Chapter 15, Problem 130AP

Interpretation Introduction

(a)

Interpretation:

The normality of the given solution is to be calculated.

Concept Introduction:

The normality of a solution is defined as the gram equivalent weight of a solute dissolved in one liter of the solution. The normality of solution is given as:

$\text{N}=\frac{{\text{n}}_{\text{eq}}}{\text{V}}$

Where,

• ${\text{n}}_{\text{eq}}$ represents the number of equivalents of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 130AP

The normality of the given solution of $\text{HCl}$ is $0.823\text{N}$.

Explanation of Solution

The mass of $\text{HCl}$ dissolved in solution is $15.0\text{g}$.

The volume of the $\text{HCl}$ solution is $500\text{mL}$.

The molar mass of $\text{HCl}$ is $36.469\text{g}\cdot {\text{mol}}^{-1}$.

The dissociation reaction of $\text{HCl}$ in water is represented as:

$\text{HCl}\left(\text{a}\text{q}\right)\to {\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{Cl}}^{-}\left(\text{a}\text{q}\right)$

The number of ${\text{H}}^{+}$ ion released by $\text{HCl}$ is $1$.

Hence the equivalence factor of $\text{HCl}$ is $1\text{eq}\cdot {\text{mol}}^{-1}$.

The equivalent mass of a substance is given as:

${\text{M}}_{\text{eq}}=\frac{{\text{M}}_{\text{m}}}{\text{n}}$

Where,

• ${\text{M}}_{\text{m}}$ represents the molar mass of the substance.
• $\text{n}$ represents the equivalence factor of the substance.

Substitute the value of ${\text{M}}_{\text{m}}$ and $\text{n}$ in the above equation.

$\begin{array}{c}{\text{M}}_{\text{eq}}=\frac{36.469\text{g}\cdot {\text{mol}}^{-1}}{1\text{eq}\cdot {\text{mol}}^{-1}}\\ =36.469\text{g}\cdot {\text{eq}}^{-1}\end{array}$

The normality of a solution is given as:

$\text{N}=\frac{\text{m}}{\text{V}{\text{M}}_{\text{eq}}}$

Where,

• $\text{m}$ represents the mass of the solute.
• ${\text{M}}_{\text{eq}}$ represents the equivalent mass of the solute.
• $\text{V}$ represents the volume of the solution.

Substitute $\text{m}$, ${\text{M}}_{\text{eq}}$ and $\text{V}$ in the above equation.

$\begin{array}{c}\text{N}=\frac{15.0\text{g}}{\left(500\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(36.469\text{g}\cdot {\text{eq}}^{-1}\right)}\\ =\left(0.823\text{eq}\cdot {\text{L}}^{-1}\right)\left(\frac{1\text{N}}{1\text{eq}\cdot {\text{L}}^{-1}}\right)\\ =0.823\text{N}\end{array}$

Therefore, the normality of the given solution is $0.823\text{N}$.

Interpretation Introduction

(b)

Interpretation:

The normality of the given solution is to be calculated.

Concept Introduction:

The normality of a solution is defined as the gram equivalent weight of a solute dissolved in one liter of the solution. The normality of solution is given as:

$\text{N}=\frac{{\text{n}}_{\text{eq}}}{\text{V}}$

Where,

• ${\text{n}}_{\text{eq}}$ represents the number of equivalents of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 130AP

The normality of the given solution of ${\text{H}}_2{\text{SO}}_4$ is $3.9968\text{N}$.

Explanation of Solution

The mass of ${\text{H}}_2{\text{SO}}_4$ dissolved in solution is $49.0\text{g}$.

The volume of the ${\text{H}}_2{\text{SO}}_4$ solution is $250\text{mL}$.

The molar mass of ${\text{H}}_2{\text{SO}}_4$ is $98.079\text{g}\cdot {\text{mol}}^{-1}$.

The dissociation reaction of ${\text{H}}_2{\text{SO}}_4$ in water is represented as:

${\text{H}}_2{\text{SO}}_4\left(\text{a}\text{q}\right)\to 2{\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{SO}}_4^{2-}\left(\text{a}\text{q}\right)$

The number of ${\text{H}}^{+}$ ion released by ${\text{H}}_2{\text{SO}}_4$ is $2$.

Hence the equivalence factor of ${\text{H}}_2{\text{SO}}_4$ is $2\text{eq}\cdot {\text{mol}}^{-1}$.

The equivalent mass of a substance is given as:

${\text{M}}_{\text{eq}}=\frac{{\text{M}}_{\text{m}}}{\text{n}}$

Where,

• ${\text{M}}_{\text{m}}$ represents the molar mass of the substance.
• $\text{n}$ represents the equivalence factor of the substance.

Substitute the value of ${\text{M}}_{\text{m}}$ and $\text{n}$ in the above equation.

$\begin{array}{c}{\text{M}}_{\text{eq}}=\frac{98.079\text{g}\cdot {\text{mol}}^{-1}}{2\text{eq}\cdot {\text{mol}}^{-1}}\\ =49.0395\text{g}\cdot {\text{eq}}^{-1}\end{array}$

The normality of a solution is given as:

$\text{N}=\frac{\text{m}}{\text{V}{\text{M}}_{\text{eq}}}$

Where,

• $\text{m}$ represents the mass of the solute.
• ${\text{M}}_{\text{eq}}$ represents the equivalent mass of the solute.
• $\text{V}$ represents the volume of the solution.

Substitute $\text{m}$, ${\text{M}}_{\text{eq}}$ and $\text{V}$ in the above equation.

$\begin{array}{c}\text{N}=\frac{49.0\text{g}}{\left(250\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(49.0395\text{g}\cdot {\text{eq}}^{-1}\right)}\\ =\left(3.9968\text{eq}\cdot {\text{L}}^{-1}\right)\left(\frac{1\text{N}}{1\text{eq}\cdot {\text{L}}^{-1}}\right)\\ =3.9968\text{N}\end{array}$

Therefore, the normality of the given solution is $3.9968\text{N}$.

Interpretation Introduction

(c)

Interpretation:

The normality of the given solution is to be calculated.

Concept Introduction:

The normality of a solution is defined as the gram equivalent weight of a solute dissolved in one liter of the solution. The normality of solution is given as:

$\text{N}=\frac{{\text{n}}_{\text{eq}}}{\text{V}}$

Where,

• ${\text{n}}_{\text{eq}}$ represents the number of equivalents of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 130AP

The normality of the given solution of ${\text{H}}_3{\text{PO}}_4$ is $3.0614\text{N}$.

Explanation of Solution

The mass of ${\text{H}}_3{\text{PO}}_4$ dissolve in solution is $10.0\text{g}$.

The volume of the ${\text{H}}_3{\text{PO}}_4$ solution is $100\text{mL}$.

The molar mass of ${\text{H}}_3{\text{PO}}_4$ is $97.994\text{g}\cdot {\text{mol}}^{-1}$.

The dissociation reaction of ${\text{H}}_3{\text{PO}}_4$ in water is represented as:

${\text{H}}_3{\text{PO}}_4\left(\text{a}\text{q}\right)\to 3{\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{PO}}_4^{3-}\left(\text{a}\text{q}\right)$

The number of ${\text{H}}^{+}$ ion released by ${\text{H}}_3{\text{PO}}_4$ is $3$.

Hence the equivalence factor of ${\text{H}}_3{\text{PO}}_4$ is $3\text{eq}\cdot {\text{mol}}^{-1}$.

The equivalent mass of a substance is given as:

${\text{M}}_{\text{eq}}=\frac{{\text{M}}_{\text{m}}}{\text{n}}$

Where,

• ${\text{M}}_{\text{m}}$ represents the molar mass of the substance.
• $\text{n}$ represents the equivalence factor of the substance.

Substitute the value of ${\text{M}}_{\text{m}}$ and $\text{n}$ in the above equation.

$\begin{array}{c}{\text{M}}_{\text{eq}}=\frac{97.994\text{g}\cdot {\text{mol}}^{-1}}{3\text{eq}\cdot {\text{mol}}^{-1}}\\ =32.6647\text{g}\cdot {\text{eq}}^{-1}\end{array}$

The normality of a solution is given as:

$\text{N}=\frac{\text{m}}{\text{V}{\text{M}}_{\text{eq}}}$

Where,

• $\text{m}$ represents the mass of the solute.
• ${\text{M}}_{\text{eq}}$ represents the equivalent mass of the solute.
• $\text{V}$ represents the volume of the solution.

Substitute $\text{m}$, ${\text{M}}_{\text{eq}}$ and $\text{V}$ in the above equation.

$\begin{array}{c}\text{N}=\frac{10.0\text{g}}{\left(100\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(32.6647\text{g}\cdot {\text{eq}}^{-1}\right)}\\ =\left(3.0614\text{eq}\cdot {\text{L}}^{-1}\right)\left(\frac{1\text{N}}{1\text{eq}\cdot {\text{L}}^{-1}}\right)\\ =3.0614\text{N}\end{array}$

Therefore, the normality of the given solution is $3.0614\text{N}$.