When aniline, C 6 H 5 NH 2 ( K b = 7.4 × 10 − 10 ) , reacts with a strong acid, its conjugate acid, C 6 H 5 NH 3 + , is formed. Calculate the pH of a 0.100 M solution of C 6 H 5 NH 3 + and compare it with the pH of acetic acid ( K a = 1.86 × 10 − 5 ) .

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Answer 1

Textbook Question

Chapter 22, Problem 26QAP

When aniline, C₆H₅NH₂ ( K b = 7.4 × 10 − 10 ) , reacts with a strong acid, its conjugate acid, C₆H₅NH₃⁺, is formed. Calculate the pH of a 0.100 M solution of C₆H₅NH₃+ and compare it with the pH of acetic acid ( K a = 1.86 × 10 − 5 ) .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The pH of 0.100 M solution of C6H4NH3+ is to be obtained and compared with the pH of acetic acid.

Concept introduction:

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

aA + bB⇄cC + dD

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Every concentration term is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

Kc = [C]c[D]d[A]a[B]b

pH scale is a type of measurement of H+ ion concentration in the solution. It is taken as a negative log of H+ ion concentration as:

pH=−log1o[H+]

For the equilibrium conditions:

ka×kb=kw=10−14

Where, k_a is acid dissociation constant, k_b is base dissociation constant and k_w is ionic product of water whose value is fixed as 10-14.

Answer to Problem 26QAP

The required pH=2.93 which is comparable to the pH of acetic acid pH=2.87.

Explanation of Solution

Given Information:

The aniline C6H4NH2(kb=7.4×10−10) reacts with a strong acid. Its conjugate acid is C6H4NH3+.

For, acetic acid CH3COOH(ka=1.86×10−5).

The conjugate acid of aniline is a weak acid which dissociates as:

C6H4NH3+(aq)+H2O⇌H3O+(aq)+C6H4NH2(aq)

Let’s discuss the main reaction in one direction and creating ICE table as:

C6H4NH3+(aq)+H2O⇌H3O+(aq)+C6H4NH2(aq)I 0.100M 0 0C −x x xE (0.100M−x) x x

Hence, we have:

The equilibrium constant expression will be:

Ka=[H3O+][C6H4NH2][C6H4NH3+]

Putting the standard value of K_a :

ka=kwkb=10−147.4×10−10=1.35×10−5

Now, the equilibrium expression is:

1.35×10−5=x×x[0.100M−x]0.100−x≈0.100x2=1.35×10−5×0.100=1.35×10−6x=1.162×10−3

Further, pH is calculated as:

pH=−log[H+] =−log[1.162×10−3]=2.93

Further the pH of acetic acid is calculated by its dissociation as:

CH3COOH(aq)+H2O(l)⇌H3O+(aq)+CH3COO−(aq)I 0.100M. 0 0C −x x xE (0.100M−x) x x

Hence, we have:

The equilibrium constant expression will be:

Ka=[H3O+][CH3COO−][CH3COOH]

Putting the standard value of K_a the equilibrium expression is:

1.8×10−5=x×x[0.100M−x]0.100−x≈0.100x2=1.8×10−5×0.100=1.8×10−6x=1.34×10−3

Further, pH is calculated as:

pH=−log[H+] =−log[1.34×10−3]=2.87

Conclusion

The required pH=2.93 which is comparable to the pH of acetic acid pH=2.87.

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Answer 2

Textbook Question

Chapter 22, Problem 26QAP

When aniline, C₆H₅NH₂ ( K b = 7.4 × 10 − 10 ) , reacts with a strong acid, its conjugate acid, C₆H₅NH₃⁺, is formed. Calculate the pH of a 0.100 M solution of C₆H₅NH₃+ and compare it with the pH of acetic acid ( K a = 1.86 × 10 − 5 ) .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The pH of 0.100 M solution of C6H4NH3+ is to be obtained and compared with the pH of acetic acid.

Concept introduction:

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

aA + bB⇄cC + dD

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Every concentration term is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

Kc = [C]c[D]d[A]a[B]b

pH scale is a type of measurement of H+ ion concentration in the solution. It is taken as a negative log of H+ ion concentration as:

pH=−log1o[H+]

For the equilibrium conditions:

ka×kb=kw=10−14

Where, k_a is acid dissociation constant, k_b is base dissociation constant and k_w is ionic product of water whose value is fixed as 10-14.

Answer to Problem 26QAP

The required pH=2.93 which is comparable to the pH of acetic acid pH=2.87.

Explanation of Solution

Given Information:

The aniline C6H4NH2(kb=7.4×10−10) reacts with a strong acid. Its conjugate acid is C6H4NH3+.

For, acetic acid CH3COOH(ka=1.86×10−5).

The conjugate acid of aniline is a weak acid which dissociates as:

C6H4NH3+(aq)+H2O⇌H3O+(aq)+C6H4NH2(aq)

Let’s discuss the main reaction in one direction and creating ICE table as:

C6H4NH3+(aq)+H2O⇌H3O+(aq)+C6H4NH2(aq)I 0.100M 0 0C −x x xE (0.100M−x) x x

Hence, we have:

The equilibrium constant expression will be:

Ka=[H3O+][C6H4NH2][C6H4NH3+]

Putting the standard value of K_a :

ka=kwkb=10−147.4×10−10=1.35×10−5

Now, the equilibrium expression is:

1.35×10−5=x×x[0.100M−x]0.100−x≈0.100x2=1.35×10−5×0.100=1.35×10−6x=1.162×10−3

Further, pH is calculated as:

pH=−log[H+] =−log[1.162×10−3]=2.93

Further the pH of acetic acid is calculated by its dissociation as:

CH3COOH(aq)+H2O(l)⇌H3O+(aq)+CH3COO−(aq)I 0.100M. 0 0C −x x xE (0.100M−x) x x

Hence, we have:

The equilibrium constant expression will be:

Ka=[H3O+][CH3COO−][CH3COOH]

Putting the standard value of K_a the equilibrium expression is:

1.8×10−5=x×x[0.100M−x]0.100−x≈0.100x2=1.8×10−5×0.100=1.8×10−6x=1.34×10−3

Further, pH is calculated as:

pH=−log[H+] =−log[1.34×10−3]=2.87

Conclusion

The required pH=2.93 which is comparable to the pH of acetic acid pH=2.87.

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Answer 3

Textbook Question

Chapter 22, Problem 26QAP

When aniline, C₆H₅NH₂ ( K b = 7.4 × 10 − 10 ) , reacts with a strong acid, its conjugate acid, C₆H₅NH₃⁺, is formed. Calculate the pH of a 0.100 M solution of C₆H₅NH₃+ and compare it with the pH of acetic acid ( K a = 1.86 × 10 − 5 ) .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The pH of 0.100 M solution of C6H4NH3+ is to be obtained and compared with the pH of acetic acid.

Concept introduction:

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

aA + bB⇄cC + dD

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Every concentration term is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

Kc = [C]c[D]d[A]a[B]b

pH scale is a type of measurement of H+ ion concentration in the solution. It is taken as a negative log of H+ ion concentration as:

pH=−log1o[H+]

For the equilibrium conditions:

ka×kb=kw=10−14

Where, k_a is acid dissociation constant, k_b is base dissociation constant and k_w is ionic product of water whose value is fixed as 10-14.

Answer to Problem 26QAP

The required pH=2.93 which is comparable to the pH of acetic acid pH=2.87.

Explanation of Solution

Given Information:

The aniline C6H4NH2(kb=7.4×10−10) reacts with a strong acid. Its conjugate acid is C6H4NH3+.

For, acetic acid CH3COOH(ka=1.86×10−5).

The conjugate acid of aniline is a weak acid which dissociates as:

C6H4NH3+(aq)+H2O⇌H3O+(aq)+C6H4NH2(aq)

Let’s discuss the main reaction in one direction and creating ICE table as:

C6H4NH3+(aq)+H2O⇌H3O+(aq)+C6H4NH2(aq)I 0.100M 0 0C −x x xE (0.100M−x) x x

Hence, we have:

The equilibrium constant expression will be:

Ka=[H3O+][C6H4NH2][C6H4NH3+]

Putting the standard value of K_a :

ka=kwkb=10−147.4×10−10=1.35×10−5

Now, the equilibrium expression is:

1.35×10−5=x×x[0.100M−x]0.100−x≈0.100x2=1.35×10−5×0.100=1.35×10−6x=1.162×10−3

Further, pH is calculated as:

pH=−log[H+] =−log[1.162×10−3]=2.93

Further the pH of acetic acid is calculated by its dissociation as:

CH3COOH(aq)+H2O(l)⇌H3O+(aq)+CH3COO−(aq)I 0.100M. 0 0C −x x xE (0.100M−x) x x

Hence, we have:

The equilibrium constant expression will be:

Ka=[H3O+][CH3COO−][CH3COOH]

Putting the standard value of K_a the equilibrium expression is:

1.8×10−5=x×x[0.100M−x]0.100−x≈0.100x2=1.8×10−5×0.100=1.8×10−6x=1.34×10−3

Further, pH is calculated as:

pH=−log[H+] =−log[1.34×10−3]=2.87

Conclusion

The required pH=2.93 which is comparable to the pH of acetic acid pH=2.87.

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Answer 4

Textbook Question

Chapter 22, Problem 26QAP

When aniline, C₆H₅NH₂ ( K b = 7.4 × 10 − 10 ) , reacts with a strong acid, its conjugate acid, C₆H₅NH₃⁺, is formed. Calculate the pH of a 0.100 M solution of C₆H₅NH₃+ and compare it with the pH of acetic acid ( K a = 1.86 × 10 − 5 ) .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The pH of 0.100 M solution of C6H4NH3+ is to be obtained and compared with the pH of acetic acid.

Concept introduction:

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

aA + bB⇄cC + dD

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Every concentration term is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

Kc = [C]c[D]d[A]a[B]b

pH scale is a type of measurement of H+ ion concentration in the solution. It is taken as a negative log of H+ ion concentration as:

pH=−log1o[H+]

For the equilibrium conditions:

ka×kb=kw=10−14

Where, k_a is acid dissociation constant, k_b is base dissociation constant and k_w is ionic product of water whose value is fixed as 10-14.

Answer to Problem 26QAP

The required pH=2.93 which is comparable to the pH of acetic acid pH=2.87.

Explanation of Solution

Given Information:

The aniline C6H4NH2(kb=7.4×10−10) reacts with a strong acid. Its conjugate acid is C6H4NH3+.

For, acetic acid CH3COOH(ka=1.86×10−5).

The conjugate acid of aniline is a weak acid which dissociates as:

C6H4NH3+(aq)+H2O⇌H3O+(aq)+C6H4NH2(aq)

Let’s discuss the main reaction in one direction and creating ICE table as:

C6H4NH3+(aq)+H2O⇌H3O+(aq)+C6H4NH2(aq)I 0.100M 0 0C −x x xE (0.100M−x) x x

Hence, we have:

The equilibrium constant expression will be:

Ka=[H3O+][C6H4NH2][C6H4NH3+]

Putting the standard value of K_a :

ka=kwkb=10−147.4×10−10=1.35×10−5

Now, the equilibrium expression is:

1.35×10−5=x×x[0.100M−x]0.100−x≈0.100x2=1.35×10−5×0.100=1.35×10−6x=1.162×10−3

Further, pH is calculated as:

pH=−log[H+] =−log[1.162×10−3]=2.93

Further the pH of acetic acid is calculated by its dissociation as:

CH3COOH(aq)+H2O(l)⇌H3O+(aq)+CH3COO−(aq)I 0.100M. 0 0C −x x xE (0.100M−x) x x

Hence, we have:

The equilibrium constant expression will be:

Ka=[H3O+][CH3COO−][CH3COOH]

Putting the standard value of K_a the equilibrium expression is:

1.8×10−5=x×x[0.100M−x]0.100−x≈0.100x2=1.8×10−5×0.100=1.8×10−6x=1.34×10−3

Further, pH is calculated as:

pH=−log[H+] =−log[1.34×10−3]=2.87

Conclusion

The required pH=2.93 which is comparable to the pH of acetic acid pH=2.87.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The pH of 0.100 M solution of C6H4NH3+ is to be obtained and compared with the pH of acetic acid.

Concept introduction:

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

aA + bB⇄cC + dD

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Every concentration term is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

Kc = [C]c[D]d[A]a[B]b

pH scale is a type of measurement of H+ ion concentration in the solution. It is taken as a negative log of H+ ion concentration as:

pH=−log1o[H+]

For the equilibrium conditions:

ka×kb=kw=10−14

Where, k_a is acid dissociation constant, k_b is base dissociation constant and k_w is ionic product of water whose value is fixed as 10-14.

Answer to Problem 26QAP

The required pH=2.93 which is comparable to the pH of acetic acid pH=2.87.

Explanation of Solution

Given Information:

The aniline C6H4NH2(kb=7.4×10−10) reacts with a strong acid. Its conjugate acid is C6H4NH3+.

For, acetic acid CH3COOH(ka=1.86×10−5).

The conjugate acid of aniline is a weak acid which dissociates as:

C6H4NH3+(aq)+H2O⇌H3O+(aq)+C6H4NH2(aq)

Let’s discuss the main reaction in one direction and creating ICE table as:

C6H4NH3+(aq)+H2O⇌H3O+(aq)+C6H4NH2(aq)I 0.100M 0 0C −x x xE (0.100M−x) x x

Hence, we have:

The equilibrium constant expression will be:

Ka=[H3O+][C6H4NH2][C6H4NH3+]

Putting the standard value of K_a :

ka=kwkb=10−147.4×10−10=1.35×10−5

Now, the equilibrium expression is:

1.35×10−5=x×x[0.100M−x]0.100−x≈0.100x2=1.35×10−5×0.100=1.35×10−6x=1.162×10−3

Further, pH is calculated as:

pH=−log[H+] =−log[1.162×10−3]=2.93

Further the pH of acetic acid is calculated by its dissociation as:

CH3COOH(aq)+H2O(l)⇌H3O+(aq)+CH3COO−(aq)I 0.100M. 0 0C −x x xE (0.100M−x) x x

Hence, we have:

The equilibrium constant expression will be:

Ka=[H3O+][CH3COO−][CH3COOH]

Putting the standard value of K_a the equilibrium expression is:

1.8×10−5=x×x[0.100M−x]0.100−x≈0.100x2=1.8×10−5×0.100=1.8×10−6x=1.34×10−3

Further, pH is calculated as:

pH=−log[H+] =−log[1.34×10−3]=2.87

Conclusion

The required pH=2.93 which is comparable to the pH of acetic acid pH=2.87.

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The pH of 0.100 M solution of C6H4NH3+ is to be obtained and compared with the pH of acetic acid.

Concept introduction:

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

aA + bB⇄cC + dD

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Every concentration term is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

Kc = [C]c[D]d[A]a[B]b

pH scale is a type of measurement of H+ ion concentration in the solution. It is taken as a negative log of H+ ion concentration as:

pH=−log1o[H+]

For the equilibrium conditions:

ka×kb=kw=10−14

Where, k_a is acid dissociation constant, k_b is base dissociation constant and k_w is ionic product of water whose value is fixed as 10-14.

Answer to Problem 26QAP

The required pH=2.93 which is comparable to the pH of acetic acid pH=2.87.

Explanation of Solution

Given Information:

The aniline C6H4NH2(kb=7.4×10−10) reacts with a strong acid. Its conjugate acid is C6H4NH3+.

For, acetic acid CH3COOH(ka=1.86×10−5).

The conjugate acid of aniline is a weak acid which dissociates as:

C6H4NH3+(aq)+H2O⇌H3O+(aq)+C6H4NH2(aq)

Let’s discuss the main reaction in one direction and creating ICE table as:

C6H4NH3+(aq)+H2O⇌H3O+(aq)+C6H4NH2(aq)I 0.100M 0 0C −x x xE (0.100M−x) x x

Hence, we have:

The equilibrium constant expression will be:

Ka=[H3O+][C6H4NH2][C6H4NH3+]

Putting the standard value of K_a :

ka=kwkb=10−147.4×10−10=1.35×10−5

Now, the equilibrium expression is:

1.35×10−5=x×x[0.100M−x]0.100−x≈0.100x2=1.35×10−5×0.100=1.35×10−6x=1.162×10−3

Further, pH is calculated as:

pH=−log[H+] =−log[1.162×10−3]=2.93

Further the pH of acetic acid is calculated by its dissociation as:

CH3COOH(aq)+H2O(l)⇌H3O+(aq)+CH3COO−(aq)I 0.100M. 0 0C −x x xE (0.100M−x) x x

Hence, we have:

The equilibrium constant expression will be:

Ka=[H3O+][CH3COO−][CH3COOH]

Putting the standard value of K_a the equilibrium expression is:

1.8×10−5=x×x[0.100M−x]0.100−x≈0.100x2=1.8×10−5×0.100=1.8×10−6x=1.34×10−3

Further, pH is calculated as:

pH=−log[H+] =−log[1.34×10−3]=2.87

Conclusion

The required pH=2.93 which is comparable to the pH of acetic acid pH=2.87.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation:

The pH of 0.100 M solution of C6H4NH3+ is to be obtained and compared with the pH of acetic acid.

Concept introduction:

The relationship between the concentration of products and reactants at equilibrium for a general reaction:

aA + bB⇄cC + dD

Where A, B, C, and D represents chemical species and a, b, c, and d are the coefficients for balanced reaction.

The equilibrium expression, K_c for reversible reaction is determined by multiplying the concentrations of products together and divided by the concentrations of the reactants. Every concentration term is raised to the power that is equal to the coefficient in the balanced reaction. So, the expression is:

Kc = [C]c[D]d[A]a[B]b