The initial pH value of the weak base before adding the acid is to be determined. Concept introduction: A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

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Answer 1

Question

Chapter 8, Problem 167AE

(a)

Interpretation Introduction

Interpretation:

The initial pH value of the weak base before adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(a)

Expert Solution

Answer to Problem 167AE

The pH of the initial weak base solution is 9.67.

Explanation of Solution

Initial pH of the analyte solution; HONH2 is a weak base that forms an equilibrium when dissolved in water. The equilibrium is as follows.

HONH2+H2O⇌HONH3++OH−

The amount of weak base at the beginning =0.2mol/L×100.0×10−3L=20×10−3mol . By constructing an ICE table, the amount of lactate ion in the solution after the acid dissociation can be determined.

Reaction	Weak base	Conjugate acid	OH-
Initial	0.2	0	0
Change	-x	+x	+x
Equilibrium	(0.2-x)	x	x

Kb=[HONH3+][OH−][HONH2]1.1×10−8=[x][x][0.2−x]=x2[0.2−x]

For small base dissociation constant value, the value of x can be neglected from the denominator.

Thus,

1.1×10−8=x20.2x=[OH−]=4.7×10−5

Then the pH of the initial solution can be determined.

pOH=−log[OH−]=−log[4.7×10−5]=4.33pH=14.00−4.33=9.67

(b)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(b)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.88.

Explanation of Solution

Addition of 25.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.02	0	0	0
Add	0	0.0025
Change	-0.0025	-0.0025	0.0025	0.0025
Equilibrium	0.0175	0	0.0025	0.0025

The base dissociation constant value is 1.1×10−8

Thus,

pKb=−logKb=−log(1.1×10−8)=7.96

Since, volume is same thus, number of moles can be used instead of concentrations.

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[2.50][17.5]=7.12

Now, pH can be calculated as follows:

pH=14−pOH=14−7.12=6.88

Thus, the pH of the solution is 6.88.

(c)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (70 mL HCl) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(c)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.31.

Explanation of Solution

Addition of 70.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×70.0×10−3L=70×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.0200	0	0	0
Add	0	0.0070
Change	-0.0070	-0.0070	0.0070	0.0070
Equilibrium	0.0130	0	0.0070	0.0070

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[0.007][0.013]=7.69pH=14−7.69=6.31

Thus, the pH of the solution is 6.31.

(d)

Interpretation Introduction

Interpretation:

The pH value of the weak base at equivalent point needs to be determined.

Concept introduction:

At equivalent point the concentration of acid and base (in which one of the species is weak) is equal in the solution. Thus, the pH will depend on the acidic or basic salt formed.

(d)

Expert Solution

Answer to Problem 167AE

The pH of the solution at the equivalence point is 3.60.

Explanation of Solution

At the equivalence point, The amount of acid added =20×10−3mol

The volume of acid added =20×10−3mol0.1mol/L=0.2L=200.0mL

At this point, there is no excess acid or base. Therefore, the only possible reaction here is the dissociation of the conjugate acid of the ammonia (that is ammonium ion).

HONH3++H2O⇌HONH2+H3O+

Thereafter, using the Ka value for this weak base, the amount of hydrogen ions in the solution can be determined to get the pH value at this point.

Now, total volume of the solution is 300 mL thus, molarity of HONH₃+ solution will be:

M=nV=0.02 mol300 mL=0.0667 M

Reaction	HONH₃+	HONH2	H+
Initial	0.0667	0	0
Change	-x	x	x
Equilibrium	(0.0667-x)	x	x

Then the pH can be calculated as follows:

Ka=[HONH2][H+][HONH3+]9.09×10−7=[x][x]0.0667=x20.0067x=2.5×10−4 M

This value is equal to the amount of hydrogen ions in the solution.

pH=−log[H+]=−log[2.5×10−4]=3.60

Thus, the pH value at equivalence point is 3.60.

(e)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (300 mL) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(e)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding acid is 1.60 .

Explanation of Solution

Addition of 300.0mLacid, The amount of acid added =0.1mol/L×300.0×10−3L=0.03mol

At this point, there is excess acid in the solution, thus, the pH can be calculated as follows:

Excess amount of acid in the solution =0.03−0.02=0.01mol

The amount of hydrogen ions =0.01mol

The concentration of hydrogen ions =0.01mol(100.0+300.0)×10−3L=0.025mol/L

pH=−log[H+]=−log[0.025]=1.60

(f)

Interpretation Introduction

Interpretation:

The volume of HCl needs to be determined that is added to the solution if pH is 6.04.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(f)

Expert Solution

Answer to Problem 167AE

100 mL

Explanation of Solution

The pKb value for NH₂OH is 7.96. From the given pH value, pOH of the solution can be calculated as follows:

pOH=14−pH=14−6.04=7.96

The Henderson Hesselbalch equation can be represented as follows:

pOH=pKb+log[salt][base]

Since, the volume is same thus, concentration can be replaced by number of moles.

7.96=7.96+log[salt][base]

Thus, the number of moles of salt and base will be same that is NH3OH+ and NH₂OH respectively.

This is the condition which is halfway to the equivalence point. For the equivalence point, volume of HCl required was 200 mL thus, at halfway equivalence point the volume will be half to it. Therefore, volume of HCl is 100 mL.

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Answer 2

Question

Chapter 8, Problem 167AE

(a)

Interpretation Introduction

Interpretation:

The initial pH value of the weak base before adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(a)

Expert Solution

Answer to Problem 167AE

The pH of the initial weak base solution is 9.67.

Explanation of Solution

Initial pH of the analyte solution; HONH2 is a weak base that forms an equilibrium when dissolved in water. The equilibrium is as follows.

HONH2+H2O⇌HONH3++OH−

The amount of weak base at the beginning =0.2mol/L×100.0×10−3L=20×10−3mol . By constructing an ICE table, the amount of lactate ion in the solution after the acid dissociation can be determined.

Reaction	Weak base	Conjugate acid	OH-
Initial	0.2	0	0
Change	-x	+x	+x
Equilibrium	(0.2-x)	x	x

Kb=[HONH3+][OH−][HONH2]1.1×10−8=[x][x][0.2−x]=x2[0.2−x]

For small base dissociation constant value, the value of x can be neglected from the denominator.

Thus,

1.1×10−8=x20.2x=[OH−]=4.7×10−5

Then the pH of the initial solution can be determined.

pOH=−log[OH−]=−log[4.7×10−5]=4.33pH=14.00−4.33=9.67

(b)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(b)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.88.

Explanation of Solution

Addition of 25.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.02	0	0	0
Add	0	0.0025
Change	-0.0025	-0.0025	0.0025	0.0025
Equilibrium	0.0175	0	0.0025	0.0025

The base dissociation constant value is 1.1×10−8

Thus,

pKb=−logKb=−log(1.1×10−8)=7.96

Since, volume is same thus, number of moles can be used instead of concentrations.

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[2.50][17.5]=7.12

Now, pH can be calculated as follows:

pH=14−pOH=14−7.12=6.88

Thus, the pH of the solution is 6.88.

(c)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (70 mL HCl) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(c)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.31.

Explanation of Solution

Addition of 70.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×70.0×10−3L=70×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.0200	0	0	0
Add	0	0.0070
Change	-0.0070	-0.0070	0.0070	0.0070
Equilibrium	0.0130	0	0.0070	0.0070

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[0.007][0.013]=7.69pH=14−7.69=6.31

Thus, the pH of the solution is 6.31.

(d)

Interpretation Introduction

Interpretation:

The pH value of the weak base at equivalent point needs to be determined.

Concept introduction:

At equivalent point the concentration of acid and base (in which one of the species is weak) is equal in the solution. Thus, the pH will depend on the acidic or basic salt formed.

(d)

Expert Solution

Answer to Problem 167AE

The pH of the solution at the equivalence point is 3.60.

Explanation of Solution

At the equivalence point, The amount of acid added =20×10−3mol

The volume of acid added =20×10−3mol0.1mol/L=0.2L=200.0mL

At this point, there is no excess acid or base. Therefore, the only possible reaction here is the dissociation of the conjugate acid of the ammonia (that is ammonium ion).

HONH3++H2O⇌HONH2+H3O+

Thereafter, using the Ka value for this weak base, the amount of hydrogen ions in the solution can be determined to get the pH value at this point.

Now, total volume of the solution is 300 mL thus, molarity of HONH₃+ solution will be:

M=nV=0.02 mol300 mL=0.0667 M

Reaction	HONH₃+	HONH2	H+
Initial	0.0667	0	0
Change	-x	x	x
Equilibrium	(0.0667-x)	x	x

Then the pH can be calculated as follows:

Ka=[HONH2][H+][HONH3+]9.09×10−7=[x][x]0.0667=x20.0067x=2.5×10−4 M

This value is equal to the amount of hydrogen ions in the solution.

pH=−log[H+]=−log[2.5×10−4]=3.60

Thus, the pH value at equivalence point is 3.60.

(e)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (300 mL) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(e)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding acid is 1.60 .

Explanation of Solution

Addition of 300.0mLacid, The amount of acid added =0.1mol/L×300.0×10−3L=0.03mol

At this point, there is excess acid in the solution, thus, the pH can be calculated as follows:

Excess amount of acid in the solution =0.03−0.02=0.01mol

The amount of hydrogen ions =0.01mol

The concentration of hydrogen ions =0.01mol(100.0+300.0)×10−3L=0.025mol/L

pH=−log[H+]=−log[0.025]=1.60

(f)

Interpretation Introduction

Interpretation:

The volume of HCl needs to be determined that is added to the solution if pH is 6.04.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(f)

Expert Solution

Answer to Problem 167AE

100 mL

Explanation of Solution

The pKb value for NH₂OH is 7.96. From the given pH value, pOH of the solution can be calculated as follows:

pOH=14−pH=14−6.04=7.96

The Henderson Hesselbalch equation can be represented as follows:

pOH=pKb+log[salt][base]

Since, the volume is same thus, concentration can be replaced by number of moles.

7.96=7.96+log[salt][base]

Thus, the number of moles of salt and base will be same that is NH3OH+ and NH₂OH respectively.

This is the condition which is halfway to the equivalence point. For the equivalence point, volume of HCl required was 200 mL thus, at halfway equivalence point the volume will be half to it. Therefore, volume of HCl is 100 mL.

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Answer 3

Question

Chapter 8, Problem 167AE

(a)

Interpretation Introduction

Interpretation:

The initial pH value of the weak base before adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(a)

Expert Solution

Answer to Problem 167AE

The pH of the initial weak base solution is 9.67.

Explanation of Solution

Initial pH of the analyte solution; HONH2 is a weak base that forms an equilibrium when dissolved in water. The equilibrium is as follows.

HONH2+H2O⇌HONH3++OH−

The amount of weak base at the beginning =0.2mol/L×100.0×10−3L=20×10−3mol . By constructing an ICE table, the amount of lactate ion in the solution after the acid dissociation can be determined.

Reaction	Weak base	Conjugate acid	OH-
Initial	0.2	0	0
Change	-x	+x	+x
Equilibrium	(0.2-x)	x	x

Kb=[HONH3+][OH−][HONH2]1.1×10−8=[x][x][0.2−x]=x2[0.2−x]

For small base dissociation constant value, the value of x can be neglected from the denominator.

Thus,

1.1×10−8=x20.2x=[OH−]=4.7×10−5

Then the pH of the initial solution can be determined.

pOH=−log[OH−]=−log[4.7×10−5]=4.33pH=14.00−4.33=9.67

(b)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(b)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.88.

Explanation of Solution

Addition of 25.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.02	0	0	0
Add	0	0.0025
Change	-0.0025	-0.0025	0.0025	0.0025
Equilibrium	0.0175	0	0.0025	0.0025

The base dissociation constant value is 1.1×10−8

Thus,

pKb=−logKb=−log(1.1×10−8)=7.96

Since, volume is same thus, number of moles can be used instead of concentrations.

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[2.50][17.5]=7.12

Now, pH can be calculated as follows:

pH=14−pOH=14−7.12=6.88

Thus, the pH of the solution is 6.88.

(c)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (70 mL HCl) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(c)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.31.

Explanation of Solution

Addition of 70.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×70.0×10−3L=70×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.0200	0	0	0
Add	0	0.0070
Change	-0.0070	-0.0070	0.0070	0.0070
Equilibrium	0.0130	0	0.0070	0.0070

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[0.007][0.013]=7.69pH=14−7.69=6.31

Thus, the pH of the solution is 6.31.

(d)

Interpretation Introduction

Interpretation:

The pH value of the weak base at equivalent point needs to be determined.

Concept introduction:

At equivalent point the concentration of acid and base (in which one of the species is weak) is equal in the solution. Thus, the pH will depend on the acidic or basic salt formed.

(d)

Expert Solution

Answer to Problem 167AE

The pH of the solution at the equivalence point is 3.60.

Explanation of Solution

At the equivalence point, The amount of acid added =20×10−3mol

The volume of acid added =20×10−3mol0.1mol/L=0.2L=200.0mL

At this point, there is no excess acid or base. Therefore, the only possible reaction here is the dissociation of the conjugate acid of the ammonia (that is ammonium ion).

HONH3++H2O⇌HONH2+H3O+

Thereafter, using the Ka value for this weak base, the amount of hydrogen ions in the solution can be determined to get the pH value at this point.

Now, total volume of the solution is 300 mL thus, molarity of HONH₃+ solution will be:

M=nV=0.02 mol300 mL=0.0667 M

Reaction	HONH₃+	HONH2	H+
Initial	0.0667	0	0
Change	-x	x	x
Equilibrium	(0.0667-x)	x	x

Then the pH can be calculated as follows:

Ka=[HONH2][H+][HONH3+]9.09×10−7=[x][x]0.0667=x20.0067x=2.5×10−4 M

This value is equal to the amount of hydrogen ions in the solution.

pH=−log[H+]=−log[2.5×10−4]=3.60

Thus, the pH value at equivalence point is 3.60.

(e)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (300 mL) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(e)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding acid is 1.60 .

Explanation of Solution

Addition of 300.0mLacid, The amount of acid added =0.1mol/L×300.0×10−3L=0.03mol

At this point, there is excess acid in the solution, thus, the pH can be calculated as follows:

Excess amount of acid in the solution =0.03−0.02=0.01mol

The amount of hydrogen ions =0.01mol

The concentration of hydrogen ions =0.01mol(100.0+300.0)×10−3L=0.025mol/L

pH=−log[H+]=−log[0.025]=1.60

(f)

Interpretation Introduction

Interpretation:

The volume of HCl needs to be determined that is added to the solution if pH is 6.04.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(f)

Expert Solution

Answer to Problem 167AE

100 mL

Explanation of Solution

The pKb value for NH₂OH is 7.96. From the given pH value, pOH of the solution can be calculated as follows:

pOH=14−pH=14−6.04=7.96

The Henderson Hesselbalch equation can be represented as follows:

pOH=pKb+log[salt][base]

Since, the volume is same thus, concentration can be replaced by number of moles.

7.96=7.96+log[salt][base]

Thus, the number of moles of salt and base will be same that is NH3OH+ and NH₂OH respectively.

This is the condition which is halfway to the equivalence point. For the equivalence point, volume of HCl required was 200 mL thus, at halfway equivalence point the volume will be half to it. Therefore, volume of HCl is 100 mL.

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Answer 4

Question

Chapter 8, Problem 167AE

(a)

Interpretation Introduction

Interpretation:

The initial pH value of the weak base before adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(a)

Expert Solution

Answer to Problem 167AE

The pH of the initial weak base solution is 9.67.

Explanation of Solution

Initial pH of the analyte solution; HONH2 is a weak base that forms an equilibrium when dissolved in water. The equilibrium is as follows.

HONH2+H2O⇌HONH3++OH−

The amount of weak base at the beginning =0.2mol/L×100.0×10−3L=20×10−3mol . By constructing an ICE table, the amount of lactate ion in the solution after the acid dissociation can be determined.

Reaction	Weak base	Conjugate acid	OH-
Initial	0.2	0	0
Change	-x	+x	+x
Equilibrium	(0.2-x)	x	x

Kb=[HONH3+][OH−][HONH2]1.1×10−8=[x][x][0.2−x]=x2[0.2−x]

For small base dissociation constant value, the value of x can be neglected from the denominator.

Thus,

1.1×10−8=x20.2x=[OH−]=4.7×10−5

Then the pH of the initial solution can be determined.

pOH=−log[OH−]=−log[4.7×10−5]=4.33pH=14.00−4.33=9.67

(b)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(b)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.88.

Explanation of Solution

Addition of 25.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.02	0	0	0
Add	0	0.0025
Change	-0.0025	-0.0025	0.0025	0.0025
Equilibrium	0.0175	0	0.0025	0.0025

The base dissociation constant value is 1.1×10−8

Thus,

pKb=−logKb=−log(1.1×10−8)=7.96

Since, volume is same thus, number of moles can be used instead of concentrations.

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[2.50][17.5]=7.12

Now, pH can be calculated as follows:

pH=14−pOH=14−7.12=6.88

Thus, the pH of the solution is 6.88.

(c)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (70 mL HCl) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(c)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.31.

Explanation of Solution

Addition of 70.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×70.0×10−3L=70×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.0200	0	0	0
Add	0	0.0070
Change	-0.0070	-0.0070	0.0070	0.0070
Equilibrium	0.0130	0	0.0070	0.0070

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[0.007][0.013]=7.69pH=14−7.69=6.31

Thus, the pH of the solution is 6.31.

(d)

Interpretation Introduction

Interpretation:

The pH value of the weak base at equivalent point needs to be determined.

Concept introduction:

At equivalent point the concentration of acid and base (in which one of the species is weak) is equal in the solution. Thus, the pH will depend on the acidic or basic salt formed.

(d)

Expert Solution

Answer to Problem 167AE

The pH of the solution at the equivalence point is 3.60.

Explanation of Solution

At the equivalence point, The amount of acid added =20×10−3mol

The volume of acid added =20×10−3mol0.1mol/L=0.2L=200.0mL

At this point, there is no excess acid or base. Therefore, the only possible reaction here is the dissociation of the conjugate acid of the ammonia (that is ammonium ion).

HONH3++H2O⇌HONH2+H3O+

Thereafter, using the Ka value for this weak base, the amount of hydrogen ions in the solution can be determined to get the pH value at this point.

Now, total volume of the solution is 300 mL thus, molarity of HONH₃+ solution will be:

M=nV=0.02 mol300 mL=0.0667 M

Reaction	HONH₃+	HONH2	H+
Initial	0.0667	0	0
Change	-x	x	x
Equilibrium	(0.0667-x)	x	x

Then the pH can be calculated as follows:

Ka=[HONH2][H+][HONH3+]9.09×10−7=[x][x]0.0667=x20.0067x=2.5×10−4 M

This value is equal to the amount of hydrogen ions in the solution.

pH=−log[H+]=−log[2.5×10−4]=3.60

Thus, the pH value at equivalence point is 3.60.

(e)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (300 mL) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(e)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding acid is 1.60 .

Explanation of Solution

Addition of 300.0mLacid, The amount of acid added =0.1mol/L×300.0×10−3L=0.03mol

At this point, there is excess acid in the solution, thus, the pH can be calculated as follows:

Excess amount of acid in the solution =0.03−0.02=0.01mol

The amount of hydrogen ions =0.01mol

The concentration of hydrogen ions =0.01mol(100.0+300.0)×10−3L=0.025mol/L

pH=−log[H+]=−log[0.025]=1.60

(f)

Interpretation Introduction

Interpretation:

The volume of HCl needs to be determined that is added to the solution if pH is 6.04.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(f)

Expert Solution

Answer to Problem 167AE

100 mL

Explanation of Solution

The pKb value for NH₂OH is 7.96. From the given pH value, pOH of the solution can be calculated as follows:

pOH=14−pH=14−6.04=7.96

The Henderson Hesselbalch equation can be represented as follows:

pOH=pKb+log[salt][base]

Since, the volume is same thus, concentration can be replaced by number of moles.

7.96=7.96+log[salt][base]

Thus, the number of moles of salt and base will be same that is NH3OH+ and NH₂OH respectively.

This is the condition which is halfway to the equivalence point. For the equivalence point, volume of HCl required was 200 mL thus, at halfway equivalence point the volume will be half to it. Therefore, volume of HCl is 100 mL.

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Answer 5

Chapter 8, Problem 167AE

(a)

Interpretation Introduction

Interpretation:

The initial pH value of the weak base before adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(a)

Expert Solution

Answer to Problem 167AE

The pH of the initial weak base solution is 9.67.

Explanation of Solution

Initial pH of the analyte solution; HONH2 is a weak base that forms an equilibrium when dissolved in water. The equilibrium is as follows.

HONH2+H2O⇌HONH3++OH−

The amount of weak base at the beginning =0.2mol/L×100.0×10−3L=20×10−3mol . By constructing an ICE table, the amount of lactate ion in the solution after the acid dissociation can be determined.

Reaction	Weak base	Conjugate acid	OH-
Initial	0.2	0	0
Change	-x	+x	+x
Equilibrium	(0.2-x)	x	x

Kb=[HONH3+][OH−][HONH2]1.1×10−8=[x][x][0.2−x]=x2[0.2−x]

For small base dissociation constant value, the value of x can be neglected from the denominator.

Thus,

1.1×10−8=x20.2x=[OH−]=4.7×10−5

Then the pH of the initial solution can be determined.

pOH=−log[OH−]=−log[4.7×10−5]=4.33pH=14.00−4.33=9.67

(b)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(b)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.88.

Explanation of Solution

Addition of 25.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.02	0	0	0
Add	0	0.0025
Change	-0.0025	-0.0025	0.0025	0.0025
Equilibrium	0.0175	0	0.0025	0.0025

The base dissociation constant value is 1.1×10−8

Thus,

pKb=−logKb=−log(1.1×10−8)=7.96

Since, volume is same thus, number of moles can be used instead of concentrations.

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[2.50][17.5]=7.12

Now, pH can be calculated as follows:

pH=14−pOH=14−7.12=6.88

Thus, the pH of the solution is 6.88.

(c)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (70 mL HCl) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(c)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.31.

Explanation of Solution

Addition of 70.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×70.0×10−3L=70×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.0200	0	0	0
Add	0	0.0070
Change	-0.0070	-0.0070	0.0070	0.0070
Equilibrium	0.0130	0	0.0070	0.0070

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[0.007][0.013]=7.69pH=14−7.69=6.31

Thus, the pH of the solution is 6.31.

(d)

Interpretation Introduction

Interpretation:

The pH value of the weak base at equivalent point needs to be determined.

Concept introduction:

At equivalent point the concentration of acid and base (in which one of the species is weak) is equal in the solution. Thus, the pH will depend on the acidic or basic salt formed.

(d)

Expert Solution

Answer to Problem 167AE

The pH of the solution at the equivalence point is 3.60.

Explanation of Solution

At the equivalence point, The amount of acid added =20×10−3mol

The volume of acid added =20×10−3mol0.1mol/L=0.2L=200.0mL

At this point, there is no excess acid or base. Therefore, the only possible reaction here is the dissociation of the conjugate acid of the ammonia (that is ammonium ion).

HONH3++H2O⇌HONH2+H3O+

Thereafter, using the Ka value for this weak base, the amount of hydrogen ions in the solution can be determined to get the pH value at this point.

Now, total volume of the solution is 300 mL thus, molarity of HONH₃+ solution will be:

M=nV=0.02 mol300 mL=0.0667 M

Reaction	HONH₃+	HONH2	H+
Initial	0.0667	0	0
Change	-x	x	x
Equilibrium	(0.0667-x)	x	x

Then the pH can be calculated as follows:

Ka=[HONH2][H+][HONH3+]9.09×10−7=[x][x]0.0667=x20.0067x=2.5×10−4 M

This value is equal to the amount of hydrogen ions in the solution.

pH=−log[H+]=−log[2.5×10−4]=3.60

Thus, the pH value at equivalence point is 3.60.

(e)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (300 mL) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(e)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding acid is 1.60 .

Explanation of Solution

Addition of 300.0mLacid, The amount of acid added =0.1mol/L×300.0×10−3L=0.03mol

At this point, there is excess acid in the solution, thus, the pH can be calculated as follows:

Excess amount of acid in the solution =0.03−0.02=0.01mol

The amount of hydrogen ions =0.01mol

The concentration of hydrogen ions =0.01mol(100.0+300.0)×10−3L=0.025mol/L

pH=−log[H+]=−log[0.025]=1.60

(f)

Interpretation Introduction

Interpretation:

The volume of HCl needs to be determined that is added to the solution if pH is 6.04.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(f)

Expert Solution

Answer to Problem 167AE

100 mL

Explanation of Solution

The pKb value for NH₂OH is 7.96. From the given pH value, pOH of the solution can be calculated as follows:

pOH=14−pH=14−6.04=7.96

The Henderson Hesselbalch equation can be represented as follows:

pOH=pKb+log[salt][base]

Since, the volume is same thus, concentration can be replaced by number of moles.

7.96=7.96+log[salt][base]

Thus, the number of moles of salt and base will be same that is NH3OH+ and NH₂OH respectively.

This is the condition which is halfway to the equivalence point. For the equivalence point, volume of HCl required was 200 mL thus, at halfway equivalence point the volume will be half to it. Therefore, volume of HCl is 100 mL.

Answer 6

Chapter 8, Problem 167AE

(a)

Interpretation Introduction

Interpretation:

The initial pH value of the weak base before adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(a)

Expert Solution

Answer to Problem 167AE

The pH of the initial weak base solution is 9.67.

Explanation of Solution

Initial pH of the analyte solution; HONH2 is a weak base that forms an equilibrium when dissolved in water. The equilibrium is as follows.

HONH2+H2O⇌HONH3++OH−

The amount of weak base at the beginning =0.2mol/L×100.0×10−3L=20×10−3mol . By constructing an ICE table, the amount of lactate ion in the solution after the acid dissociation can be determined.

Reaction	Weak base	Conjugate acid	OH-
Initial	0.2	0	0
Change	-x	+x	+x
Equilibrium	(0.2-x)	x	x

Kb=[HONH3+][OH−][HONH2]1.1×10−8=[x][x][0.2−x]=x2[0.2−x]

For small base dissociation constant value, the value of x can be neglected from the denominator.

Thus,

1.1×10−8=x20.2x=[OH−]=4.7×10−5

Then the pH of the initial solution can be determined.

pOH=−log[OH−]=−log[4.7×10−5]=4.33pH=14.00−4.33=9.67

Answer 7

Chapter 8, Problem 167AE

(a)

Interpretation Introduction

Interpretation:

The initial pH value of the weak base before adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer 8

(a)

Expert Solution

Answer to Problem 167AE

The pH of the initial weak base solution is 9.67.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Initial pH of the analyte solution; HONH2 is a weak base that forms an equilibrium when dissolved in water. The equilibrium is as follows.

HONH2+H2O⇌HONH3++OH−

The amount of weak base at the beginning =0.2mol/L×100.0×10−3L=20×10−3mol . By constructing an ICE table, the amount of lactate ion in the solution after the acid dissociation can be determined.

Reaction	Weak base	Conjugate acid	OH-
Initial	0.2	0	0
Change	-x	+x	+x
Equilibrium	(0.2-x)	x	x

Kb=[HONH3+][OH−][HONH2]1.1×10−8=[x][x][0.2−x]=x2[0.2−x]

For small base dissociation constant value, the value of x can be neglected from the denominator.

Thus,

1.1×10−8=x20.2x=[OH−]=4.7×10−5

Then the pH of the initial solution can be determined.

pOH=−log[OH−]=−log[4.7×10−5]=4.33pH=14.00−4.33=9.67

Answer 13

(b)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(b)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.88.

Explanation of Solution

Addition of 25.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.02	0	0	0
Add	0	0.0025
Change	-0.0025	-0.0025	0.0025	0.0025
Equilibrium	0.0175	0	0.0025	0.0025

The base dissociation constant value is 1.1×10−8

Thus,

pKb=−logKb=−log(1.1×10−8)=7.96

Since, volume is same thus, number of moles can be used instead of concentrations.

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[2.50][17.5]=7.12

Now, pH can be calculated as follows:

pH=14−pOH=14−7.12=6.88

Thus, the pH of the solution is 6.88.

Answer 14

(b)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer 15

(b)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.88.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Addition of 25.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×25.0×10−3L=25×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.02	0	0	0
Add	0	0.0025
Change	-0.0025	-0.0025	0.0025	0.0025
Equilibrium	0.0175	0	0.0025	0.0025

The base dissociation constant value is 1.1×10−8

Thus,

pKb=−logKb=−log(1.1×10−8)=7.96

Since, volume is same thus, number of moles can be used instead of concentrations.

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[2.50][17.5]=7.12

Now, pH can be calculated as follows:

pH=14−pOH=14−7.12=6.88

Thus, the pH of the solution is 6.88.

Answer 20

(c)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (70 mL HCl) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(c)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.31.

Explanation of Solution

Addition of 70.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×70.0×10−3L=70×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.0200	0	0	0
Add	0	0.0070
Change	-0.0070	-0.0070	0.0070	0.0070
Equilibrium	0.0130	0	0.0070	0.0070

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[0.007][0.013]=7.69pH=14−7.69=6.31

Thus, the pH of the solution is 6.31.

Answer 21

(c)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (70 mL HCl) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer 22

(c)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.31.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Addition of 70.0mLofacid:

Total amount of base to be neutralized =0.2mol/L×100.0×10−3L=20×10−3mol

Amount of base added =0.100mol/L×70.0×10−3L=70×10−4mol

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.0200	0	0	0
Add	0	0.0070
Change	-0.0070	-0.0070	0.0070	0.0070
Equilibrium	0.0130	0	0.0070	0.0070

Applying the Henderson-Hasselbalch equation,

pOH=pKb+log[Conjugate acid][weak base]pOH=7.96+log[0.007][0.013]=7.69pH=14−7.69=6.31

Thus, the pH of the solution is 6.31.

Answer 27

(d)

Interpretation Introduction

Interpretation:

The pH value of the weak base at equivalent point needs to be determined.

Concept introduction:

At equivalent point the concentration of acid and base (in which one of the species is weak) is equal in the solution. Thus, the pH will depend on the acidic or basic salt formed.

(d)

Expert Solution

Answer to Problem 167AE

The pH of the solution at the equivalence point is 3.60.

Explanation of Solution

At the equivalence point, The amount of acid added =20×10−3mol

The volume of acid added =20×10−3mol0.1mol/L=0.2L=200.0mL

At this point, there is no excess acid or base. Therefore, the only possible reaction here is the dissociation of the conjugate acid of the ammonia (that is ammonium ion).

HONH3++H2O⇌HONH2+H3O+

Thereafter, using the Ka value for this weak base, the amount of hydrogen ions in the solution can be determined to get the pH value at this point.

Now, total volume of the solution is 300 mL thus, molarity of HONH₃+ solution will be:

M=nV=0.02 mol300 mL=0.0667 M

Reaction	HONH₃+	HONH2	H+
Initial	0.0667	0	0
Change	-x	x	x
Equilibrium	(0.0667-x)	x	x

Then the pH can be calculated as follows:

Ka=[HONH2][H+][HONH3+]9.09×10−7=[x][x]0.0667=x20.0067x=2.5×10−4 M

This value is equal to the amount of hydrogen ions in the solution.

pH=−log[H+]=−log[2.5×10−4]=3.60

Thus, the pH value at equivalence point is 3.60.

Answer 28

(d)

Interpretation Introduction

Interpretation:

The pH value of the weak base at equivalent point needs to be determined.

Concept introduction:

At equivalent point the concentration of acid and base (in which one of the species is weak) is equal in the solution. Thus, the pH will depend on the acidic or basic salt formed.

Answer 29

(d)

Expert Solution

Answer to Problem 167AE

The pH of the solution at the equivalence point is 3.60.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

At the equivalence point, The amount of acid added =20×10−3mol

The volume of acid added =20×10−3mol0.1mol/L=0.2L=200.0mL

At this point, there is no excess acid or base. Therefore, the only possible reaction here is the dissociation of the conjugate acid of the ammonia (that is ammonium ion).

HONH3++H2O⇌HONH2+H3O+

Thereafter, using the Ka value for this weak base, the amount of hydrogen ions in the solution can be determined to get the pH value at this point.

Now, total volume of the solution is 300 mL thus, molarity of HONH₃+ solution will be:

M=nV=0.02 mol300 mL=0.0667 M

Reaction	HONH₃+	HONH2	H+
Initial	0.0667	0	0
Change	-x	x	x
Equilibrium	(0.0667-x)	x	x

Then the pH can be calculated as follows:

Ka=[HONH2][H+][HONH3+]9.09×10−7=[x][x]0.0667=x20.0067x=2.5×10−4 M

This value is equal to the amount of hydrogen ions in the solution.

pH=−log[H+]=−log[2.5×10−4]=3.60

Thus, the pH value at equivalence point is 3.60.

Answer 34

(e)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (300 mL) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(e)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding acid is 1.60 .

Explanation of Solution

Addition of 300.0mLacid, The amount of acid added =0.1mol/L×300.0×10−3L=0.03mol

At this point, there is excess acid in the solution, thus, the pH can be calculated as follows:

Excess amount of acid in the solution =0.03−0.02=0.01mol

The amount of hydrogen ions =0.01mol

The concentration of hydrogen ions =0.01mol(100.0+300.0)×10−3L=0.025mol/L

pH=−log[H+]=−log[0.025]=1.60

Answer 35

(e)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (300 mL) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer 36

(e)

Expert Solution

Answer to Problem 167AE

The pH of the solution after adding acid is 1.60 .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Addition of 300.0mLacid, The amount of acid added =0.1mol/L×300.0×10−3L=0.03mol

At this point, there is excess acid in the solution, thus, the pH can be calculated as follows:

Excess amount of acid in the solution =0.03−0.02=0.01mol

The amount of hydrogen ions =0.01mol

The concentration of hydrogen ions =0.01mol(100.0+300.0)×10−3L=0.025mol/L

pH=−log[H+]=−log[0.025]=1.60

Answer 41

(f)

Interpretation Introduction

Interpretation:

The volume of HCl needs to be determined that is added to the solution if pH is 6.04.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

(f)

Expert Solution

Answer to Problem 167AE

100 mL

Explanation of Solution

The pKb value for NH₂OH is 7.96. From the given pH value, pOH of the solution can be calculated as follows:

pOH=14−pH=14−6.04=7.96

The Henderson Hesselbalch equation can be represented as follows:

pOH=pKb+log[salt][base]

Since, the volume is same thus, concentration can be replaced by number of moles.

7.96=7.96+log[salt][base]

Thus, the number of moles of salt and base will be same that is NH3OH+ and NH₂OH respectively.

This is the condition which is halfway to the equivalence point. For the equivalence point, volume of HCl required was 200 mL thus, at halfway equivalence point the volume will be half to it. Therefore, volume of HCl is 100 mL.

Answer 42

(f)

Interpretation Introduction

Interpretation:

The volume of HCl needs to be determined that is added to the solution if pH is 6.04.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer 43

(f)

Expert Solution

Answer to Problem 167AE

100 mL

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

The pKb value for NH₂OH is 7.96. From the given pH value, pOH of the solution can be calculated as follows:

pOH=14−pH=14−6.04=7.96

The Henderson Hesselbalch equation can be represented as follows:

pOH=pKb+log[salt][base]

Since, the volume is same thus, concentration can be replaced by number of moles.

7.96=7.96+log[salt][base]

Thus, the number of moles of salt and base will be same that is NH3OH+ and NH₂OH respectively.

This is the condition which is halfway to the equivalence point. For the equivalence point, volume of HCl required was 200 mL thus, at halfway equivalence point the volume will be half to it. Therefore, volume of HCl is 100 mL.

Answer 48

Ch. 8 - Prob. 1DQ Ch. 8 - Prob. 2DQ Ch. 8 - Mixing together solutions of acetic acid and...Ch. 8 - Sketch two pH curves, one for the titration of a...Ch. 8 - Sketch a pH curve for the titration of a weak acid...Ch. 8 - You have a solution of the weak acid HA and add...Ch. 8 - You have a solution of the weak acid HA and add...Ch. 8 - Prob. 8DQ Ch. 8 - You are browsing through the Handbook of...Ch. 8 - A friend tells you: "The constant Ksp of a salt is...

The initial pH value of the weak base before adding the acid is to be determined. Concept introduction: A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Concept explainers

Videos

Interpretation:

The initial pH value of the weak base before adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

Explanation of Solution

Interpretation:

The pH value of the weak base after adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

Explanation of Solution

Interpretation:

The pH value of the weak base after adding the acid (70 mL HCl) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

Explanation of Solution

Interpretation:

The pH value of the weak base at equivalent point needs to be determined.

Concept introduction:

At equivalent point the concentration of acid and base (in which one of the species is weak) is equal in the solution. Thus, the pH will depend on the acidic or basic salt formed.

Answer to Problem 167AE

Explanation of Solution

Interpretation:

The pH value of the weak base after adding the acid (300 mL) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

Explanation of Solution

Interpretation:

The volume of HCl needs to be determined that is added to the solution if pH is 6.04.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

Explanation of Solution

Want to see more full solutions like this?

Chapter 8 Solutions

The initial pH value of the weak base before adding the acid is to be determined. Concept introduction: A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Concept explainers

Videos

Interpretation: The initial pH value of the weak base before adding the acid is to be determined. Concept introduction: A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

Explanation of Solution

Interpretation: The pH value of the weak base after adding the acid is to be determined. Concept introduction: A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

Explanation of Solution

Interpretation: The pH value of the weak base after adding the acid (70 mL HCl) is to be determined. Concept introduction: A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

Explanation of Solution

Interpretation: The pH value of the weak base at equivalent point needs to be determined. Concept introduction: At equivalent point the concentration of acid and base (in which one of the species is weak) is equal in the solution. Thus, the pH will depend on the acidic or basic salt formed.

Answer to Problem 167AE

Explanation of Solution

Interpretation: The pH value of the weak base after adding the acid (300 mL) is to be determined. Concept introduction: A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

Explanation of Solution

Interpretation: The volume of HCl needs to be determined that is added to the solution if pH is 6.04. Concept introduction: A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

Explanation of Solution

Want to see more full solutions like this?

Chapter 8 Solutions

Interpretation:

The initial pH value of the weak base before adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Interpretation:

The pH value of the weak base after adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Interpretation:

The pH value of the weak base after adding the acid (70 mL HCl) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Interpretation:

The pH value of the weak base at equivalent point needs to be determined.

Concept introduction:

At equivalent point the concentration of acid and base (in which one of the species is weak) is equal in the solution. Thus, the pH will depend on the acidic or basic salt formed.

Interpretation:

The pH value of the weak base after adding the acid (300 mL) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Interpretation:

The volume of HCl needs to be determined that is added to the solution if pH is 6.04.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.