Chapter 11, Problem 38E

(a)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$\text{Normality(N)=}\frac{\text{Number}\text{of}\text{gram}\text{equivalent(in}\text{g)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$

Answer to Problem 38E

The normality of $\text{HCl}$ is $\text{0}\text{.25N}$ $\text{Equivalent}\text{mass}\text{=}\text{molar}\text{mass}\text{of}\text{HCl}\text{=36}\text{.46}\text{g}$

Explanation of Solution

Record the given data

Gram equivalent of $\text{HCl}$ = $0.25$

Molarity of $\text{HCl}$ = $\text{0}\text{.25}\text{M}$

To calculate the normality of $\text{HCl}$

$\text{HCl}$ has one acidic protons because it is a monoprotic acid Hence, its molarity can be one time

Therefore, the normality of $\text{HCl}$ is given by,

$\begin{array}{l}\text{Normality}\text{=}\frac{\text{0}\text{.25}\text{mol}\text{HCl}}{\text{L}}\text{×}\frac{\text{1}\text{equivalent}}{\text{mol}\text{HCl}}\text{=}\frac{\text{0}\text{.25}\text{equivalents}}{\text{L}}\\ \text{Equivalent}\text{mass}\text{=}\text{molar}\text{mass}\text{of}\text{HCl}\text{=36}\text{.46}\text{g}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$\text{Normality(N)=}\frac{\text{Number}\text{of}\text{gram}\text{equivalent(in}\text{g)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$

Answer to Problem 38E

The normality of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{0}\text{.21N}$

Explanation of Solution

Record the given data

Gram equivalent of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ = $0.21$

Molarity of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ = $\text{0}\text{.105}\text{M}$

The gram equivalent and molarity of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ are recorded as shown above.

To calculate the normality of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ has two acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is given by,

$\begin{array}{l}\text{Normality}\text{=}\frac{\text{0}\text{.105}\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{L}}\text{×}\frac{\text{2}\text{equivalent}}{\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{=}\frac{\text{0}\text{.210}\text{equivalents}}{\text{L}}\\ \text{Equivalent}\text{mass}\text{=}\frac{\text{1}}{\text{2}}\text{(molar}\text{mass}\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{)}\text{=}\frac{\text{1}}{\text{2}}\text{(98}\text{.09)=49}\text{.05}\text{g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$\text{Normality(N)=}\frac{\text{Number}\text{of}\text{gram}\text{equivalent(in}\text{g)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$

Answer to Problem 38E

The normality of ${\text{H}}_3{\text{PO}}_{\text{4}}$ is $\text{0}\text{.16N}$

Explanation of Solution

Record the given data

Gram equivalent of ${\text{H}}_3{\text{PO}}_{\text{4}}$ = $0.16$

Molarity of ${\text{H}}_3{\text{PO}}_{\text{4}}$ = $\text{5}{\text{.3×10}}^{\text{-2}}\text{M}$

To calculate the normality of ${\text{H}}_3{\text{PO}}_{\text{4}}$

${\text{H}}_3{\text{PO}}_{\text{4}}$ has three acidic protons because it is a diprotic acid Hence, its molarity can be three times

Therefore, the normality of ${\text{H}}_3{\text{PO}}_{\text{4}}$ is given by,

$\begin{array}{l}\text{Normality}\text{=}\frac{\text{5}{\text{.3× 10}}^{\text{-2}}\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{L}}\text{×}\frac{\text{3}\text{equivalent}}{\text{mol}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}\text{=}\frac{\text{0}\text{.16}\text{equivalents}}{\text{L}}\\ \text{Equivalent}\text{mass}\text{=}\frac{\text{1}}{\text{3}}\text{(molar}\text{mass}\text{of}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{)}\text{=}\frac{\text{1}}{\text{3}}\text{(97}\text{.09)=32}\text{.66}\text{g}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$\text{Normality(N)=}\frac{\text{Number}\text{of}\text{gram}\text{equivalent(in}\text{g)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$

Answer to Problem 38E

The normality of $\text{NaOH}$ is $\text{0}\text{.134N}$

Explanation of Solution

Record the given data

Gram equivalent of $\text{NaOH}$ = $0.134$

Molarity of $\text{NaOH}$ = $\text{0}\text{.134}\text{M}$

To calculate the normality of $\text{NaOH}$

$\text{NaOH}$ has mono acidic protons because it is a monoprotic acid Hence, its molarity can be one times

Therefore, the normality of $\text{NaOH}$ is given by,

$\begin{array}{l}\text{Normality}\text{=}\frac{\text{0}\text{.134}\text{mol}\text{NaOH}}{\text{L}}\text{×}\frac{\text{1}\text{equivalent}}{\text{mol}\text{NaOH}}\text{=}\frac{\text{0}\text{.134}\text{equivalents}}{\text{L}}\\ \text{Equivalent}\text{mass}\text{= molar mass of NaOH 40}\text{g}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$\text{Normality(N)=}\frac{\text{Number}\text{of}\text{gram}\text{equivalent(in}\text{g)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$

Answer to Problem 38E

The normality of ${\text{Ca(OH)}}_{\text{2}}$ is $\text{0}\text{.002605 N}$

Explanation of Solution

Record the given data

Gram equivalent of ${\text{Ca(OH)}}_{\text{2}}$ = $0.0104$

Molarity of ${\text{Ca(OH)}}_{\text{2}}$ = $\text{0}\text{.00521}\text{M}$

To calculate the normality of ${\text{Ca(OH)}}_{\text{2}}$

$\text{NaOH}$ has di acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of ${\text{Ca(OH)}}_{\text{2}}$ is given by,

$\begin{array}{l}\text{Normality}\text{=}\frac{\text{0}\text{.00521}\text{mol}{\text{Ca(OH)}}_{\text{2}}}{\text{L}}\text{×}\frac{\text{2}\text{equivalent}}{\text{mol}{\text{Ca(OH)}}_{\text{2}}}\text{=}\frac{\text{0}\text{.0104}\text{equivalents}}{\text{L}}\\ \text{Equivalent}\text{mass}\text{= }\frac{\text{1}}{\text{2}}{\text{(molar mass of Ca(OH)}}_{\text{2}}\text{)=}\frac{\text{1}}{\text{2}}\text{(74}\text{.10)=37}\text{.05}\text{g}\end{array}$