefore, the pH of 0.10M aqueous solution of Mg(CH3COO)2 is 9.02.

Answer 25

(d)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of Mg(CH3COO)2 given below has to be calculated.

Answer 26

(d)

Expert Solution

Explanation of Solution

The concentration of CH3COO− can be calculated as shown below.

[CH3COO−]=0.10 MMg(CH3COO)2×2 mol CH3COO−1 mol Mg(CH3COO)2=0.20 M.

The equilibrium reaction is given below.

CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)

The base ionization constant can be written as given below.

Kb=[CH3COOH][OH−][CH3COO−]

A table can be set up as shown below.

CH3COO−(aq)CH3COOH(aq)OH−(aq)Initial conc.(M)0.2001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.20−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.20−x)=5.6×10−10

Assuming x is very small, it can be written 0.20−x≅0.20.

Then,

Kb=(x)(x)(0.20−x)=5.6×10−10⇒x2=(5.6×10−10)×(0.20)⇒x=(5.6×10−10)×(0.20)=1.1×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(1.1×10−5)=4.98.pH=14.00−pOH=14.00−4.98=9.02.

Therefore, the pH of 0.10M aqueous solution of Mg(CH3COO)2 is 9.02.

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

(e)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of BaO given below has to be calculated.

(e)

Expert Solution

Explanation of Solution

The reaction of BaO with water is given below.

O2−(aq)+H2O(l)→2OH−(aq)

The concentration of OH− can be calculated as shown below.

[OH−]=0.10 M O2−×2 mol OH−1 mol O2−=0.20 M.

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(0.20)=0.70.pH=14.00−pOH=14.00−0.70=13.30.

Therefore, the pH of 0.10M aqueous solution of BaO is 13.30.

Answer 31

(e)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of BaO given below has to be calculated.

Answer 32

(e)

Expert Solution

Explanation of Solution

The reaction of BaO with water is given below.

O2−(aq)+H2O(l)→2OH−(aq)

The concentration of OH− can be calculated as shown below.

[OH−]=0.10 M O2−×2 mol OH−1 mol O2−=0.20 M.

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(0.20)=0.70.pH=14.00−pOH=14.00−0.70=13.30.

Therefore, the pH of 0.10M aqueous solution of BaO is 13.30.

Answer 33

(e)

Expert Solution

Answer 34

(e)

Expert Solution

Answer 35

Expert Solution

Answer 36

(f)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of KHSO4 given below has to be calculated.

(f)

Expert Solution

Explanation of Solution

HSO4− acts as both acid and base.

HSO4−(aq)+H2O(aq)⇌H3O+(aq)+SO42−(aq)Ka=[H3O+][SO42−][HSO4−]=1.1×10−2HSO4−(aq)+H2O(aq)⇌H3O+(aq)+H2SO4(aq)Kb=[H2SO4][OH−][HSO4−]=very small

HSO4− is more acidic than basic.

A table can be set up as shown below.

HSO4−(aq)H3O+(aq)SO42−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=1.1×10−2

Then,

x2=(1.1×10−2)×(0.10−x)⇒x2+1.1×10−2x−1.1×10−3=0⇒x=2.8×10−2M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(2.8×10−2)=1.55.

Therefore, the pH of 0.10M aqueous solution of KHSO4 is 1.55.

Answer 37

(f)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of KHSO4 given below has to be calculated.

Answer 38

(f)

Expert Solution

Explanation of Solution

HSO4− acts as both acid and base.

HSO4−(aq)+H2O(aq)⇌H3O+(aq)+SO42−(aq)Ka=[H3O+][SO42−][HSO4−]=1.1×10−2HSO4−(aq)+H2O(aq)⇌H3O+(aq)+H2SO4(aq)Kb=[H2SO4][OH−][HSO4−]=very small

HSO4− is more acidic than basic.

A table can be set up as shown below.

HSO4−(aq)H3O+(aq)SO42−(aq)Initial conc.(M)0.101×10−70Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of H3O+ comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10−x)=1.1×10−2

Then,

x2=(1.1×10−2)×(0.10−x)⇒x2+1.1×10−2x−1.1×10−3=0⇒x=2.8×10−2M=[H3O+]

The pH of the solution can be calculated as shown below.

pH=−log[H3O+]=−log(2.8×10−2)=1.55.

Therefore, the pH of 0.10M aqueous solution of KHSO4 is 1.55.

Answer 39

(f)

Expert Solution

Answer 40

(f)

Expert Solution

Answer 41

Expert Solution

Answer 42

(g)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaHCO3 given below has to be calculated.

(g)

Expert Solution

Explanation of Solution

HCO3− acts as both acid and base.

HCO3−(aq)+H2O(aq)⇌H3O+(aq)+CO32−(aq)Ka=[H3O+][CO32−][HCO3−]=4.7×10−11HCO3−(aq)+H2O(aq)⇌OH−(aq)+H2CO3(aq)Kb=[H2CO3][OH−][HCO3−]=2.3×10−8

HCO3− is more basic than acidic.

A table can be set up as shown below.

HCO3−(aq)H2CO3(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=2.3×10−8

Then,

Assuming x is very small, it can be written 0.20−x≅0.20.

Kb=(x)(x)(0.10−x)=2.3×10−8⇒x2=(2.3×10−8)×(0.10)⇒x=(2.3×10−8)×(0.10)=4.8×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(4.8×10−5)=4.32.pH=14.00−pOH=14.00−4.32=9.68.

Therefore, the pH of 0.10M aqueous solution of NaHCO3 is 9.68.

Answer 43

(g)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions of NaHCO3 given below has to be calculated.

Answer 44

(g)

Expert Solution

Explanation of Solution

HCO3− acts as both acid and base.

HCO3−(aq)+H2O(aq)⇌H3O+(aq)+CO32−(aq)Ka=[H3O+][CO32−][HCO3−]=4.7×10−11HCO3−(aq)+H2O(aq)⇌OH−(aq)+H2CO3(aq)Kb=[H2CO3][OH−][HCO3−]=2.3×10−8

HCO3− is more basic than acidic.

A table can be set up as shown below.

HCO3−(aq)H2CO3(aq)OH−(aq)Initial conc.(M)0.1001×10−7Change in conc.(M)−x+x+xEquilibrium conc.(M)0.10−xxx

The concentration of OH− comes from its autoionization that is 1.0×10−7. It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(0.10−x)=2.3×10−8

Then,

Assuming x is very small, it can be written 0.20−x≅0.20.

Kb=(x)(x)(0.10−x)=2.3×10−8⇒x2=(2.3×10−8)×(0.10)⇒x=(2.3×10−8)×(0.10)=4.8×10−5M=[OH−]

The pH of the solution can be calculated as shown below.

pOH=−log[OH−]=−log(4.8×10−5)=4.32.pH=14.00−pOH=14.00−4.32=9.68.

Therefore, the pH of 0.10M aqueous solution of NaHCO3 is 9.68.

Answer 45

(g)

Expert Solution

Answer 46

(g)

Expert Solution

Answer 47

Expert Solution

Answer 48

(h)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions BaCl2 given below has to be calculated.

(h)

Expert Solution

Explanation of Solution

BaCl2→Ba2++2Cl−

Neither of these ions affect the pH of a water solution, because Ba2+ is the cation of Ba(OH)2 and Cl– is a weak base. So pH is 7.00 and no calculation is needed.

Answer 49

(h)

Interpretation Introduction

Interpretation:

The pH of 0.10 M aqueous solutions BaCl2 given below has to be calculated.

Answer 50

(h)

Expert Solution

Explanation of Solution

BaCl2→Ba2++2Cl−

Neither of these ions affect the pH of a water solution, because Ba2+ is the cation of Ba(OH)2 and Cl– is a weak base. So pH is 7.00 and no calculation is needed.

Answer 51

(h)

Expert Solution

Answer 52

(h)

Expert Solution

Answer 53

Expert Solution

Answer 54

Ch. 14.1 - Identify each molecule or ion as a Brnsted-Lowry...Ch. 14.1 - Using Le Chatelier’s Principle Use Le Chatelier’s...Ch. 14.1 - Prob. 14.3E Ch. 14.1 - Complete the table. Ch. 14.1 - Prob. 14.4E Ch. 14.1 - Prob. 14.5E Ch. 14.2 - Prob. 14.6CE Ch. 14.2 - Prob. 14.7E Ch. 14.3 - Prob. 14.2PSP Ch. 14.3 - Prob. 14.3PSP

The p H of 0.10 M aqueous solutions of H N O 2 given below has to be calculated.

Concept explainers

Videos

Interpretation:

The pH of 0.10 M aqueous solutions of HNO2 given below has to be calculated.

Explanation of Solution

Interpretation:

The pH of 0.10 M aqueous solutions of NH4Cl given below has to be calculated.

Explanation of Solution

Interpretation:

The pH of 0.10 M aqueous solutions of NaF given below has to be calculated.

Explanation of Solution

Interpretation:

The pH of 0.10 M aqueous solutions of Mg(CH3COO)2 given below has to be calculated.

Explanation of Solution

Interpretation:

The pH of 0.10 M aqueous solutions of BaO given below has to be calculated.

Explanation of Solution

Interpretation:

The pH of 0.10 M aqueous solutions of KHSO4 given below has to be calculated.

Explanation of Solution

Interpretation:

The pH of 0.10 M aqueous solutions of NaHCO3 given below has to be calculated.

Explanation of Solution

Interpretation:

The pH of 0.10 M aqueous solutions BaCl2 given below has to be calculated.

Explanation of Solution

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Chapter 14 Solutions

The p H of 0.10 M aqueous solutions of H N O 2 given below has to be calculated.

Concept explainers

Videos

Interpretation: The pH of 0.10 M aqueous solutions of HNO2 given below has to be calculated.

Explanation of Solution

Interpretation: The pH of 0.10 M aqueous solutions of NH4Cl given below has to be calculated.

Explanation of Solution

Interpretation: The pH of 0.10 M aqueous solutions of NaF given below has to be calculated.

Explanation of Solution

Interpretation: The pH of 0.10 M aqueous solutions of Mg(CH3COO)2 given below has to be calculated.

Explanation of Solution

Interpretation: The pH of 0.10 M aqueous solutions of BaO given below has to be calculated.

Explanation of Solution

Interpretation: The pH of 0.10 M aqueous solutions of KHSO4 given below has to be calculated.

Explanation of Solution

Interpretation: The pH of 0.10 M aqueous solutions of NaHCO3 given below has to be calculated.

Explanation of Solution

Interpretation: The pH of 0.10 M aqueous solutions BaCl2 given below has to be calculated.

Explanation of Solution

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Chapter 14 Solutions

Interpretation:

The pH of 0.10 M aqueous solutions of HNO2 given below has to be calculated.

Interpretation:

The pH of 0.10 M aqueous solutions of NH4Cl given below has to be calculated.

Interpretation:

The pH of 0.10 M aqueous solutions of NaF given below has to be calculated.

Interpretation:

The pH of 0.10 M aqueous solutions of Mg(CH3COO)2 given below has to be calculated.

Interpretation:

The pH of 0.10 M aqueous solutions of BaO given below has to be calculated.

Interpretation:

The pH of 0.10 M aqueous solutions of KHSO4 given below has to be calculated.

Interpretation:

The pH of 0.10 M aqueous solutions of NaHCO3 given below has to be calculated.

Interpretation:

The pH of 0.10 M aqueous solutions BaCl2 given below has to be calculated.