Chapter 9, Problem 9.JE

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the 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,

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

For base:

Consider a basic reaction:

${\text{BH}}^{\text{+}}\stackrel{{\text{K}}_{\text{a}}{\text{=K}}_{\text{w}}{\text{/K}}_{\text{b}}}{\rightleftharpoons }{\text{B+H}}^{\text{+}}$

The Henderson- Hasselbalch equation for a basic reaction can be given as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{\text{+}}\right]}$

Where,

${\text{pK}}_{\text{a}}$= acid dissociation constant for weak acid ${\text{BH}}^{\text{+}}$

To calculate the $\text{pH}$ of the solution

Answer to Problem 9.JE

The $\text{pH}$ of the solution is found to be $\text{8}\text{.21}$.

Explanation of Solution

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

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

Volume of solution = $\text{0}\text{.100}\text{L}$

$\text{pH}$ of the solution can be calculated as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{\text{+}}\right]}$

$\begin{array}{l}\text{pH=8}\text{.04+log}\frac{\left[\left(\text{1}\text{.00}\text{g}\right)\text{/}\left(\text{74}\text{.08}\text{g/mol}\right)\right]}{\left[\left(\text{1}\text{.00}\text{g}\right)\text{/}\left(\text{110}\text{.54}\text{g/mol}\right)\right]}\\ \\ \text{pH=8}\text{.04+log}\left(\text{1}\text{.48}\right)\\ \\ \text{pH=8}\text{.21}\end{array}$

The $\text{pH}$ of the solution = $\text{8}\text{.21}$.

(b)

Interpretation Introduction

Interpretation:

The number of grams of Glycine amide to be added 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,

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

For base:

Consider a basic reaction:

${\text{BH}}^{\text{+}}\stackrel{{\text{K}}_{\text{a}}{\text{=K}}_{\text{w}}{\text{/K}}_{\text{b}}}{\rightleftharpoons }{\text{B+H}}^{\text{+}}$

The Henderson- Hasselbalch equation for a basic reaction can be given as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{\text{+}}\right]}$

Where,

${\text{pK}}_{\text{a}}$= acid dissociation constant for weak acid ${\text{BH}}^{\text{+}}$

To calculate the number of grams of Glycine amide to be added

Answer to Problem 9.JE

The number of grams of Glycine amide to be added is found to be $\text{0}\text{.611}\text{g}$

Explanation of Solution

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

Volume of solution = $\text{100}\text{mL}$

$\text{pH=8}$

The number of grams of Glycine amide to be added can be calculated as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{\text{+}}\right]}$

$\begin{array}{l}{\text{pH=pK}}_{\text{a}}\text{+log}\frac{\text{mol}\text{B}}{\text{mol}{\text{BH}}^{\text{+}}}\\ \\ \text{8}\text{.00=8}\text{.04+log}\frac{\text{mol}\text{B}}{\left(\text{1}\text{.00}\text{g}\right)\text{/}\left(\text{110}\text{.54}\text{g/mol}\right)}\\ \\ \text{mol}\text{B=0}\text{.00825}\\ \\ \text{B}\text{(in}\text{grams)=0}\text{.611}\text{g}\\ \end{array}$

The number of grams of Glycine amide to be added = $\text{0}\text{.611}\text{g}$

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the 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,

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

For base:

Consider a basic reaction:

${\text{BH}}^{\text{+}}\stackrel{{\text{K}}_{\text{a}}{\text{=K}}_{\text{w}}{\text{/K}}_{\text{b}}}{\rightleftharpoons }{\text{B+H}}^{\text{+}}$

The Henderson- Hasselbalch equation for a basic reaction can be given as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{\text{+}}\right]}$

Where,

${\text{pK}}_{\text{a}}$= acid dissociation constant for weak acid ${\text{BH}}^{\text{+}}$

To calculate the $\text{pH}$ of the solution

Answer to Problem 9.JE

The $\text{pH}$ of the solution is found to be $\text{8}\text{.17}$.

Explanation of Solution

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

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

Volume of solution = $\text{0}\text{.100}\text{L}$

Volume of $\text{HCl}$ = $\text{5}\text{.00}\text{mL}$

Molarity of $\text{HCl}$= $\text{0}\text{.100}\text{M}$

The equilibrium constant ratio can be calculated as,

$\begin{array}{l}\begin{array}{cccc}& \text{B}\text{+}& {\text{H}}^{\text{+}}\stackrel{}{\to }& {\text{BH}}^{\text{+}}\end{array}\\ \\ \text{Initial}\text{moles}\text{0}\text{.013499}\text{0}\text{.000500}\text{0}\text{.0009046}\\ \\ \text{Final}\text{moles}\text{0}\text{.012999}\text{-}\text{0}\text{.0009546}\end{array}$

$\text{pH}$ of the solution can be calculated as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{\text{+}}\right]}$

$\begin{array}{l}\text{pH=8}\text{.04+log}\left(\frac{\text{0}\text{.012999}}{\text{0}\text{.009546}}\right)\\ \\ \text{pH=8}\text{.040+log}\left(\text{1}\text{.3617}\right)\\ \\ \text{pH=8}\text{.17}\end{array}$

The $\text{pH}$ of the solution = $\text{8}\text{.17}$

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the 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,

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

For base:

Consider a basic reaction:

${\text{BH}}^{\text{+}}\stackrel{{\text{K}}_{\text{a}}{\text{=K}}_{\text{w}}{\text{/K}}_{\text{b}}}{\rightleftharpoons }{\text{B+H}}^{\text{+}}$

The Henderson- Hasselbalch equation for a basic reaction can be given as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{\text{+}}\right]}$

Where,

${\text{pK}}_{\text{a}}$= acid dissociation constant for weak acid ${\text{BH}}^{\text{+}}$

To calculate the $\text{pH}$ of the solution

Answer to Problem 9.JE

The $\text{pH}$ of the solution is found to be $\text{8}\text{.25}$.

Explanation of Solution

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

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

Volume of solution = $\text{0}\text{.100}\text{L}$

Volume of $\text{NaOH}$ = $\text{10}\text{.00}\text{mL}$

Molarity of $\text{NaOH}$= $\text{0}\text{.100}\text{M}$

The equilibrium constant ratio can be calculated as,

$\begin{array}{l}\begin{array}{cccc}& {\text{BH}}^{\text{+}}\text{+}& {\text{OH}}^{\text{-}}\stackrel{}{\to }& \text{B}\end{array}\\ \\ \text{Initial}\text{moles}\text{0}\text{.009546}\text{0}\text{.001000}\text{0}\text{.012999}\\ \\ \text{Final}\text{moles}\text{0}\text{.008546}\text{-}\text{0}\text{.013999}\end{array}$

$\text{pH}$ of the solution can be calculated as,

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\left[\text{B}\right]}{\left[{\text{BH}}^{\text{+}}\right]}$

$\begin{array}{l}\text{pH=8}\text{.04+log}\left(\frac{\text{0}\text{.013999}}{\text{0}\text{.008546}}\right)\\ \\ \text{pH=8}\text{.040+log}\left(\text{1}\text{.638}\right)\\ \\ \text{pH=8}\text{.25}\end{array}$

The $\text{pH}$ of the solution = $\text{8}\text{.25}$

(e)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

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.

The value of ${\text{K}}_{\text{w}}$ is calculated by the formula,

${\text{K}}_{\text{w}}{\text{=K}}_{\text{a}}{\text{×K}}_{\text{b}}$

The $\text{pH}$ of weak base is calculated using the equation,

$\begin{array}{l}\text{pH=}\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{OH}}^{\text{-}}\right]}\\ \\ \text{pH=-log}\left[{\text{H}}^{\text{+}}\right]\end{array}$

Answer to Problem 9.JE

The $\text{pH}$ of the solution is found to be $\text{10}\text{.56}$ .

Explanation of Solution

Given,

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

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

Volume of solution = $\text{0}\text{.100}\text{L}$

Volume of $\text{NaOH=0}\text{.100}\text{M}$

Molarity of $\text{NaOH=90}\text{.46}\text{mL}$

$\text{9}\text{.046}\text{mmol}$ of Glycine amide hydrochloride and $\text{13}\text{.499}\text{mmol}$ of Glycine amide is present in solution.

Upon adding $\text{9}\text{.046}\text{mmol}$ of ${\text{OH}}^{\text{-}}$ , conversion of  Glycine amide hydrochloride into Glycine amide takes place. Therefore, the number of moles present is new solution of Glycine amide is $\text{9}\text{.046+13}\text{.499=22}\text{.545}\text{mmol}$ in $\text{190}\text{.46}\text{mL}$ .

Concentration of Glycine amine is $\left(\text{22}\text{.545}\text{mmol}\right)\text{/}\left(\text{190}\text{.46}\text{mL}\right)\text{=0}\text{.1184M}$

$\text{pH}$ is determined by hydrolysis of Glycine amide,

$\begin{array}{l}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.1184-x}}{\text{=K}}_{\text{b}}\\ \\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{10}}^{\text{-14}\text{.00}}}{{\text{10}}^{\text{-8}\text{.04}}}\\ \\ {\text{K}}_{\text{b}}\text{=1}{\text{.10×10}}^{\text{-6}}\text{M}\\ \\ \text{x=3}{\text{.60×10}}^{\text{-4}}\text{M}\end{array}$

$\begin{array}{l}\text{pH=-log}\left(\frac{{\text{K}}_{\text{w}}}{\text{x}}\right)\\ \\ \text{pH=-log}\left(\frac{\text{1}{\text{.011×10}}^{\text{-14}}}{\text{3}{\text{.60×10}}^{\text{-4}}}\right)\\ \\ \text{pH=10}\text{.56}\end{array}$

The $\text{pH}$ of the solution = $\text{10}\text{.56}$ .