The mass of CH 3 NH 2 present in a 100.0 mL solution, which has the pH value of 10.64 , is to be determined. Concept introduction: pH is the measure of acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution. The formula to calculate [ H 3 O + ] in the solution from the pH is: pH = − log [ H 3 O + ] = − log [ H + ] as [ H 3 O + ] = [ H + ] [ H 3 O + ] = 10 − pH …… (1) The relationship between [ H 3 O + ] and [ OH − ] is: K w = [ H 3 O + ] [ OH − ] …… (2) The value of K w is 1.0 × 10 − 14 . The ionization of the weak base takes place as: B ( a q ) + H 2 O ( l ) ⇌ BH + ( a q ) + OH − ( a q ) K b is the measure of ionization of a base and is known as base-ionization constant, which is specific at a particular temperature: K b = [ BH + ] [ OH − ] [ B ] …… (3) The number of moles ( n ) of any substance from its given mass ( m ) and molar mass ( M ) is calculated as: n = m M …… (4) Here, m is the given mass and M is the molar mass. Molarity ( M ) of a solution is calculated as: M = n V …… (5) Here, n is the number of moles and V is the volume.

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Answer 1

Question

Chapter 16, Problem 161AP

Interpretation Introduction

Interpretation:

The mass of $CH3NH2$ present in a $100.0 mL$ solution, which has the pH value of $10.64$ , is to be determined.

Concept introduction:

pH is the measure of acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution.

The formula to calculate $[H3O+]$ in the solution from the pH is:

$pH=−log[H3O+]=−log[H+]$ as $[H3O+]=[H+]$

$[H3O+]=10−pH$ …… (1)

The relationship between $[H3O+]$ and $[OH−]$ is:

$Kw=[H3O+][OH−]$ …… (2)

The value of $Kw$ is $1.0×10−14$ .

The ionization of the weak base takes place as:

$B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)$

$Kb$ is the measure of ionization of a base and is known as base-ionization constant, which is specific at a particular temperature:

$Kb=[BH+][OH−][B]$ …… (3)

The number of moles $(n)$ of any substance from its given mass $(m)$ and molar mass $(M)$ is calculated as:

$n=mM$ …… (4)

Here, $m$ is the given mass and $M$ is the molar mass.

Molarity $(M)$ of a solution is calculated as:

$M=nV$ …… (5)

Here, $n$ is the number of moles and $V$ is the volume.

Expert Solution & Answer

Answer to Problem 161AP

Solution: $2.7×10−3 g$ .

Explanation of Solution

Given information:

The pH of methylamine $CH3NH2$ solution is $10.64$ . Also, the volume of the solution is given as $100.0 mL$ .

Use equation (1) to calculate the concentration of hydronium ions in the solution with the use of pH as:

$[H3O+]=10−pH$

$[H3O+]=10−10.64=2.29×10−11M$

Now, use equation (2) to calculate the value of $[OH−]$ , and substitute the value of $[H3O+]$ and $Kw$ as follows:

$Kw=[H3O+][OH−]$

$1.0×10−14=(2.29×10−11)[OH−][OH−]=1.0×10−142.29×10−11[OH−]=4.4×10−4M$

$CH3NH2$ is a weak base and when a weak base is dissolved in water, only partial ionization takes place into $OH−$ ions and its conjugate acid. Thus, the concentration of $CH3NH2$ is determined by $Kb$ of the base.

Refer to table $16.7$ for the $Kb$ value as $4.4×10−4$ .

The reaction of methylamine is depicted as:

$CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as shown below.

$CH3NH2(aq)H2O(l)CH3NH3+(aq)OH−(aq)Initial concentration(M)x−00Change in concentration(M)−4.4×10−4−+4.4×10−4+4.4×10−4Equilibrium concentration(M)x−(4.4×10−4)−4.4×10−44.4×10−4$

Now, substitute these concentrations in equation (3) as follows:

$Kb=[BH+][OH−][B]$

$Kb=(4.4×10−4)(4.4×10−4)(x−(4.4×10−4))$

Now, substitute the value of $Kb$ in the above equation as follows:

$4.4×10−4=(4.4×10−4)(4.4×10−4)(x−(4.4×10−4))x−(4.4×10−4)=(4.4×10−4)(4.4×10−4)(4.4×10−4)x=8.8×10−4$

Thus,

$[CH3NH2]=8.8×10−4 M$ .

The molarity of methylamine is $8.8×10−4 M$ . Now, use equation (5) to calculate the number of moles of $CH3NH2$ for $100.0 mL$ :

$M=nV$

Here, $n$ is the number of moles and $V$ is the volume.

$8.8×10−4 M=n0.100 Ln=(8.8×10−4 M)(0.100 L)n=8.8×10−5 mol$

The molar mass of $CH3NH2$ is $31.06 g/mol$ . Now, use equation (4) to calculate the mass of $CH3NH2$ as follows:

$n=mM$

Here, $m$ is the given mass and $M$ is the molar mass.

$8.8×10−5 mol=m31.06 g/molm=(8.8×10−5 mol)(31.06 g/mol)m=2.7×10−3 g$

Conclusion

The mass for the methylamine solution in water is $2.7×10−3 g$ .

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Answer 2

Question

Chapter 16, Problem 161AP

Interpretation Introduction

Interpretation:

The mass of $CH3NH2$ present in a $100.0 mL$ solution, which has the pH value of $10.64$ , is to be determined.

Concept introduction:

pH is the measure of acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution.

The formula to calculate $[H3O+]$ in the solution from the pH is:

$pH=−log[H3O+]=−log[H+]$ as $[H3O+]=[H+]$

$[H3O+]=10−pH$ …… (1)

The relationship between $[H3O+]$ and $[OH−]$ is:

$Kw=[H3O+][OH−]$ …… (2)

The value of $Kw$ is $1.0×10−14$ .

The ionization of the weak base takes place as:

$B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)$

$Kb$ is the measure of ionization of a base and is known as base-ionization constant, which is specific at a particular temperature:

$Kb=[BH+][OH−][B]$ …… (3)

The number of moles $(n)$ of any substance from its given mass $(m)$ and molar mass $(M)$ is calculated as:

$n=mM$ …… (4)

Here, $m$ is the given mass and $M$ is the molar mass.

Molarity $(M)$ of a solution is calculated as:

$M=nV$ …… (5)

Here, $n$ is the number of moles and $V$ is the volume.

Expert Solution & Answer

Answer to Problem 161AP

Solution: $2.7×10−3 g$ .

Explanation of Solution

Given information:

The pH of methylamine $CH3NH2$ solution is $10.64$ . Also, the volume of the solution is given as $100.0 mL$ .

Use equation (1) to calculate the concentration of hydronium ions in the solution with the use of pH as:

$[H3O+]=10−pH$

$[H3O+]=10−10.64=2.29×10−11M$

Now, use equation (2) to calculate the value of $[OH−]$ , and substitute the value of $[H3O+]$ and $Kw$ as follows:

$Kw=[H3O+][OH−]$

$1.0×10−14=(2.29×10−11)[OH−][OH−]=1.0×10−142.29×10−11[OH−]=4.4×10−4M$

$CH3NH2$ is a weak base and when a weak base is dissolved in water, only partial ionization takes place into $OH−$ ions and its conjugate acid. Thus, the concentration of $CH3NH2$ is determined by $Kb$ of the base.

Refer to table $16.7$ for the $Kb$ value as $4.4×10−4$ .

The reaction of methylamine is depicted as:

$CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as shown below.

$CH3NH2(aq)H2O(l)CH3NH3+(aq)OH−(aq)Initial concentration(M)x−00Change in concentration(M)−4.4×10−4−+4.4×10−4+4.4×10−4Equilibrium concentration(M)x−(4.4×10−4)−4.4×10−44.4×10−4$

Now, substitute these concentrations in equation (3) as follows:

$Kb=[BH+][OH−][B]$

$Kb=(4.4×10−4)(4.4×10−4)(x−(4.4×10−4))$

Now, substitute the value of $Kb$ in the above equation as follows:

$4.4×10−4=(4.4×10−4)(4.4×10−4)(x−(4.4×10−4))x−(4.4×10−4)=(4.4×10−4)(4.4×10−4)(4.4×10−4)x=8.8×10−4$

Thus,

$[CH3NH2]=8.8×10−4 M$ .

The molarity of methylamine is $8.8×10−4 M$ . Now, use equation (5) to calculate the number of moles of $CH3NH2$ for $100.0 mL$ :

$M=nV$

Here, $n$ is the number of moles and $V$ is the volume.

$8.8×10−4 M=n0.100 Ln=(8.8×10−4 M)(0.100 L)n=8.8×10−5 mol$

The molar mass of $CH3NH2$ is $31.06 g/mol$ . Now, use equation (4) to calculate the mass of $CH3NH2$ as follows:

$n=mM$

Here, $m$ is the given mass and $M$ is the molar mass.

$8.8×10−5 mol=m31.06 g/molm=(8.8×10−5 mol)(31.06 g/mol)m=2.7×10−3 g$

Conclusion

The mass for the methylamine solution in water is $2.7×10−3 g$ .

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Answer 3

Question

Chapter 16, Problem 161AP

Interpretation Introduction

Interpretation:

The mass of $CH3NH2$ present in a $100.0 mL$ solution, which has the pH value of $10.64$ , is to be determined.

Concept introduction:

pH is the measure of acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution.

The formula to calculate $[H3O+]$ in the solution from the pH is:

$pH=−log[H3O+]=−log[H+]$ as $[H3O+]=[H+]$

$[H3O+]=10−pH$ …… (1)

The relationship between $[H3O+]$ and $[OH−]$ is:

$Kw=[H3O+][OH−]$ …… (2)

The value of $Kw$ is $1.0×10−14$ .

The ionization of the weak base takes place as:

$B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)$

$Kb$ is the measure of ionization of a base and is known as base-ionization constant, which is specific at a particular temperature:

$Kb=[BH+][OH−][B]$ …… (3)

The number of moles $(n)$ of any substance from its given mass $(m)$ and molar mass $(M)$ is calculated as:

$n=mM$ …… (4)

Here, $m$ is the given mass and $M$ is the molar mass.

Molarity $(M)$ of a solution is calculated as:

$M=nV$ …… (5)

Here, $n$ is the number of moles and $V$ is the volume.

Expert Solution & Answer

Answer to Problem 161AP

Solution: $2.7×10−3 g$ .

Explanation of Solution

Given information:

The pH of methylamine $CH3NH2$ solution is $10.64$ . Also, the volume of the solution is given as $100.0 mL$ .

Use equation (1) to calculate the concentration of hydronium ions in the solution with the use of pH as:

$[H3O+]=10−pH$

$[H3O+]=10−10.64=2.29×10−11M$

Now, use equation (2) to calculate the value of $[OH−]$ , and substitute the value of $[H3O+]$ and $Kw$ as follows:

$Kw=[H3O+][OH−]$

$1.0×10−14=(2.29×10−11)[OH−][OH−]=1.0×10−142.29×10−11[OH−]=4.4×10−4M$

$CH3NH2$ is a weak base and when a weak base is dissolved in water, only partial ionization takes place into $OH−$ ions and its conjugate acid. Thus, the concentration of $CH3NH2$ is determined by $Kb$ of the base.

Refer to table $16.7$ for the $Kb$ value as $4.4×10−4$ .

The reaction of methylamine is depicted as:

$CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as shown below.

$CH3NH2(aq)H2O(l)CH3NH3+(aq)OH−(aq)Initial concentration(M)x−00Change in concentration(M)−4.4×10−4−+4.4×10−4+4.4×10−4Equilibrium concentration(M)x−(4.4×10−4)−4.4×10−44.4×10−4$

Now, substitute these concentrations in equation (3) as follows:

$Kb=[BH+][OH−][B]$

$Kb=(4.4×10−4)(4.4×10−4)(x−(4.4×10−4))$

Now, substitute the value of $Kb$ in the above equation as follows:

$4.4×10−4=(4.4×10−4)(4.4×10−4)(x−(4.4×10−4))x−(4.4×10−4)=(4.4×10−4)(4.4×10−4)(4.4×10−4)x=8.8×10−4$

Thus,

$[CH3NH2]=8.8×10−4 M$ .

The molarity of methylamine is $8.8×10−4 M$ . Now, use equation (5) to calculate the number of moles of $CH3NH2$ for $100.0 mL$ :

$M=nV$

Here, $n$ is the number of moles and $V$ is the volume.

$8.8×10−4 M=n0.100 Ln=(8.8×10−4 M)(0.100 L)n=8.8×10−5 mol$

The molar mass of $CH3NH2$ is $31.06 g/mol$ . Now, use equation (4) to calculate the mass of $CH3NH2$ as follows:

$n=mM$

Here, $m$ is the given mass and $M$ is the molar mass.

$8.8×10−5 mol=m31.06 g/molm=(8.8×10−5 mol)(31.06 g/mol)m=2.7×10−3 g$

Conclusion

The mass for the methylamine solution in water is $2.7×10−3 g$ .

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Answer 4

Question

Chapter 16, Problem 161AP

Interpretation Introduction

Interpretation:

The mass of $CH3NH2$ present in a $100.0 mL$ solution, which has the pH value of $10.64$ , is to be determined.

Concept introduction:

pH is the measure of acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution.

The formula to calculate $[H3O+]$ in the solution from the pH is:

$pH=−log[H3O+]=−log[H+]$ as $[H3O+]=[H+]$

$[H3O+]=10−pH$ …… (1)

The relationship between $[H3O+]$ and $[OH−]$ is:

$Kw=[H3O+][OH−]$ …… (2)

The value of $Kw$ is $1.0×10−14$ .

The ionization of the weak base takes place as:

$B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)$

$Kb$ is the measure of ionization of a base and is known as base-ionization constant, which is specific at a particular temperature:

$Kb=[BH+][OH−][B]$ …… (3)

The number of moles $(n)$ of any substance from its given mass $(m)$ and molar mass $(M)$ is calculated as:

$n=mM$ …… (4)

Here, $m$ is the given mass and $M$ is the molar mass.

Molarity $(M)$ of a solution is calculated as:

$M=nV$ …… (5)

Here, $n$ is the number of moles and $V$ is the volume.

Expert Solution & Answer

Answer to Problem 161AP

Solution: $2.7×10−3 g$ .

Explanation of Solution

Given information:

The pH of methylamine $CH3NH2$ solution is $10.64$ . Also, the volume of the solution is given as $100.0 mL$ .

Use equation (1) to calculate the concentration of hydronium ions in the solution with the use of pH as:

$[H3O+]=10−pH$

$[H3O+]=10−10.64=2.29×10−11M$

Now, use equation (2) to calculate the value of $[OH−]$ , and substitute the value of $[H3O+]$ and $Kw$ as follows:

$Kw=[H3O+][OH−]$

$1.0×10−14=(2.29×10−11)[OH−][OH−]=1.0×10−142.29×10−11[OH−]=4.4×10−4M$

$CH3NH2$ is a weak base and when a weak base is dissolved in water, only partial ionization takes place into $OH−$ ions and its conjugate acid. Thus, the concentration of $CH3NH2$ is determined by $Kb$ of the base.

Refer to table $16.7$ for the $Kb$ value as $4.4×10−4$ .

The reaction of methylamine is depicted as:

$CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as shown below.

$CH3NH2(aq)H2O(l)CH3NH3+(aq)OH−(aq)Initial concentration(M)x−00Change in concentration(M)−4.4×10−4−+4.4×10−4+4.4×10−4Equilibrium concentration(M)x−(4.4×10−4)−4.4×10−44.4×10−4$

Now, substitute these concentrations in equation (3) as follows:

$Kb=[BH+][OH−][B]$

$Kb=(4.4×10−4)(4.4×10−4)(x−(4.4×10−4))$

Now, substitute the value of $Kb$ in the above equation as follows:

$4.4×10−4=(4.4×10−4)(4.4×10−4)(x−(4.4×10−4))x−(4.4×10−4)=(4.4×10−4)(4.4×10−4)(4.4×10−4)x=8.8×10−4$

Thus,

$[CH3NH2]=8.8×10−4 M$ .

The molarity of methylamine is $8.8×10−4 M$ . Now, use equation (5) to calculate the number of moles of $CH3NH2$ for $100.0 mL$ :

$M=nV$

Here, $n$ is the number of moles and $V$ is the volume.

$8.8×10−4 M=n0.100 Ln=(8.8×10−4 M)(0.100 L)n=8.8×10−5 mol$

The molar mass of $CH3NH2$ is $31.06 g/mol$ . Now, use equation (4) to calculate the mass of $CH3NH2$ as follows:

$n=mM$

Here, $m$ is the given mass and $M$ is the molar mass.

$8.8×10−5 mol=m31.06 g/molm=(8.8×10−5 mol)(31.06 g/mol)m=2.7×10−3 g$

Conclusion

The mass for the methylamine solution in water is $2.7×10−3 g$ .

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Answer 5

Chapter 16, Problem 161AP

Interpretation Introduction

Interpretation:

The mass of $CH3NH2$ present in a $100.0 mL$ solution, which has the pH value of $10.64$ , is to be determined.

Concept introduction:

pH is the measure of acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution.

The formula to calculate $[H3O+]$ in the solution from the pH is:

$pH=−log[H3O+]=−log[H+]$ as $[H3O+]=[H+]$

$[H3O+]=10−pH$ …… (1)

The relationship between $[H3O+]$ and $[OH−]$ is:

$Kw=[H3O+][OH−]$ …… (2)

The value of $Kw$ is $1.0×10−14$ .

The ionization of the weak base takes place as:

$B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)$

$Kb$ is the measure of ionization of a base and is known as base-ionization constant, which is specific at a particular temperature:

$Kb=[BH+][OH−][B]$ …… (3)

The number of moles $(n)$ of any substance from its given mass $(m)$ and molar mass $(M)$ is calculated as:

$n=mM$ …… (4)

Here, $m$ is the given mass and $M$ is the molar mass.

Molarity $(M)$ of a solution is calculated as:

$M=nV$ …… (5)

Here, $n$ is the number of moles and $V$ is the volume.

Expert Solution & Answer

Answer to Problem 161AP

Solution: $2.7×10−3 g$ .

Explanation of Solution

Given information:

The pH of methylamine $CH3NH2$ solution is $10.64$ . Also, the volume of the solution is given as $100.0 mL$ .

Use equation (1) to calculate the concentration of hydronium ions in the solution with the use of pH as:

$[H3O+]=10−pH$

$[H3O+]=10−10.64=2.29×10−11M$

Now, use equation (2) to calculate the value of $[OH−]$ , and substitute the value of $[H3O+]$ and $Kw$ as follows:

$Kw=[H3O+][OH−]$

$1.0×10−14=(2.29×10−11)[OH−][OH−]=1.0×10−142.29×10−11[OH−]=4.4×10−4M$

$CH3NH2$ is a weak base and when a weak base is dissolved in water, only partial ionization takes place into $OH−$ ions and its conjugate acid. Thus, the concentration of $CH3NH2$ is determined by $Kb$ of the base.

Refer to table $16.7$ for the $Kb$ value as $4.4×10−4$ .

The reaction of methylamine is depicted as:

$CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as shown below.

$CH3NH2(aq)H2O(l)CH3NH3+(aq)OH−(aq)Initial concentration(M)x−00Change in concentration(M)−4.4×10−4−+4.4×10−4+4.4×10−4Equilibrium concentration(M)x−(4.4×10−4)−4.4×10−44.4×10−4$

Now, substitute these concentrations in equation (3) as follows:

$Kb=[BH+][OH−][B]$

$Kb=(4.4×10−4)(4.4×10−4)(x−(4.4×10−4))$

Now, substitute the value of $Kb$ in the above equation as follows:

$4.4×10−4=(4.4×10−4)(4.4×10−4)(x−(4.4×10−4))x−(4.4×10−4)=(4.4×10−4)(4.4×10−4)(4.4×10−4)x=8.8×10−4$

Thus,

$[CH3NH2]=8.8×10−4 M$ .

The molarity of methylamine is $8.8×10−4 M$ . Now, use equation (5) to calculate the number of moles of $CH3NH2$ for $100.0 mL$ :

$M=nV$

Here, $n$ is the number of moles and $V$ is the volume.

$8.8×10−4 M=n0.100 Ln=(8.8×10−4 M)(0.100 L)n=8.8×10−5 mol$

The molar mass of $CH3NH2$ is $31.06 g/mol$ . Now, use equation (4) to calculate the mass of $CH3NH2$ as follows:

$n=mM$

Here, $m$ is the given mass and $M$ is the molar mass.

$8.8×10−5 mol=m31.06 g/molm=(8.8×10−5 mol)(31.06 g/mol)m=2.7×10−3 g$

Conclusion

The mass for the methylamine solution in water is $2.7×10−3 g$ .

Answer 6

Chapter 16, Problem 161AP

Interpretation Introduction

Interpretation:

The mass of $CH3NH2$ present in a $100.0 mL$ solution, which has the pH value of $10.64$ , is to be determined.

Concept introduction:

pH is the measure of acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution.

The formula to calculate $[H3O+]$ in the solution from the pH is:

$pH=−log[H3O+]=−log[H+]$ as $[H3O+]=[H+]$

$[H3O+]=10−pH$ …… (1)

The relationship between $[H3O+]$ and $[OH−]$ is:

$Kw=[H3O+][OH−]$ …… (2)

The value of $Kw$ is $1.0×10−14$ .

The ionization of the weak base takes place as:

$B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)$

$Kb$ is the measure of ionization of a base and is known as base-ionization constant, which is specific at a particular temperature:

$Kb=[BH+][OH−][B]$ …… (3)

The number of moles $(n)$ of any substance from its given mass $(m)$ and molar mass $(M)$ is calculated as:

$n=mM$ …… (4)

Here, $m$ is the given mass and $M$ is the molar mass.

Molarity $(M)$ of a solution is calculated as:

$M=nV$ …… (5)

Here, $n$ is the number of moles and $V$ is the volume.

Expert Solution & Answer

Answer to Problem 161AP

Solution: $2.7×10−3 g$ .

Explanation of Solution

Given information:

The pH of methylamine $CH3NH2$ solution is $10.64$ . Also, the volume of the solution is given as $100.0 mL$ .

Use equation (1) to calculate the concentration of hydronium ions in the solution with the use of pH as:

$[H3O+]=10−pH$

$[H3O+]=10−10.64=2.29×10−11M$

Now, use equation (2) to calculate the value of $[OH−]$ , and substitute the value of $[H3O+]$ and $Kw$ as follows:

$Kw=[H3O+][OH−]$

$1.0×10−14=(2.29×10−11)[OH−][OH−]=1.0×10−142.29×10−11[OH−]=4.4×10−4M$

$CH3NH2$ is a weak base and when a weak base is dissolved in water, only partial ionization takes place into $OH−$ ions and its conjugate acid. Thus, the concentration of $CH3NH2$ is determined by $Kb$ of the base.

Refer to table $16.7$ for the $Kb$ value as $4.4×10−4$ .

The reaction of methylamine is depicted as:

$CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as shown below.

$CH3NH2(aq)H2O(l)CH3NH3+(aq)OH−(aq)Initial concentration(M)x−00Change in concentration(M)−4.4×10−4−+4.4×10−4+4.4×10−4Equilibrium concentration(M)x−(4.4×10−4)−4.4×10−44.4×10−4$

Now, substitute these concentrations in equation (3) as follows:

$Kb=[BH+][OH−][B]$

$Kb=(4.4×10−4)(4.4×10−4)(x−(4.4×10−4))$

Now, substitute the value of $Kb$ in the above equation as follows:

$4.4×10−4=(4.4×10−4)(4.4×10−4)(x−(4.4×10−4))x−(4.4×10−4)=(4.4×10−4)(4.4×10−4)(4.4×10−4)x=8.8×10−4$

Thus,

$[CH3NH2]=8.8×10−4 M$ .

The molarity of methylamine is $8.8×10−4 M$ . Now, use equation (5) to calculate the number of moles of $CH3NH2$ for $100.0 mL$ :

$M=nV$

Here, $n$ is the number of moles and $V$ is the volume.

$8.8×10−4 M=n0.100 Ln=(8.8×10−4 M)(0.100 L)n=8.8×10−5 mol$

The molar mass of $CH3NH2$ is $31.06 g/mol$ . Now, use equation (4) to calculate the mass of $CH3NH2$ as follows:

$n=mM$

Here, $m$ is the given mass and $M$ is the molar mass.

$8.8×10−5 mol=m31.06 g/molm=(8.8×10−5 mol)(31.06 g/mol)m=2.7×10−3 g$

Conclusion

The mass for the methylamine solution in water is $2.7×10−3 g$ .

Answer 7

Chapter 16, Problem 161AP

Interpretation Introduction

Interpretation:

The mass of $CH3NH2$ present in a $100.0 mL$ solution, which has the pH value of $10.64$ , is to be determined.

Concept introduction:

pH is the measure of acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution.

The formula to calculate $[H3O+]$ in the solution from the pH is:

$pH=−log[H3O+]=−log[H+]$ as $[H3O+]=[H+]$

$[H3O+]=10−pH$ …… (1)

The relationship between $[H3O+]$ and $[OH−]$ is:

$Kw=[H3O+][OH−]$ …… (2)

The value of $Kw$ is $1.0×10−14$ .

The ionization of the weak base takes place as:

$B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)$

$Kb$ is the measure of ionization of a base and is known as base-ionization constant, which is specific at a particular temperature:

$Kb=[BH+][OH−][B]$ …… (3)

The number of moles $(n)$ of any substance from its given mass $(m)$ and molar mass $(M)$ is calculated as:

$n=mM$ …… (4)

Here, $m$ is the given mass and $M$ is the molar mass.

Molarity $(M)$ of a solution is calculated as:

$M=nV$ …… (5)

Here, $n$ is the number of moles and $V$ is the volume.

Answer 8

Expert Solution & Answer

Answer to Problem 161AP

Solution: $2.7×10−3 g$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The pH of methylamine $CH3NH2$ solution is $10.64$ . Also, the volume of the solution is given as $100.0 mL$ .

Use equation (1) to calculate the concentration of hydronium ions in the solution with the use of pH as:

$[H3O+]=10−pH$

$[H3O+]=10−10.64=2.29×10−11M$

Now, use equation (2) to calculate the value of $[OH−]$ , and substitute the value of $[H3O+]$ and $Kw$ as follows:

$Kw=[H3O+][OH−]$

$1.0×10−14=(2.29×10−11)[OH−][OH−]=1.0×10−142.29×10−11[OH−]=4.4×10−4M$

$CH3NH2$ is a weak base and when a weak base is dissolved in water, only partial ionization takes place into $OH−$ ions and its conjugate acid. Thus, the concentration of $CH3NH2$ is determined by $Kb$ of the base.

Refer to table $16.7$ for the $Kb$ value as $4.4×10−4$ .

The reaction of methylamine is depicted as:

$CH3NH2(aq)+H2O(l)⇌CH3NH3+(aq)+OH−(aq)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as shown below.

$CH3NH2(aq)H2O(l)CH3NH3+(aq)OH−(aq)Initial concentration(M)x−00Change in concentration(M)−4.4×10−4−+4.4×10−4+4.4×10−4Equilibrium concentration(M)x−(4.4×10−4)−4.4×10−44.4×10−4$