Chapter 9, Problem 9.41P

a)

Interpretation Introduction

Interpretation:

The chemical reaction with equilibrium constants ${\text{K}}_{\text{b}}$ and ${\text{K}}_{\text{a}}$ for Imidazole and Imidazole hydrochloride have to be written.

Concept Introduction:

Acid dissociation constant:

Consider a weak-acid equilibrium reaction,

$\text{HA}\stackrel{{\text{K}}_{\text{a}}}{\rightleftharpoons }{\text{H}}^{\text{+}}{\text{+A}}^{\text{-}}$

The acid dissociation constant ${\text{K}}_{\text{a}}$ can be given as,

${\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{A}}^{\text{-}}\right]}{\left[\text{HA}\right]}$

A weak acid is one that doesn’t undergo completion.

Base dissociation constant:

Consider a reaction of weak base,

${\text{B+H}}_{\text{2}}\text{O}\stackrel{{\text{K}}_{\text{b}}}{\rightleftharpoons }{\text{BH}}^{\text{+}}{\text{+OH}}^{\text{-}}$

The value of ${\text{K}}_{\text{b}}$ can be calculated as,

${\text{K}}_{\text{b}}\text{=}\frac{\left[{\text{BH}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[\text{B}\right]}$

Where ${\text{K}}_{\text{b}}$ is called as base dissociation constant.

To write the chemical reaction with equilibrium constants ${\text{K}}_{\text{b}}$ and ${\text{K}}_{\text{a}}$ for Imidazole and Imidazole hydrochloride

Explanation of Solution

The chemical reaction with equilibrium constants ${\text{K}}_{\text{b}}$ and ${\text{K}}_{\text{a}}$ for Imidazole and Imidazole hydrochloride can be written as,

b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of a solution has to be calculated.

Concept Introduction:

Henderson-Hasselbalch equation:

Henderson- Hasselbalch equation is rearranged form of acid dissociation expression.

For acid:

Consider an acidic reaction:

$\text{HA}\stackrel{{\text{K}}_{\text{a}}}{\rightleftharpoons }{\text{H}}^{\text{+}}{\text{+A}}^{\text{-}}$

The Henderson- Hasselbalch equation for an acidic reaction can be given as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[{\text{A}}^{\text{-}}\right]}{\left[\text{HA}\right]}$

Answer to Problem 9.41P

The $\text{pH}$ of a solution is $\text{7}\text{.18}$ .

Explanation of Solution

Given,

Mass of Imidazole = $\text{1}\text{.00}\text{g}$

Mass of Imidazole hydrochloride = $\text{1}\text{.00}\text{g}$

Volume to be diluted = $\text{1}\text{.00}\text{ml}$

Formula mass of Imidazole = $\text{68}\text{.08}\text{g}$

Formula mass of Imidazole hydrochloride = $\text{104}\text{.54}$

${\text{pK}}_{\text{a}}\text{=6}\text{.993}$

The $\text{pH}$ can be calculated as,

$\begin{array}{l}{\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[{\text{A}}^{\text{-}}\right]}{\left[\text{HA}\right]}\\ \\ \text{pH=6}\text{.993+log}\frac{\text{1}\text{.00/68}\text{.08}}{\text{1}\text{.00/104}\text{.54}}\\ \\ \text{pH=7}\text{.18}\end{array}$

The $\text{pH}$ of a solution = $\text{7}\text{.18}$

c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of a solution has to be calculated.

Concept Introduction:

Henderson-Hasselbalch equation:

Henderson- Hasselbalch equation is rearranged form of acid dissociation expression.

For acid:

Consider an acidic reaction:

$\text{HA}\stackrel{{\text{K}}_{\text{a}}}{\rightleftharpoons }{\text{H}}^{\text{+}}{\text{+A}}^{\text{-}}$

The Henderson- Hasselbalch equation for an acidic reaction can be given as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[{\text{A}}^{\text{-}}\right]}{\left[\text{HA}\right]}$

Answer to Problem 9.41P

The $\text{pH}$ of a solution is $\text{7}\text{.00}$ .

Explanation of Solution

Given,

Mass of Imidazole = $\text{1}\text{.00}\text{g}$

Mass of Imidazole hydrochloride = $\text{1}\text{.00}\text{g}$

Volume to be diluted = $\text{1}\text{.00}\text{ml}$

Volume and Molarity of ${\text{HClO}}_{\text{4}}$ = $\text{2}\text{.30}\text{ml}\text{and}\text{1}\text{.07}\text{M}$

Formula mass of Imidazole = $\text{68}\text{.08}\text{g}$

Formula mass of Imidazole hydrochloride = $\text{104}\text{.54}$

The $\text{pH}$ can be calculated as,

$\begin{array}{l}\underset{\_}{\begin{array}{ccc}\text{B}\text{+}& {\text{H}}^{\text{+}}\stackrel{}{\to }& {\text{BH}}^{\text{+}}\end{array}}\\ \text{Initial}\text{moles}\text{14}\text{.69}\text{2}\text{.46}\text{9}\text{.57}\\ \\ \text{Final}\text{moles}\text{12}\text{.23}\text{-}\text{12}\text{.03}\end{array}$

$\begin{array}{l}{\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[{\text{A}}^{\text{-}}\right]}{\left[\text{HA}\right]}\\ \\ \text{pH=6}\text{.993+log}\frac{\text{12}\text{.23}}{\text{12}\text{.03}}\\ \\ \text{pH=7}\text{.00}\end{array}$

The $\text{pH}$ of a solution = $\text{7}\text{.00}$

d)

Interpretation Introduction

Interpretation:

The volume in millimetres that is required to be add has to be given.

To give the required volume to be added in millimetres.

Answer to Problem 9.41P

The required volume that is to be added is $\text{6}\text{.86}\text{ml}$.

Explanation of Solution

In order to obtain ${\text{pH=pK}}_{\text{a}}$ , the Imidazole should be half neutralized.

Because there are $\text{14}\text{.69}\text{mmol}$ of Imidazole, the volume that is required of $\text{1}\text{.07}\text{M}{\text{HClO}}_{\text{4}}$ to be added to Imidazole is calculated as,

Volume of ${\text{HClO}}_{\text{4}}$ = $\frac{\text{1}}{\text{2}}\left(\text{14}\text{.69}\right)\text{=7}\text{.34}\text{mmol=6}\text{.86}\text{ml}$

The volume that is required of $\text{1}\text{.07}\text{M}{\text{HClO}}_{\text{4}}$ to be added to Imidazole = $\text{6}\text{.86}\text{ml}$