Chapter 15, Problem 73QAP

Interpretation Introduction

(a)

Interpretation:

The volume of $1.00\text{M}$

$\text{NaOH}$ required to neutralize the given solution is to be calculated.

$25.0\text{mL}$ of $\text{0}\text{.154}\text{M}$ acetic acid.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\left(\text{L}\right)}$.

Answer to Problem 73QAP

The volume of $1.00\text{M}$

$\text{NaOH}$ required to neutralize the given solution is $3.85\text{mL}$.

Explanation of Solution

The value of ${\text{M}}_1,{\text{V}}_1$ and ${\text{M}}_2$ is given to be $0.154\text{M}$, $25.0\text{mL}$ and $1.00\text{M}$ respectively.

The balanced equation when $\text{NaOH}$ reacts with ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ is shown below.

$\text{NaOH}+{\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\to {\text{H}}_{\text{2}}\text{O}+{\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\text{Na}$

The above reaction indicates that one equivalent of $\text{NaOH}$ is required to neutralize one equivalent of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$.

The relationship between concentration and volume of $\text{NaOH}$ and ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ solutions is shown below.

${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Where,

• ${\text{M}}_1$ is the molarity of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ solution.
• ${\text{V}}_1$ is the volume of ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ solution.
• ${\text{M}}_2$ is the molarity of $\text{NaOH}$ solution.
• ${\text{V}}_2$ is the volume of $\text{NaOH}$ solution.

Rearrange an above expression to calculate ${\text{V}}_2$ which is needed to neutralize ${\text{HC}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}$ solution.

$$

${\text{V}}_2=\frac{{\text{M}}_1\times {\text{V}}_1}{{\text{M}}_2}$

Substitute the value of ${\text{M}}_1,{\text{M}}_2$ and ${\text{V}}_1$ in the above expression.

$\begin{array}{c}{\text{V}}_2=\frac{0.154\text{M}\times 25.0\text{mL}}{1.00\text{M}}\\ =3.85\text{mL}\end{array}$

Therefore, the volume of $1.00\text{M}$

$\text{NaOH}$ required to neutralize the given solution is $3.85\text{mL}$.

Interpretation Introduction

(b)

Interpretation:

The volume of $1.00\text{M}$

$\text{NaOH}$ required to neutralize the given solution is to be calculated.

$\text{35}\text{.0}\text{mL}$ of $0.102\text{M}$ hydrofluoric acid

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Liters}\text{of}\text{solution}}$.

Answer to Problem 73QAP

The volume of $1.00\text{M}$

$\text{NaOH}$ required to neutralize the given solution is $3.57\text{mL}$.

Explanation of Solution

The value of ${\text{M}}_1,{\text{V}}_1$ and ${\text{M}}_2$ is given to be $0.102\text{M}$, $35.0\text{mL}$ and $1.00\text{M}$ respectively.

The balanced equation when $\text{NaOH}$ reacts with $\text{HF}$ is shown below.

$\text{NaOH}+\text{HF}\to {\text{H}}_{\text{2}}\text{O}+\text{NaF}$

The above reaction indicates that one equivalent of $\text{NaOH}$ is required to neutralize one equivalent of $\text{HF}$.

The relationship between concentration and volume of $\text{NaOH}$ and $\text{HF}$ solutions is shown below.

${\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Where,

• ${\text{M}}_1$ is the molarity of $\text{HF}$ solution.
• ${\text{V}}_1$ is the volume of $\text{HF}$ solution.
• ${\text{M}}_2$ is the molarity of $\text{NaOH}$ solution.
• ${\text{V}}_2$ is the volume of $\text{NaOH}$ solution.

Rearrange an above expression to calculate ${\text{V}}_2$ which is needed to neutralize $\text{HF}$ solution.

$$

${\text{V}}_2=\frac{{\text{M}}_1\times {\text{V}}_1}{{\text{M}}_2}$

Substitute the value of ${\text{M}}_1,{\text{M}}_2$ and ${\text{V}}_1$ in the above expression.

$\begin{array}{c}{\text{V}}_2=\frac{0.102\text{M}\times 35.0\text{mL}}{1.00\text{M}}\\ =3.57\text{mL}\end{array}$

Therefore, the volume of $1.00\text{M}$

$\text{NaOH}$ required to neutralize the given solution is $3.57\text{mL}$.

Interpretation Introduction

(c)

Interpretation:

The volume of $1.00\text{M}$

$\text{NaOH}$ required to neutralize the given solution is to be calculated.

$10.0\text{mL}$ of $0.143\text{M}$ phosphoric acid.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Liters}\text{of}\text{solution}}$.

Answer to Problem 73QAP

The volume of $1.00\text{M}$

$\text{NaOH}$ required to neutralize the given solution is $4.29\text{mL}$.

Explanation of Solution

The value of ${\text{M}}_1,{\text{V}}_1$ and ${\text{M}}_2$ is given to be $0.143\text{M}$, $10.0\text{mL}$ and $1.00\text{M}$ respectively.

The balanced equation when $\text{NaOH}$ reacts with ${\text{H}}_3{\text{PO}}_{\text{4}}$ is shown below.

$\text{3NaOH}+{\text{H}}_3{\text{PO}}_{\text{4}}\to 3{\text{H}}_{\text{2}}\text{O}+{\text{Na}}_3{\text{PO}}_{\text{4}}$

The above reaction indicates that three equivalents of $\text{NaOH}$ required to neutralize one equivalent of ${\text{H}}_3{\text{PO}}_{\text{4}}$.

The relationship between concentration and volume of $\text{NaOH}$ and ${\text{H}}_3{\text{PO}}_{\text{4}}$ solutions is shown below.

$3\times {\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Where,

• ${\text{M}}_1$ is the molarity of ${\text{H}}_3{\text{PO}}_{\text{4}}$ solution.
• ${\text{V}}_1$ is the volume of ${\text{H}}_3{\text{PO}}_{\text{4}}$ solution.
• ${\text{M}}_2$ is the molarity of $\text{NaOH}$ solution.
• ${\text{V}}_2$ is the volume of $\text{NaOH}$ solution.

Rearrange an above expression to calculate ${\text{V}}_2$ which is needed to neutralize ${\text{H}}_3{\text{PO}}_{\text{4}}$ solution.

$$

${\text{V}}_2=3\times \frac{{\text{M}}_1\times {\text{V}}_1}{{\text{M}}_2}$

Substitute the value of ${\text{M}}_1,{\text{M}}_2$ and ${\text{V}}_1$ in the above expression.

$\begin{array}{c}{\text{V}}_2=3\times \frac{0.143\text{M}\times 10.0\text{mL}}{1.00\text{M}}\\ =3\times \frac{1.43}{1.00}\text{mL}\\ =4.29\text{mL}\end{array}$

Therefore, the volume of $1.00\text{M}$

$\text{NaOH}$ required to neutralize the given solution is $4.29\text{mL}$.

Interpretation Introduction

(d)

Interpretation:

The volume of $1.00\text{M}$

$\text{NaOH}$ required to neutralize the given solution is to be calculated.

$35.0\text{mL}$ of $0.220\text{M}$ sulfuric acid

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

The molarity is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Liters}\text{of}\text{solution}}$.

Answer to Problem 73QAP

The volume of $1.00\text{M}$

$\text{NaOH}$ required to neutralize the given solution is $15.4\text{mL}$.

Explanation of Solution

The value of ${\text{M}}_1,{\text{V}}_1$ and ${\text{M}}_2$ is given to be $0.220\text{M}$, $35.0\text{mL}$ and $1.00\text{M}$ respectively.

The balanced equation when $\text{NaOH}$ reacts with ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is shown below.

$2\text{NaOH}+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to 2{\text{H}}_{\text{2}}\text{O}+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

The above reaction indicates that two equivalents of $\text{NaOH}$ required to neutralize one equivalent of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

The relationship between concentration and volume of $\text{NaOH}$ and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solutions is shown below.

$2\times {\text{M}}_1{\text{V}}_1={\text{M}}_2{\text{V}}_2$

Where,

• ${\text{M}}_1$ is the molarity of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution.
• ${\text{V}}_1$ is the volume of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution.
• ${\text{M}}_2$ is the molarity of $\text{NaOH}$ solution.
• ${\text{V}}_2$ is the volume of $\text{NaOH}$ solution.

Rearrange an above expression to calculate ${\text{V}}_2$ which is needed to neutralize ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution.

$$

${\text{V}}_2=2\times \frac{{\text{M}}_1\times {\text{V}}_1}{{\text{M}}_2}$

Substitute the value of ${\text{M}}_1,{\text{M}}_2$ and ${\text{V}}_1$ in the above expression.

$\begin{array}{c}{\text{V}}_2=2\times \frac{0.220\text{M}\times 35.0\text{mL}}{1.00\text{M}}\\ =2\times \frac{7.7}{1.00}\text{mL}\\ =15.4\text{mL}\end{array}$

Therefore, the volume of $1.00\text{M}$

$\text{NaOH}$ required to neutralize the given solution is $15.4\text{mL}$.