Chapter 4, Problem 61E

(a)

Interpretation Introduction

Interpretation: The volume of $0.100\text{M}$

  $\text{HCl}$ that reacts completely with $50.0\text{mL}$ of $0.100\text{M}$

  $\text{NaOH}$ is to be calculated.

Concept introduction: The molarity of any solution is obtained from the moles of solute exist in per unit volume of that solution.The formula for the molarity equation, when neutralization reaction between acid and base is studied in terms of the molarities and volumes is as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Answer to Problem 61E

The volume of $0.100\text{M}$

  $\text{HCl}$ reacts completely with given base is $50.0\text{mL}$ .

Explanation of Solution

The given volume of $\text{NaOH}$ solution is $50.0\text{mL}$ .

The given molarityof $\text{NaOH}$ solution is $0.100\text{}\text{M}$ .

The given molarity of $\text{HCl}$ solution is $0.100\text{}\text{M}$ .

The required balanced chemical equation between two solutions is given below:

  $\text{NaOH}+\text{HCl}\to \text{NaCl}+{\text{H}}_2\text{O}$

Thus, the n-factor for the base NaOH is 1 as it has one ${\text{OH}}^{-}$ and the n-factor of the acid $\text{HCl}$ is also 1 as it has only one ${\text{H}}^{+}$ ion.

The molarity equation formula is as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Where,

• $M_1$ is molarity of hydrochloric acid which equals to $0.100\text{}\text{M}$ .
• $V_1$is volume of hydrochloric acid to be calculated.
• $n_1$isn-factor of hydrochloric acid.
• $M_2$ is molarity of sodium hydroxide which equals to $0.100\text{}\text{M}$ .
• $V_2$ is volume of sodium hydroxide that is $50.0\text{mL}$ .
• $n_2$isn-factor of sodium hydroxide.

Substitute these values in above formula as shown below:

  $\begin{array}{c}1\times 0.100\text{M}\times V_1=1\times 0.100\text{M}\times 50.0\text{mL}\\ V_1=\frac{0.100\text{M}\times 50.0\text{mL}\times 1}{1\times 0.100\text{M}}\\ V_1=50.0\text{mL}\end{array}$

Therefore, the volume of $0.100\text{M}$

  $\text{HCl}$ reacts completely is $50.0\text{mL}$ .

(b)

Interpretation Introduction

Interpretation: The volume of $0.100\text{M}$

  ${\text{H}}_2{\text{SO}}_3$ that reacts completely with $50.0\text{mL}$ of $0.100\text{M}$

  $\text{NaOH}$ is to be calculated.

Concept introduction: The molarity of any solution is obtained from the moles of solute exist in per unit volume of that solution.The formula for the molarity equation, when neutralization reaction between acid and base is studied in terms of the molarities and volumes is as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Answer to Problem 61E

The volume of $0.100\text{M}$

  ${\text{H}}_2{\text{SO}}_3$ reacts completely with given base is $25.0\text{mL}$ .

Explanation of Solution

The given volume of $\text{NaOH}$ solution is $50.0\text{mL}$ .

The given molarityof $\text{NaOH}$ solution is $0.100\text{}\text{M}$ .

The given molarity of ${\text{H}}_2{\text{SO}}_3$ solution is $0.100\text{}\text{M}$ .

The required balanced chemical equation between two solutions is given below:

  $\text{2NaOH}+{\text{H}}_2{\text{SO}}_3\to {\text{Na}}_2{\text{SO}}_3+2{\text{H}}_2\text{O}$

Thus, the n-factor for the base, $\text{NaOH}$ is $1$ as it has one ${\text{OH}}^{-}$ and the n-factor of the acid ${\text{H}}_2{\text{SO}}_3$ is $2$ as it has two ${\text{H}}^{+}$ ion.

The molarity equation formula is as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Where,

• $M_1$ is molarity of ${\text{H}}_2{\text{SO}}_3$ which equals to $0.100\text{}\text{M}$ .
• $V_1$ is volume of ${\text{H}}_2{\text{SO}}_3$to be calculated.
• $n_1$ isn-factor of ${\text{H}}_2{\text{SO}}_3$ .
• $M_2$ is molarity of sodium hydroxide which equals to $0.100\text{}\text{M}$ .
• $V_2$ is volume of sodium hydroxide that is $50.0\text{mL}$ .
• $n_2$isn-factor sodium hydroxide.

Substitute these values in above formula as shown below:

  $\begin{array}{c}2\times 0.100\text{M}\times V_1=1\times 0.100\text{M}\times 50.0\text{mL}\\ V_1=\frac{1\times 0.100\text{M}\times 50.0\text{mL}}{\text{2 }\times 0.100\text{M}}\\ V_1=25.0\text{mL}\end{array}$

Therefore, the volume of $0.100\text{M}$

  ${\text{H}}_2{\text{SO}}_3$ reacts completely is $25.0\text{mL}$ .

(c)

Interpretation Introduction

Interpretation: The volume of $0.200\text{}\text{M}$

  ${\text{H}}_3{\text{PO}}_4$ thatreacts completely with $50.0\text{mL}$ of $0.100\text{M}$

  $\text{NaOH}$ is to be calculated.

Concept introduction: The molarity of any solution is obtained from the moles of solute exist in per unit volume of that solution.

The formula for the molarity equation, when neutralization reaction between acid and base is studied in terms of the molarities and volumes is as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Answer to Problem 61E

The volume of $0.200\text{}\text{M}$

  ${\text{H}}_3{\text{PO}}_4$ reacts completely with given base is $8.33\text{mL}$ .

Explanation of Solution

The given volume of $\text{NaOH}$ solution is $50.0\text{mL}$ .

The given molarityof $\text{NaOH}$ solution is $0.100\text{}\text{M}$ .

The given molarity of ${\text{H}}_3{\text{PO}}_4$ solution is $0.200\text{}\text{M}$ .

The required balanced chemical equation between two solutions is given below:

  $\text{3NaOH}+{\text{H}}_3{\text{PO}}_4\to {\text{Na}}_3{\text{PO}}_4+3{\text{H}}_2\text{O}$

Thus, the n-factor for the base $\text{NaOH}$ is $1$ as it has one ${\text{OH}}^{-}$ and the n-factor of the acid ${\text{H}}_3{\text{PO}}_4$ is $3$ as it has three ${\text{H}}^{+}$ ions.

The molarity equation formula is as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Where,

• $M_1$ is molarity of ${\text{H}}_3{\text{PO}}_4$ which equals to $0.200\text{}\text{M}$ .
• $V_1$ is volume of ${\text{H}}_3{\text{PO}}_4$to be calculated.
• $n_1$ isn-factor of ${\text{H}}_3{\text{PO}}_4$ .
• $M_2$ is molarity of sodium hydroxide which equals to $0.100\text{}\text{M}$ .
• $V_2$ is volume of sodium hydroxide that is $50.0\text{mL}$ .
• $n_2$is-factor of sodium hydroxide.

Substitute these values in above formula as shown below:

  $\begin{array}{c}3\times 0.200\text{M}\times V_1=1\times 0.100\text{M}\times 50.0\text{mL}\\ V_1=\frac{1\times 0.100\text{M}\times 50.0\text{mL}}{3\times 0.200\text{M}}\\ V_1=8.33\text{mL}\end{array}$

Therefore, the volume of $0.200\text{}\text{M}$

  ${\text{H}}_3{\text{PO}}_4$ reacts completely is $8.33\text{mL}$ .

(d)

Interpretation Introduction

Interpretation: The volume of $0.150\text{}\text{M}$

  ${\text{HNO}}_3$ thatreacts completely with $50.0\text{mL}$ of $0.100\text{M}$

  $\text{NaOH}$ is to be calculated.

Concept introduction: The molarity of any solution is obtained from the moles of solute exist in per unit volume of that solution.The formula for the molarity equation, when neutralization reaction between acid and base is studied in terms of the molarities and volumes is as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Answer to Problem 61E

The volume of $0.150\text{}\text{M}$

  ${\text{HNO}}_3$ reacts is $33.33\text{mL}$ .

Explanation of Solution

The given volume of $\text{NaOH}$ solution is $50.0\text{mL}$ .

The given molarityof $\text{NaOH}$ solution is $0.100\text{}\text{M}$ .

The given molarity of ${\text{HNO}}_3$ solution is $0.150\text{}\text{M}$ .

The required balanced chemical equation between two solutions is given below:

  $\text{NaOH}+{\text{HNO}}_3\to {\text{NaNO}}_3+{\text{H}}_2\text{O}$

Thus, the n-factor for the base NaOH is 1 as it has one ${\text{OH}}^{-}$ and the n-factor of the acid ${\text{HNO}}_3$ is also 1 as it has only one ${\text{H}}^{+}$ ion.

The molarity equation formula is as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Where,

• $M_1$ is molarity of ${\text{HNO}}_3$ which equals to $0.150\text{}\text{M}$ .
• $V_1$ is volume of ${\text{HNO}}_3$to be calculated.
• $n_1$ isn-factor of ${\text{HNO}}_3$ .
• $M_2$ is molarity of sodium hydroxide which equals to $0.100\text{}\text{M}$ .
• $V_2$ is volume of sodium hydroxide that is $50.0\text{mL}$ .
• $n_2$isn-factor of sodium hydroxide.

Substitute these values in above formula as shown below:

  $\begin{array}{c}1\times 0.150\text{M}\times V_1=1\times 0.100\text{M}\times 50.0\text{mL}\\ V_1=\frac{1\times 0.100\text{M}\times 50.0\text{mL}}{1\times 0.150\text{M}}\\ V_1=33.33\text{mL}\end{array}$

Therefore, the volume of $0.150\text{}\text{M}$

  ${\text{HNO}}_3$ reacts completely is $33.33\text{mL}$ .

(e)

Interpretation Introduction

Interpretation: The volume of $0.200\text{}\text{M}$

  ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ thatreacts completely with $50.0\text{mL}$ of $0.100\text{M}$

  $\text{NaOH}$ is to be calculated.

Concept introduction: The molarity of any solution is obtained from the moles of solute exist in per unit volume of that solution.The formula for the molarity equation, when neutralization reaction between acid and base is studied in terms of the molarities and volumes is as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Answer to Problem 61E

The volume of $0.200\text{}\text{M}$

  ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ reacts completely with given base is $25.0\text{mL}$ .

Explanation of Solution

The given volume of $\text{NaOH}$ solution is $50.0\text{mL}$ .

The given molarityof $\text{NaOH}$ solution is $0.100\text{}\text{M}$ .

The given molarity of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ solution is $0.200\text{}\text{M}$ .

The required balanced chemical equation between two solutions is given below:

  $\text{NaOH}+{\text{HC}}_2{\text{H}}_3{\text{O}}_2\to {\text{NaC}}_2{\text{H}}_3{\text{O}}_2+{\text{H}}_2\text{O}$

Thus, the n-factor for the base $\text{NaOH}$ is 1 as it has one ${\text{OH}}^{-}$ and the n-factor of the acid ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ is also 1 as it has only one ${\text{H}}^{+}$ ion.

The molarity equation formula is as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Where,

• $M_1$ is molarity of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ which equals to $0.200\text{}\text{M}$ .
• $V_1$ is volume of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$to be calculated.
• $n_1$ isn-factor of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ .
• $M_2$ is molarity of sodium hydroxide which equals to $0.100\text{}\text{M}$ .
• $V_2$ is volume of sodium hydroxide that is $50.0\text{mL}$ .
• $n_2$isn-factor of sodium hydroxide.

Substitute these values in above formula as shown below:

  $\begin{array}{c}1\times 0.200\text{M}\times V_1=1\times 0.100\text{M}\times 50.0\text{mL}\\ V_1=\frac{1\times 0.100\text{M}\times 50.0\text{mL}}{1\times 0.200\text{M}}\\ V_1=25.0\text{mL}\end{array}$

Therefore, the volume of $0.200\text{}\text{M}$

  ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ reacts completely is $25.0\text{mL}$ .

(f)

Interpretation Introduction

Interpretation: The volume of $0.300\text{}\text{M}$

  ${\text{H}}_2{\text{SO}}_4$ that reacts completely with $50.0\text{mL}$ of $0.100\text{M}$

  $\text{NaOH}$ is to be calculated.

Concept introduction: The molarity of any solution is obtained from the moles of solute exist in per unit volume of that solution.The formula for the molarity equation, when neutralization reaction between acid and base is studied in terms of the molarities and volumes is as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Answer to Problem 61E

The volume of $0.300\text{}\text{M}$

  ${\text{H}}_2{\text{SO}}_4$ reacts completely is $8.33\text{mL}$ .

Explanation of Solution

The given volume of $\text{NaOH}$ solution is $50.0\text{ml}$ .

The given molarityof $\text{NaOH}$ solution is $0.100\text{}\text{M}$ .

The given molarity of ${\text{H}}_2{\text{SO}}_4$ solution is $0.300\text{}\text{M}$ .

The required balanced chemical equation between two solutions is given below:

  $\text{2NaOH}+{\text{H}}_2{\text{SO}}_4\to {\text{Na}}_2{\text{SO}}_4+2{\text{H}}_2\text{O}$

Thus, the n-factor for the base $\text{NaOH}$ is $1$ as it has one ${\text{OH}}^{-}$ and the n-factor of the acid ${\text{H}}_2{\text{SO}}_4$ is $2$ as it has two ${\text{H}}^{+}$ ion.

The molarity equation formula is as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Where,

• $M_1$ is molarity of ${\text{H}}_2{\text{SO}}_4$ which equals to $0.300\text{}\text{M}$ .
• $V_1$ is volume of ${\text{H}}_2{\text{SO}}_4$to be calculated.
• $n_1$ isn-factor of ${\text{H}}_2{\text{SO}}_4$ .
• $M_2$ is molarity of sodium hydroxide which equals to $0.100\text{}\text{M}$ .
• $V_2$ is volume of sodium hydroxide that is $50.0\text{mL}$ .
• $n_2$isn-factor sodium hydroxide.

Substitute these values in above formula as shown below:

  $\begin{array}{c}2\times 0.300\text{M}\times V_1=1\times 0.100\text{M}\times 50.0\text{mL}\\ V_1=\frac{1\times 0.100\text{M}\times 50.0\text{mL}}{2\times 0.300\text{M}}\\ V_1=8.33\text{mL}\end{array}$

Therefore, the volume of $0.300\text{}\text{M}$

  ${\text{H}}_2{\text{SO}}_4$ reacts completely is $8.33\text{mL}$ .