Chapter 15, Problem 58A

(a)

Interpretation Introduction

Interpretation: The normality of 0.50 M acetic acid ( $C{H}_{3}COOH$ ) needs to be determined.

Concept introduction: Normality mainly used in acid-base reaction where it gives more in-depth information of the concentration of the solution.

  $\text{Normality}\left(N\right)=\frac{\text{grams equivalent of substance}}{\text{volume of the solution in litre}}$

Molecules reactive capacity is known as $\text{grams equivalent}$ weight. Simple relation between normality and molarity is as follows:

  $\text{Normality}\left(N\right)=n\text{ (number of equivalent)}\times M\left(\text{Molarity}\right)$

Answer to Problem 58A

The normality of 0.50 M acetic acid ( $C{H}_{3}COOH$ ) is 0.50 N.

Explanation of Solution

Acetic acid dissociates as follows: $C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$

Since one mol acetic acid one $H^{+}$ upon dissociation acetic acid is mono-acidic.

The normality of acetic acid ( $C{H}_{3}COOH$ ) $=(0.50\times 1)N=0.50N$

(b)

Interpretation Introduction

Interpretation: The normality of 0.00250 M sulfuric acid ( $H_{2}S{O}_4$ ) needs to be determined.

Concept introduction: Normality mainly used in acid-base reaction where it gives more in-depth information of the concentration of the solution.

  $\text{Normality}\left(N\right)=\frac{\text{grams equivalent of substance}}{\text{volume of the solution in litre}}$

Molecules reactive capacity is known as $\text{grams equivalent}$ weight. Simple relation between normality and molarity is as follows:

  $\text{Normality}\left(N\right)=n\text{ (number of equivalent)}\times M\left(\text{Molarity}\right)$

Answer to Problem 58A

The normality of 0.00250 M sulfuric acid ( $H_{2}S{O}_4$ ) is 0.005 N.

Explanation of Solution

Sulfuric acid dissociates as follows: $H_{2}S{O}_4\rightleftharpoons \text{ 2}{H}^{+}+{\text{ SO}}_4^{2-}$

Since one mol sulfuric acid gives two moles of $H^{+}$ upon dissociation and hence it is di-acidic.

The normality of sulfuric acid ( $H_{2}S{O}_4$ ) $=(0.00250\times 2)N=0.005N$

(c)

Interpretation Introduction

Interpretation: The normality of 0.10 M potassium hydroxide ( $KOH$ ) needs to be determined.

Concept introduction: Normality mainly used in acid-base reaction where it gives more in-depth information of the concentration of the solution.

  $\text{Normality}\left(N\right)=\frac{\text{grams equivalent of substance}}{\text{volume of the solution in litre}}$

Molecules reactive capacity is known as $\text{grams equivalent}$ weight. Simple relation between normality and molarity is as follows:

  $\text{Normality}\left(N\right)=n\text{ (number of equivalent)}\times M\left(\text{Molarity}\right)$

Answer to Problem 58A

The normality of 0.10 M potassium hydroxide ( $KOH$ ) is 0.10 N.

Explanation of Solution

Potassium hydroxide dissociates as follows: $KOH\rightleftharpoons K^{+}+O{H}^{-}$

Since one mol of potassium hydroxide gives one mole of $O{H}^{-}$ upon dissociation and hence it is mono-basic.

The normality of potassium hydroxide ( $KOH$ ) $=(0.10\times 1)N=0.10N$