Chapter 14, Problem 65E

 (a)

Interpretation Introduction

Interpretation: The halfway point and the equivalence point $\text{pH}$ for each of the given titrations are to be calculated.

Concept introduction: The $\text{pH}$ value is defined as a figure that is used to express the acidity or basicity of a solution by using the logarithmic scale on which 7 is a neutral. The values lower than 7 shows acidic or greater than 7 shows basic behavior of the solution.

To determine: The halfway point and the equivalence point $\text{pH}$ for the given titration.

Answer to Problem 65E

Answer

The $\text{pH}$ at the half-way equivalence point is $\underset{\_}{4.20}$.

The $\text{pH}$ at equivalence point is $\underset{\_}{8.45}$.

Explanation of Solution

Explanation

Given

Concentration of ${\text{HC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}$ is $0.10\text{M}$.

Volume of the solution is $100\text{mL}$.

Concentration of $\text{NaOH}$ is $0.10\text{M}$

According to the half-way titration, the volume of the base consumed is half of the equivalence point. Volume of the base is calculated by using the expression,

${\text{M}}_{\text{1}}{\text{V}}_{\text{1}}={\text{M}}_{\text{2}}{\text{V}}_{\text{2}}$

Where,

• ${\text{M}}_{\text{1}}$ is the concentration of ${\text{HC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}$.
• ${\text{M}}_2$ is the concentration of $\text{NaOH}$.
• ${\text{V}}_{\text{1}}$ is the volume of ${\text{HC}}_{\text{7}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}$ solution.
• ${\text{V}}_2$ is the volume of $\text{NaOH}$ solution.

Substitute the values of ${\text{M}}_{\text{1}}$, ${\text{V}}_{\text{1}}$, and ${\text{M}}_2$ in the above expression to calculate the volume of $\text{NaOH}$ solution.

$\begin{array}{c}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}={\text{M}}_{\text{2}}{\text{V}}_{\text{2}}\\ 0.10\text{M}\times \text{100}\text{mL}=0.10\text{M}\times {\text{V}}_2\\ {\text{V}}_{\text{2}}=100\text{mL}\end{array}$

The volume of the base at the equivalence point is ${\text{V}}_{\text{2}}=100\text{mL}$.

The volume of the base at the half-way point is $50\text{mL}$.

The reaction between ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ and $\text{NaOH}$ at the half-way equivalence point is,

$\begin{array}{ccccccc}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}& +& \text{NaOH}& \underset{}{\to }& {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COONa}& +& {\text{H}}_{\text{2}}\text{O}\\ 0.01\text{mole}& & 0.005\text{mole}& & 0.005\text{mole}& & \end{array}$

The $\text{pH}$ at half-way point is calculated by using the formula,

$\text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}$

The ${\text{pK}}_{\text{a}}$ of benzoic acid is $4.20$.

Substitute the values of ${\text{pK}}_{\text{a}}$ and concentration of salt and acid in the above formula to calculate the $\text{pH}$ at half-way point.

$\begin{array}{c}\text{pH}=4.20+\log \frac{\left[0.005\right]}{\left[0.005\right]}\\ =\underset{\_}{4.20}\end{array}$

Hence, the $\text{pH}$ at the half-way equivalence point is $\underset{\_}{4.20}$.

At equivalence point, the reaction between ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ and $\text{NaOH}$ is,

$\begin{array}{ccccccc}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}& +& \text{NaOH}& \underset{}{\to }& {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COONa}& +& {\text{H}}_{\text{2}}\text{O}\\ 0.01\text{mole}& & 0.01\text{mole}& & 0.01\text{mole}& & \end{array}$

Now, first ${\text{K}}_{\text{b}}$ is calculated by using the formula,

${\text{K}}_{\text{b}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}$

Where,

• ${\text{K}}_{\text{b}}$ is the dissociation constant of base.
• ${\text{K}}_{\text{a}}$ is the dissociation constant of an acid.
• ${\text{K}}_{\text{w}}$ is the self-ionization constant of water.

The value of ${\text{K}}_{\text{a}}=6.4\times {10}^{-5}$

The value of ${\text{K}}_{\text{w}}=1.0\times {10}^{-14}$

Substitute the value of ${\text{K}}_{\text{a}}$ and ${\text{K}}_{\text{w}}$ in the above formula.

$\begin{array}{c}{\text{K}}_{\text{b}}=\frac{1.0\times {10}^{-14}}{6.4\times {10}^{-5}}\\ =1.56\times {10}^{-10}\end{array}$

The molar concentration of $\left[{\text{OH}}^{-}\right]$ is calculated by using the formula,

$\left[{\text{OH}}^{-}\right]=\sqrt{{\text{K}}_{\text{b}}\times \text{C}}$

Where, the concentration of $\left({\text{OH}}^{-}\right)$ C is $\frac{0.01\text{mol}}{\left(100+100\right)\times {10}^{-3}\text{L}}=0.05$

Substitute the values of ${\text{K}}_{\text{b}}$ and C in the above formula to calculate $\left[{\text{OH}}^{-}\right]$.

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\sqrt{{\text{K}}_{\text{b}}\times \text{C}}\\ =\sqrt{1.56\times {10}^{-10}\times 0.05}\\ =2.79\times {10}^{-6}\end{array}$

The $\text{pH}$ is calculated by using the formula,

$\text{pH}=-\log \left[{\text{OH}}^{-}\right]$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above formula.

$\begin{array}{c}\text{pH}=-\log \left[2.79\times {10}^{-6}\right]\\ =5.55\end{array}$

The $\text{pH}$ at equivalence point is $14-5.55=\underset{\_}{8.45}$.

Hence, the $\text{pH}$ at equivalence point is $\underset{\_}{8.45}$.

(b)

Interpretation Introduction

Interpretation: The halfway point and the equivalence point $\text{pH}$ for each of the given titrations are to be calculated.

Concept introduction: The $\text{pH}$ value is defined as a figure that is used to express the acidity or basicity of a solution by using the logarithmic scale on which 7 is a neutral. The values lower than 7 shows acidic or greater than 7 shows basic behavior of the solution.

To determine: The halfway point and the equivalence point $\text{pH}$ for the given titration.

Answer to Problem 65E

Answer

The $\text{pH}$ at the half-way equivalence point is $\underset{\_}{10.748}$.

The $\text{pH}$ at equivalence point is $\underset{\_}{5.96}$.

Explanation of Solution

Explanation

Given

Concentration of ${\text{C}}_2{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ is $0.10\text{M}$.

Volume of the solution is $100\text{mL}$.

Concentration of ${\text{HNO}}_3$ is $0.20\text{M}$.

According to the half-way titration, the volume of the base consumed is half of the equivalence point. Volume of the base is calculated by using the expression,

${\text{M}}_{\text{1}}{\text{V}}_{\text{1}}={\text{M}}_{\text{2}}{\text{V}}_{\text{2}}$

Where,

• ${\text{M}}_{\text{1}}$ is the concentration of ${\text{C}}_2{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$.
• ${\text{M}}_2$ is the concentration of ${\text{HNO}}_3$.
• ${\text{V}}_{\text{1}}$ is the volume of ${\text{C}}_2{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ solution.
• ${\text{V}}_2$ is the volume of ${\text{HNO}}_3$ solution.

Substitute the values of ${\text{M}}_{\text{1}}$, ${\text{V}}_{\text{1}}$, and ${\text{M}}_2$ in the above expression to calculate the volume of ${\text{HNO}}_3$ solution.

$\begin{array}{c}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}={\text{M}}_{\text{2}}{\text{V}}_{\text{2}}\\ 0.10\text{M}\times \text{100}\text{mL}=0.20\text{M}\times {\text{V}}_2\\ {\text{V}}_{\text{2}}=50\text{mL}\end{array}$

The volume of the acid at the equivalence point is ${\text{V}}_{\text{2}}=50\text{mL}$.

The volume of the acid at the half-way point is $25\text{mL}$.

The reaction between ${\text{C}}_2{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ and ${\text{HNO}}_3$ at the half-way equivalence point is,

$\begin{array}{ccccccc}{\text{C}}_2{\text{H}}_{\text{5}}{\text{NH}}_2& +& {\text{HNO}}_3& \underset{}{\to }& {\text{C}}_2{\text{H}}_{\text{5}}{\text{NH}}_3^{+}& +& {\text{NO}}_3^{-}\\ 0.01\text{mole}& & 0.005\text{mole}& & 0.005\text{mole}& & \end{array}$

The $\text{pH}$ at half-way point is calculated by using the formula,

$\text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}$

The ${\text{pK}}_{\text{a}}$ of ${\text{C}}_2{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ is $10.748$.

Substitute the values of ${\text{pK}}_{\text{a}}$ and concentration of salt and acid in the above formula to calculate the $\text{pH}$ at half-way point.

$\begin{array}{c}\text{pH}=10.748+\log \frac{\left[0.005\right]}{\left[0.005\right]}\\ =\underset{\_}{10.748}\end{array}$

Hence, the $\text{pH}$ at the half-way equivalence point is $\underset{\_}{10.748}$.

At equivalence point, the reaction between ${\text{C}}_2{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ and ${\text{HNO}}_3$ is,

$\begin{array}{ccccccc}{\text{C}}_2{\text{H}}_{\text{5}}{\text{NH}}_2& +& {\text{HNO}}_3& \underset{}{\to }& {\text{C}}_2{\text{H}}_{\text{5}}{\text{NH}}_3^{+}& +& {\text{NO}}_3^{-}\\ 0.01\text{mole}& & 0.01\text{mole}& & 0.01\text{mole}& & \end{array}$

Now, first ${\text{K}}_{\text{a}}$ is calculated by using the formula,

${\text{K}}_{\text{a}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}$

Where,

• ${\text{K}}_{\text{b}}$ is the dissociation constant of base.
• ${\text{K}}_{\text{a}}$ is the dissociation constant of an acid.
• ${\text{K}}_{\text{w}}$ is the self-ionization constant of water.

The value of ${\text{K}}_{\text{b}}=5.6\times {10}^{-4}$

The value of ${\text{K}}_{\text{w}}=1.0\times {10}^{-14}$

Substitute the value of ${\text{K}}_{\text{b}}$ and ${\text{K}}_{\text{w}}$ in the above formula.

$\begin{array}{c}{\text{K}}_{\text{a}}=\frac{1.0\times {10}^{-14}}{5.6\times {10}^{-4}}\\ =1.78\times {10}^{-11}\end{array}$

The molar concentration of $\left[{\text{H}}^{+}\right]$ is calculated by using the formula,

$\left[{\text{H}}^{+}\right]=\sqrt{{\text{K}}_{\text{a}}\times \text{C}}$

Where, the concentration of $\left({\text{H}}^{+}\right)$ C is $\frac{0.01\text{mol}}{\left(100+50\right)\times {10}^{-3}\text{L}}=0.066$

Substitute the values of ${\text{K}}_{\text{a}}$ and C in the above formula to calculate $\left[{\text{H}}^{+}\right]$.

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\sqrt{{\text{K}}_{\text{a}}\times \text{C}}\\ =\sqrt{1.78\times {10}^{-11}\times 0.066}\\ =1.091\times {10}^{-6}\end{array}$

The $\text{pH}$ is calculated by using the formula,

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above formula.

$\begin{array}{c}\text{pH}=-\log \left[1.091\times {10}^{-6}\right]\\ =\underset{\_}{5.96}\end{array}$

Hence, the $\text{pH}$ at equivalence point is $\underset{\_}{5.96}$.

(c)

Interpretation Introduction

Interpretation: The halfway point and the equivalence point $\text{pH}$ for each of the given titrations are to be calculated.

Concept introduction: The $\text{pH}$ value is defined as a figure that is used to express the acidity or basicity of a solution by using the logarithmic scale on which 7 is a neutral. The values lower than 7 shows acidic or greater than 7 shows basic behavior of the solution.

To determine: The halfway point and the equivalence point $\text{pH}$ for the given titration.

Answer to Problem 65E

Answer

The $\text{pH}$ at the half-way equivalence point is $\underset{\_}{0.90}$.

The $\text{pH}$ at equivalence point is $\underset{\_}{7}$

Explanation of Solution

Explanation

Given

Concentration of $\text{NaOH}$ is $0.25\text{M}$.

Volume of the $\text{HCl}$ solution is $0.1\text{L}$.

Concentration of $\text{HCl}$ is $0.50\text{M}$.

According to the half-way titration, the volume of the base consumed is half of the equivalence point. Volume of the base is calculated by using the expression,

${\text{M}}_{\text{1}}{\text{V}}_{\text{1}}={\text{M}}_{\text{2}}{\text{V}}_{\text{2}}$

Where,

• ${\text{M}}_{\text{1}}$ is the concentration of $\text{HCl}$.
• ${\text{M}}_2$ is the concentration of $\text{NaOH}$.
• ${\text{V}}_{\text{1}}$ is the volume of $\text{HCl}$ solution.
• ${\text{V}}_2$ is the volume of $\text{NaOH}$ solution.

Substitute the values of ${\text{M}}_{\text{1}}$, ${\text{V}}_{\text{1}}$, and ${\text{M}}_2$ in the above expression to calculate the volume of $\text{NaOH}$ solution.

$\begin{array}{c}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}={\text{M}}_{\text{2}}{\text{V}}_{\text{2}}\\ 0.50\text{M}\times 0.1\text{L}=0.25\text{M}\times {\text{V}}_2\\ {\text{V}}_{\text{2}}=0.2\text{L}\end{array}$

The volume of the base at the equivalence point is ${\text{V}}_{\text{2}}=0.2\text{L}$.

The volume of the base at the half-way point is $0.1\text{L}$.

The concentration of $\left[{\text{H}}^{+}\right]$ is calculated by using the expression,

$\left[{\text{H}}^{+}\right]={\text{M}}_{\text{1}}{\text{V}}_{\text{1}}-\frac{{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}}{{\text{V}}_{\text{1}}{\text{+V}}_{\text{2}}}$

Substitute the values of ${\text{M}}_{\text{1}}$, ${\text{V}}_{\text{1}}$, ${\text{M}}_2$ and ${\text{V}}_{\text{2}}$ in the above formula to calculate $\left[{\text{H}}^{+}\right]$ at half-way point.

$\begin{array}{c}\left[{\text{H}}^{+}\right]=0.50\text{M}\times \text{0}\text{.1}\text{L}-\frac{0.25\text{M}\times \text{0}\text{.1}\text{L}}{\text{0}\text{.1}\text{L+0}\text{.1}\text{L}}\\ =0.125\text{M}\end{array}$

The $\text{pH}$ is calculated by using the formula,

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above formula.

$\begin{array}{c}\text{pH}=-\log \left[0.125\right]\\ =\underset{\_}{0.90}\end{array}$

Hence, the $\text{pH}$ at the half-way equivalence point is $\underset{\_}{0.90}$.

The reaction is between strong acid and strong base. So, complete neutralization takes place.

${\text{H}}^{\text{+}}+{\text{OH}}^{-}\underset{}{\to }{\text{H}}_{\text{2}}\text{O}$

For the complete neutralization, the concentration of acid $\left[{\text{H}}^{+}\right]$ is equal to the concentration of the base $\left[{\text{OH}}^{-}\right]$.

$\left[{\text{H}}^{\text{+}}\right]=\left[{\text{OH}}^{-}\right]$

Also, the product of $\left[{\text{H}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ is,

$\begin{array}{c}\left[{\text{H}}^{\text{+}}\right]\times \left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}\\ 2\left[{\text{H}}^{\text{+}}\right]=1.0\times {10}^{-14}\\ \left[{\text{H}}^{\text{+}}\right]=1.0\times {10}^{-7}\end{array}$

The $\text{pH}$ is calculated by using the formula,

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above formula.

$\begin{array}{c}\text{pH}=-\log \left[1.0\times {10}^{-7}\right]\\ =\underset{\_}{7}\end{array}$

Hence, the $\text{pH}$ at equivalence point is $\underset{\_}{7}$.