Chapter 18, Problem 18.184P

(a)

Interpretation Introduction

Interpretation:

$pH$ of the saturated solution of quinine in water has to be calculated.

Concept Introduction:

The self-dissociation constant for water can be written as follows,

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

Where,

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{ = water dissociation constant}\\ {\text{K}}_{\text{a}}\text{ = acid dissociation constant}\\ {\text{ K}}_{\text{b}}\text{ = base dissociation constant}\end{array}$

  ${\text{K}}_{\text{w}}={10}^{-14}$

Also, $K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]$

$pH$ can be calculated from the following equation if $[H_3{O}^{+}]$ is known.

$pH=-{\log }_{10}[H_3{O}^{+}]$

Explanation of Solution

Given,

    $p{K}_b=5.1$ for more basic nitrogen. So corresponding $K_b$ value is $K_b={10}^{-5.1}=\text{7}{\text{.94328x10}}^{\text{-6}}.$

The base dissociation reaction for quinine can be written as follows,

${\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right){\text{ + H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\rightleftarrows }{\text{OH}}^{\text{-}}\left(\text{aq}\right){\text{ + HC}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_2^{+}\left(\text{aq}\right)$

The reaction table for this reaction is shown below.

$\begin{array}{l}Initial:1.6\times {10}^{-3}M00\\ \frac{Change:-x+x+x}{Equilibrium:1.6\times {10}^{-3}-xxx}\end{array}$

The base dissociation constant $K_b$ can be written as follows,

$\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{\left[{\text{HC}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]}\\ \end{array}$

Where,

                $\begin{array}{l}\left[{\text{HC}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}^{\text{+}}\right]\text{ = 1}{\text{.6×10}}^{\text{-3}}\text{ - x}\\ \left[{\text{OH}}^{\text{-}}\right]\text{ = x}\\ \left[{\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]\text{ = x}\end{array}$

On substituting these in the above equation x is obtained as follows,

$K_b=\frac{\left[{\text{HC}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_2^{+}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]}=\frac{[x][x]}{\left[1.6\times {10}^{-3}-x\right]}=7.94328\times {10}^{-6}$

x is obtained by solving the following quadratic equation.

  $\begin{array}{l}{x}^2=\left(7.94328\times {10}^{-6}\right)\left(1.6\times {10}^{-3}-x\right)=1.27092\times {10}^{-8}-7.94328\times {10}^{-6}x\\ \\ x^2=7.94328\times {10}^{-6}x-1.270925\times {10}^{-8}=0\\ \\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \\ where,\\ a=1\\ b=7.94328\times {10}^{-6}\\ c=-1.270925\times {10}^{-8}\\ \\ x=\frac{-7.94328\times {10}^{-6}\pm \sqrt{{\left(7.94328\times {10}^{-6}\right)}^2-4(1)(-1.270925\times {10}^{-8})}}{2(1)}=1.08834\times {10}^{-4}M\end{array}$

$1.08834\times {10}^{-4}M$ is the concentration of $\left[{\text{OH}}^{\text{-}}\right]$. Concentration of $H_3{O}^{+}$ can be calculated from the following relation.

  $\begin{array}{l}{K}_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]\\ \\ \left[H_3{O}^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{10{\times }^{-14}}{1.08834\times {10}^{-4}}=9.18830513\times {10}^{-11}M\end{array}$

  $pH$ can be calculated using the following equation.

  $pH=-{\log }_{10}[H_3{O}^{+}]=-{\log }_{10}\left(9.18830513\times {10}^{-11}\right)=10.03676=10.0$

(b)

Interpretation Introduction

Interpretation:

Less contribution of aromatic N to the $pH$ has to be shown.

Concept Introduction:

The self-dissociation constant for water can be written as follows,

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

Where,

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{ = water dissociation constant}\\ {\text{K}}_{\text{a}}\text{ = acid dissociation constant}\\ {\text{ K}}_{\text{b}}\text{ = base dissociation constant}\end{array}$

  ${\text{K}}_{\text{w}}={10}^{-14}$

$pH$ can be calculated from the following equation if $[H_3{O}^{+}]$ is known.

$pH=-{\log }_{10}[H_3{O}^{+}]$

Also, $K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]$

Explanation of Solution

Given,

    $p{K}_b=9.7$ for aromatic nitrogen. So corresponding $K_b$ value is $K_b={10}^{-9.7}=\text{1}{\text{.995262x10}}^{-10}.$

The base dissociation reaction for quinine can be written as follows,

${\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right){\text{ + H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\rightleftarrows }{\text{OH}}^{\text{-}}\left(\text{aq}\right){\text{ + HC}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_2^{+}\left(\text{aq}\right)$

The reaction table for this reaction is shown below.

$\begin{array}{l}Initial:1.6\times {10}^{-3}M00\\ \frac{Change:-x+x+x}{Equilibrium:1.6\times {10}^{-3}-xxx}\end{array}$

The base dissociation constant $K_b$ can be written as follows,

${\text{K}}_{\text{b}}\text{=}\frac{\left[{\text{HC}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]}$

Where,

    $\begin{array}{l}\left[{\text{HC}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}^{\text{+}}\right]\text{ = 1}{\text{.6×10}}^{\text{-3}}\text{ - x}\\ \left[{\text{OH}}^{\text{-}}\right]\text{ = x}\\ \left[{\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]\text{ = x}\end{array}$

On substituting these in the above equation x is obtained as follows,

$K_b=\frac{\left[{\text{HC}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_2^{+}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]}=\frac{[x][x]}{\left[1.6\times {10}^{-3}-x\right]}=1.995262\times {10}^{-10}$

Assuming x is small compared to $1.6\times {10}^{-3}$ the x in the denominator can be neglected.

$\begin{array}{l}\frac{[x][x]}{\left[1.6\times {10}^{-3}\right]}=1.995262\times {10}^{-10}\\ \\ x^2=\left(1.995262\times {10}^{-10}\right)\left(1.6\times {10}^{-3}\right)\\ \\ x=5.65015\times {10}^{-7}MO{H}^{-}\end{array}$

Concentration of $H_3{O}^{+}$ can be calculated from the following relation.

$\begin{array}{l}{K}_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]\\ \\ \left[H_3{O}^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{10{\times }^{-14}}{5.65015\times {10}^{-7}}=1.769\times 10{\times }^{-8}M\end{array}$

$pH$ can be calculated using the following equation.

  $pH=-{\log }_{10}[H_3{O}^{+}]=-{\log }_{10}\left(1.769\times 10{\times }^{-8}\right)=7.75$

(c)

Interpretation Introduction

Interpretation:

$pH$ of quinine hydrochloride at given concentration has to be calculated.

Concept Introduction:

The self-dissociation constant for water can be written as follows,

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

Where,

  $\begin{array}{l}{\text{K}}_{\text{w}}\text{ = water dissociation constant}\\ {\text{K}}_{\text{a}}\text{ = acid dissociation constant}\\ {\text{ K}}_{\text{b}}\text{ = base dissociation constant}\end{array}$

  ${\text{K}}_{\text{w}}={10}^{-14}$

$pH$ can be calculated from the following equation if $[H_3{O}^{+}]$ is known.

$pH=-{\log }_{10}[H_3{O}^{+}]$

Also, $K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]$

Explanation of Solution

The acid dissociation reaction for qunine can be written as follows,

${\text{HC}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_2^{+}\left(\text{aq}\right)\text{ +}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\rightleftarrows }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right){\text{ + C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)$

The reaction table for this reaction is shown below.

$\begin{array}{l}Initial:0.33M00\\ \frac{Change:-x+x+x}{Equilibrium:0.33-xxx}\end{array}$

${\text{K}}_{\text{a}}$ is obtained from following relation.

$\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ \\ {\text{K}}_{\text{a}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}=\frac{{10}^{-14}}{7.94328\times {10}^{-6}}=1.25893\times {10}^{-9}\end{array}$

From the reaction ${\text{K}}_{\text{a}}$ is obtained as follows,

${\text{K}}_{\text{a}}=1.25893\times {10}^{-9}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]}{\left[{\text{HC}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_2^{+}\right]}=\frac{\left(\text{x}\right)\left(\text{x}\right)}{\left(0.33-\text{x}\right)}$

Assume x is very much smaller than 0.33, so x can be neglected from the denominator. And x is obtained as follows,

  $\begin{array}{l}\frac{\left(\text{x}\right)\left(\text{x}\right)}{\left(0.33\right)}=1.25893\times {10}^{-9}\\ \\ {\text{x}}^{\text{2}}\text{=}\left(1.25893\times {10}^{-9}\right)\left(0.33\right)\\ \\ {\text{x =[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] = 2}\text{.038252}\times {\text{ 10}}^{\text{-5}}M\end{array}$

$pH$ can be calculated using the following equation.

$pH=-{\log }_{10}[H_3{O}^{+}]=-{\log }_{10}\left(\text{2}\text{.038252}\times {\text{ 10}}^{\text{-5}}\right)=4.69074=4.7$

(d)

Interpretation Introduction

Interpretation:

$pH$ of 1.5% of quinine hydrochloride has to be calculated.

Concept Introduction:

Molarity of a solution can be calculated using the following equation.

$Molarity=\frac{numberofmolesofsolute}{Volumeofsolutioninlitre}$

And number of moles of solute can be calculated using the following equation.

$\text{Number of moles = }\frac{\text{Given mass of solute}}{\text{Molecular mass of the solute}}$

Therefore molarity can be calculated as follows,

$Molarity=\frac{numberofmolesofsolute}{Volumeofsolutioninlitre}=\frac{W_2\times 1000}{M_2\times V\left(ml\right)}$

Where,

  $\begin{array}{l}{\text{W}}_{\text{2}}\text{ = given mass of solute}\\ {\text{M}}_{\text{2}}\text{= molecular mass of solute}\\ \text{V}\left(\text{ml}\right)\text{ = volume of solution in ml}\end{array}$

Explanation of Solution

Use $QHCl$ to represent quinine hydrochloride.

Given that 1.5% of $QHCl$ is present in the solution. This means 1.5 g of $QHCl$ in 100 ml of solution. O molarity can be calculated as follows,

Molarity of a solution can be calculated using the following equation.

$Molarity=\frac{numberofmolesofQHCl}{Volumeofsolutioninlitre}$

And number of moles of solute can be calculated using the following equation.

$Molarity=\frac{numberofmolesofQHCl}{Volumeofsolutioninlitre}=\frac{W_2\times 1000}{M_2\times V\left(ml\right)}=\frac{1.5\times 1000}{360.87\times 100}=0.041566M$

The acid dissociation reaction for quinine can be written as follows,

${\text{HC}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_2^{+}\left(\text{aq}\right)\text{ +}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{}{\rightleftarrows }{\text{ H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right){\text{ + C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)$

The reaction table for this reaction is shown below.

$\begin{array}{l}Initial:0.041566M00\\ \frac{Change:-x+x+x}{Equilibrium:0.041566-xxx}\end{array}$

${\text{K}}_{\text{a}}$ is obtained from following relation.

$\begin{array}{l}{\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}\\ \\ {\text{K}}_{\text{a}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}=\frac{{10}^{-14}}{7.94328\times {10}^{-6}}=1.25893\times {10}^{-9}\end{array}$

From the reaction ${\text{K}}_{\text{a}}$ is obtained as follows,

${\text{K}}_{\text{a}}=1.25893\times {10}^{-9}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]}{\left[{\text{HC}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_2^{+}\right]}=\frac{\left(\text{x}\right)\left(\text{x}\right)}{\left(0.041566-\text{x}\right)}$

Assume x is very much smaller than $0.041566$, so x can be neglected from the denominator. And x is obtained as follows,

$\begin{array}{l}\frac{\left(\text{x}\right)\left(\text{x}\right)}{\left(0.041566\right)}=1.25893\times {10}^{-9}\\ \\ {\text{x}}^{\text{2}}\text{=}\left(1.25893\times {10}^{-9}\right)\left(0.041566\right)\\ \\ {\text{x =[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] = 7}\text{.233857}\times {\text{ 10}}^{\text{-6}}M\end{array}$

pH can be calculated using the following equation.

$pH=-{\log }_{10}[H_3{O}^{+}]=-{\log }_{10}\left(\text{7}\text{.233857}\times {\text{ 10}}^{\text{-6}}\right)=5.1406=5.1$