Chapter 15, Problem 74QAP

Interpretation Introduction

(a)

Interpretation:

The volume of $0.101\text{M}$

${\text{HNO}}_3$ required to neutralize the given solution is to be calculated.

$12.7\text{mL}$ of $\text{0}\text{.501}\text{M}$

$\text{NaOH}$.

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,

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

Answer to Problem 74QAP

The volume of $0.101\text{M}$

${\text{HNO}}_3$ required to neutralize the given solution is $63.0\text{mL}$.

Explanation of Solution

The value of ${\text{M}}_1,{\text{V}}_1$ and ${\text{M}}_2$ is given to be $0.501\text{M}$, $12.7\text{mL}$ and $0.101\text{M}$ respectively. ${\text{V}}_2$ is the volume which is required to neutralize the given reaction.

The balanced equation when ${\text{HNO}}_3$ reacts with $\text{NaOH}$ is shown below.

$\text{NaOH}+{\text{HNO}}_3\to {\text{H}}_{\text{2}}\text{O}+{\text{NaNO}}_3$

The above reaction indicates that one equivalent of ${\text{HNO}}_3$ required to neutralize one equivalent of $\text{NaOH}$.

The relationship between concentration and volume of $\text{NaOH}$ and ${\text{HNO}}_3$ solutions is shown below.

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

Where,

• ${\text{M}}_1$ is the molarity of $\text{NaOH}$ solution.
• ${\text{V}}_1$ is the volume of $\text{NaOH}$ solution.
• ${\text{M}}_2$ is the molarity of ${\text{HNO}}_3$ solution.
• ${\text{V}}_2$ is the volume of ${\text{HNO}}_3$ solution.

Rearrange an above expression to calculate the value of ${\text{V}}_2$.

$$

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

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

$\begin{array}{c}{\text{V}}_2=\frac{0.501\text{M}\times 12.7\text{mL}}{0.101\text{M}}\\ =63.0\text{mL}\end{array}$

Therefore, the volume of $0.101\text{M}$

${\text{HNO}}_3$ required to neutralize the given solution is $63.0\text{mL}$.

Interpretation Introduction

(b)

Interpretation:

The volume of $0.101\text{M}$

${\text{HNO}}_3$ required to neutralize the given solution is to be calculated.

$24.9\text{mL}$ of $0.00491\text{M}$

$\text{Ba}{\left(\text{OH}\right)}_2$.

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,

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

Answer to Problem 74QAP

The volume of $0.101\text{M}$

${\text{HNO}}_3$ required to neutralize the given solution is $2.42\text{mL}$.

Explanation of Solution

The value of ${\text{M}}_1,{\text{V}}_1$ and ${\text{M}}_2$ is given to be $0.00491\text{M}$, $24.9\text{mL}$ and $0.101\text{M}$ respectively. ${\text{V}}_2$ is the volume which is required to neutralize the given reaction.

The balanced equation when ${\text{HNO}}_3$ reacts with $\text{Ba}{\left(\text{OH}\right)}_2$ is shown below.

$\text{Ba}{\left(\text{OH}\right)}_2+2{\text{HNO}}_3\to 2{\text{H}}_{\text{2}}\text{O}+\text{Ba}{\left({\text{NO}}_3\right)}_2$

The above reaction indicates that two equivalents of ${\text{HNO}}_3$ required to neutralize one equivalent of $\text{Ba}{\left(\text{OH}\right)}_2$.

The relationship between concentration and volume of $\text{Ba}{\left(\text{OH}\right)}_2$ and ${\text{HNO}}_3$ solutions is shown below.

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

Where,

• ${\text{M}}_1$ is the molarity of $\text{Ba}{\left(\text{OH}\right)}_2$ solution.
• ${\text{V}}_1$ is the volume of $\text{Ba}{\left(\text{OH}\right)}_2$ solution.
• ${\text{M}}_2$ is the molarity of ${\text{HNO}}_3$ solution.
• ${\text{V}}_2$ is the volume of ${\text{HNO}}_3$ solution.

Rearrange an above expression to calculate the value of ${\text{V}}_2$.

$$

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

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

$\begin{array}{c}{\text{V}}_2=2\times \frac{0.00491\text{M}\times 24.9\text{mL}}{0.101\text{M}}\\ =2.42\text{mL}\end{array}$

Therefore, the volume of $0.101\text{M}$

${\text{HNO}}_3$ required to neutralize the given solution is $2.42\text{mL}$.

Interpretation Introduction

(c)

Interpretation:

The volume of $0.101\text{M}$

${\text{HNO}}_3$ required to neutralize the given solution is to be calculated.

$49.1\text{mL}$ of $0.103\text{M}$

${\text{NH}}_{\text{3}}$.

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,

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

Answer to Problem 74QAP

The volume of $0.101\text{M}$

${\text{HNO}}_3$ required to neutralize the given solution is $50.07\text{mL}$.

Explanation of Solution

The value of ${\text{M}}_1,{\text{V}}_1$ and ${\text{M}}_2$ is given to be $0.103\text{M}$, $49.1\text{mL}$ and $0.101\text{M}$ respectively. ${\text{V}}_2$ is the volume which is required to neutralize the given reaction.

The balanced equation when ${\text{HNO}}_3$ reacts with ${\text{NH}}_{\text{3}}$ is shown below.

${\text{NH}}_{\text{3}}+{\text{HNO}}_3\to {\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$

The above reaction indicates that one equivalent of ${\text{HNO}}_3$ reacts with one equivalent of ${\text{NH}}_{\text{3}}$.

The relationship between concentration and volume of ${\text{NH}}_{\text{3}}$ and ${\text{HNO}}_3$ solutions is shown below.

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

Where,

• ${\text{M}}_1$ is the molarity of ${\text{NH}}_{\text{3}}$ solution.
• ${\text{V}}_1$ is the volume of ${\text{NH}}_{\text{3}}$ solution.
• ${\text{M}}_2$ is the molarity of ${\text{HNO}}_3$ solution.
• ${\text{V}}_2$ is the volume of ${\text{HNO}}_3$ solution.

Rearrange an above expression to calculate the value of ${\text{V}}_2$.

$$

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

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

$\begin{array}{c}{\text{V}}_2=\frac{0.103\text{M}\times 49.1\text{mL}}{0.101\text{M}}\\ =50.07\text{mL}\end{array}$

Therefore, the volume of $0.101\text{M}$

${\text{HNO}}_3$ required to neutralize the given solution is $50.07\text{mL}$.

Interpretation Introduction

(d)

Interpretation:

The volume of $0.101\text{M}$

${\text{HNO}}_3$ required to neutralize the given solution is to be calculated.

$1.21\text{L}$ of $0.102\text{M}$

$\text{KOH}$.

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,

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

Answer to Problem 74QAP

The volume of $0.101\text{M}$

${\text{HNO}}_3$ required to neutralize the given solution is $1.22\text{L}$.

Explanation of Solution

The value of ${\text{M}}_1,{\text{V}}_1$ and ${\text{M}}_2$ is given to be $0.102\text{M}$, $1.21\text{L}$ and $0.101\text{M}$ respectively. ${\text{V}}_2$ is the volume which is required to neutralize the given reaction.

The balanced equation when ${\text{HNO}}_3$ reacts with $\text{KOH}$ is shown below.

$\text{KOH}+{\text{HNO}}_3\to {\text{KNO}}_3+{\text{H}}_{\text{2}}\text{O}$

The above reaction indicates that one equivalent of ${\text{HNO}}_3$ reacts with one equivalent of $\text{KOH}$.

The relationship between concentration and volume of $\text{KOH}$ and ${\text{HNO}}_3$ solutions is shown below.

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

Where,

• ${\text{M}}_1$ is the molarity of $\text{KOH}$ solution.
• ${\text{V}}_1$ is the volume of $\text{KOH}$ solution.
• ${\text{M}}_2$ is the molarity of ${\text{HNO}}_3$ solution.
• ${\text{V}}_2$ is the volume of ${\text{HNO}}_3$ solution.

Rearrange an above expression to calculate the value of ${\text{V}}_2$.

$$

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

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

$\begin{array}{c}{\text{V}}_2=\frac{0.102\text{M}\times 1.21\text{L}}{0.101\text{M}}\\ =1.22\text{L}\end{array}$

Therefore, the volume of $0.101\text{M}$

${\text{HNO}}_3$ required to neutralize the given solution is $1.22\text{L}$.