Chapter 16, Problem 23PE

(a)

Interpretation Introduction

Interpretation:

Percent ionization and $\text{pH}$ of solution of $\text{1}\text{.0 }M$ phenol $\left({\text{HC}}_6{\text{H}}_5\text{O}\right)$ has to be calculated.

Concept Introduction:

Consider a weak acid $\text{HA}$ that is in equilibrium with its ions and is as follows:

  $\text{HA}\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{A}}^{-}\left(aq\right)$

The equilibrium constant for ionization of weak acid is called ionization constant $K_a$ and is expressed as follows:

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

Percent ionization for a weak acid is defined as the ratio of concentration of ${\text{H}}^{\text{+}}$ or ${\text{A}}^{-}$ to initial concentration of $\text{HA}$ that is multiplied with 100 and is expressed as follows:

  $\text{Percent ionization}=\left(\frac{\left[{\text{H}}^{\text{+}}\right]\text{or}\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}\right)\cdot 100$

$\text{pH}$ is defined as the concentration of hydrogen ion. It also explains about the acidity of solution.

Explanation of Solution

The ionization of phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$ is as follows:

  ${\text{HC}}_6{\text{H}}_5\text{O}\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\left(aq\right)$

The expression for $K_a$ for given reaction is as follows:

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}{\left[{\text{HC}}_6{\text{H}}_5\text{O}\right]}$        (1)

Through chemical equation it is evident that one ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ is formed for every ${\text{H}}^{\text{+}}$. Thus concentration of ${\text{H}}^{\text{+}}$ is equal to concentration of ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$. Thus consider $\text{x}$ for ${\text{H}}^{\text{+}}$ and ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ and value of $\text{x}$ is small so concentration of ${\text{HC}}_6{\text{H}}_5\text{O}$ can be considered as $1.0M$.

Rearrange equation (1) for $\left[{\text{H}}^{\text{+}}\right]$.

  $\left[{\text{H}}^{\text{+}}\right]=\frac{K_a\left[{\text{HC}}_6{\text{H}}_5\text{O}\right]}{\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}$        (2)

Substitute $\text{x}$ for $\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]$, $\text{x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $1.3\times {10}^{-10}$ for $K_a$ and $1.0$ for $\left[{\text{HC}}_6{\text{H}}_5\text{O}\right]$ in equation (2).

  $\begin{array}{c}\text{x}=\frac{\left(1.3\times {10}^{-10}\right)\left(1.0\right)}{\text{x}}\\ {\text{x}}^2=1.3\times {10}^{-10}\end{array}$

Solve this equation for $\text{x}$.

  $\text{x}=1.14\times {10}^{-}{}^5$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in of $\text{1}\text{.0 }M$ phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$ is $1.14\times {10}^{-}{}^{5}M$.

$\text{pH}$ is defined as negative logarithm of concentration of ${\text{H}}^{+}$. It is calculated as follows:

  $\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]$        (3)

Substitute $1.14\times {10}^{-}{}^5$ for ${\text{H}}^{+}$ in equation (3).

  $\begin{array}{c}\text{pH}=-\log \left[1.14\times {10}^{-}{}^5\right]\\ =-\log 1.14+5\text{log10}\\ \text{=}-0.056+5\\ =4.94\end{array}$

Hence, $\text{pH}$ in $1.0M$ phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$ is $4.94$.

The percent ionization of phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$  is as follows:

  $\text{Percent ionization}=\left(\frac{\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{HC}}_6{\text{H}}_5\text{O}\right]}\right)100$        (4)

Substitute $1.14\times {10}^{-}{}^{5}M$ for ${\text{H}}^{+}$ and $1.0M$ phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$ in equation (4).

  $\begin{array}{c}\text{Percent ionization}=\left(\frac{1.14\times {10}^{-}{}^{5}M}{1.0M}\right)100\\ =0.0011\text{ \%}\end{array}$

Hence, percent ionization of phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$ is $0.0011\text{ \%}$.

(b)

Interpretation Introduction

Interpretation:

Percent ionization and $\text{pH}$ of solution of $\text{0}\text{.10 }M$ phenol $\left({\text{HC}}_6{\text{H}}_5\text{O}\right)$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The ionization of phenol $\left({\text{HC}}_6{\text{H}}_5\text{O}\right)$ is as follows:

  ${\text{HC}}_6{\text{H}}_5\text{O}\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\left(aq\right)$

The expression for $K_a$ for given reaction is as follows:

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}{\left[{\text{HC}}_6{\text{H}}_5\text{O}\right]}$        (1)

Through chemical equation it is evident that one ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ is formed for every ${\text{H}}^{\text{+}}$. Thus concentration of ${\text{H}}^{\text{+}}$ is equal to concentration of ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$. Thus consider $\text{x}$ for ${\text{H}}^{\text{+}}$ and ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ and value of $\text{x}$ is small so concentration of ${\text{HC}}_6{\text{H}}_5\text{O}$ can be considered as $\text{0}\text{.10 }M$.

Rearrange equation (1) for $\left[{\text{H}}^{\text{+}}\right]$.

  $\left[{\text{H}}^{\text{+}}\right]=\frac{K_a\left[{\text{HC}}_6{\text{H}}_5\text{O}\right]}{\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}$        (2)

Substitute $\text{x}$ for $\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]$, $\text{x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $1.3\times {10}^{-10}$ for $K_a$ and $\text{0}\text{.10}$ for $\left[{\text{HC}}_6{\text{H}}_5\text{O}\right]$ in equation (2).

  $\begin{array}{c}\text{x}=\frac{\left(1.3\times {10}^{-10}\right)\left(0.10\right)}{\text{x}}\\ {\text{x}}^2=1.3\times {10}^{-11}\end{array}$

Solve this equation for $\text{x}$.

  $\text{x}=3.6\times {10}^{-}{}^6$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in of $\text{0}\text{.10 }M$ phenol $\left({\text{HC}}_6{\text{H}}_5\text{O}\right)$ is $3.6\times {10}^{-}{}^{6}M$.

$\text{pH}$ is defined as negative logarithm of concentration of ${\text{H}}^{+}$. It is calculated as follows:

  $\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]$        (3)

Substitute $3.6\times {10}^{-}{}^6$ for ${\text{H}}^{+}$ in equation (3).

  $\begin{array}{c}\text{pH}=-\log \left[3.6\times {10}^{-}{}^6\right]\\ =-\log 3.6+6\text{log10}\\ \text{=}-0.556+6\\ =5.44\end{array}$

Hence, $\text{pH}$ in $0.10M$ phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$ is $5.44$.

The percent ionization of phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$  is as follows:

  $\text{Percent ionization}=\left(\frac{\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]}\right)100$        (4)

Substitute $3.6\times {10}^{-}{}^{6}M$ for ${\text{H}}^{+}$ and $0.10M$ for $\left[{\text{HC}}_6{\text{H}}_5\text{O}\right]$ in equation (4).

Hence, percent ionization of phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$ is $0.0036\text{ \%}$.

(c)

Interpretation Introduction

Interpretation:

Percent ionization and $\text{pH}$ of solution of $\text{0}\text{.010 }M$ phenol $\left({\text{HC}}_6{\text{H}}_5\text{O}\right)$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The ionization of phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$ is as follows:

  ${\text{HC}}_6{\text{H}}_5\text{O}\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\left(aq\right)$

The expression for $K_a$ for given reaction is as follows:

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}{\left[{\text{HC}}_6{\text{H}}_5\text{O}\right]}$        (1)

Through chemical equation it is evident that one ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ is formed for every ${\text{H}}^{\text{+}}$. Thus concentration of ${\text{H}}^{\text{+}}$ is equal to concentration of ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$. Thus consider $\text{x}$ for ${\text{H}}^{\text{+}}$ and ${\text{C}}_6{\text{H}}_5{\text{O}}^{-}$ and value of $\text{x}$ is small so concentration of ${\text{HC}}_6{\text{H}}_5\text{O}$ can be considered as $\text{0}\text{.010 }M$.

Rearrange equation (1) for $\left[{\text{H}}^{\text{+}}\right]$.

  $\left[{\text{H}}^{\text{+}}\right]=\frac{K_a\left[{\text{HC}}_6{\text{H}}_5\text{O}\right]}{\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]}$        (2)

Substitute $\text{x}$ for $\left[{\text{C}}_6{\text{H}}_5{\text{O}}^{-}\right]$, $\text{x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $1.3\times {10}^{-10}$ for $K_a$ and $\text{0}\text{.010}$ for $\left[{\text{HC}}_6{\text{H}}_5\text{O}\right]$ in equation (2).

  $\begin{array}{c}\text{x}=\frac{\left(1.3\times {10}^{-10}\right)\left(0.010\right)}{\text{x}}\\ {\text{x}}^2=3\times {10}^{-13}\end{array}$

Solve this equation for $\text{x}$.

  $\text{x}=5.48\times {10}^{-7}$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in of $\text{0}\text{.010 }M$ phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$ is $5.48\times {10}^{-7}M$.

$\text{pH}$ is defined as negative logarithm of concentration of ${\text{H}}^{+}$. It is calculated as follows:

  $\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]$        (3)

Substitute $5.48\times {10}^{-7}$ for ${\text{H}}^{+}$ in equation (3).

  $\begin{array}{c}\text{pH}=-\log \left[5.48\times {10}^{-7}\right]\\ =-\log 5.48+7\text{log10}\\ \text{=}-0.739+7\\ =6.261\end{array}$

Hence, $\text{pH}$ in $0.010M$ phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$ is $6.261$.

The percent ionization of phenol, ${\text{HC}}_6{\text{H}}_5\text{O}$  is as follows:

  $\text{Percent ionization}=\left(\frac{\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]}\right)100$        (4)

Substitute $5.48\times {10}^{-7}M$ for ${\text{H}}^{+}$ and $0.010M$ for $\left[{\text{HC}}_6{\text{H}}_5\text{O}\right]$  in equation (4).

  $\begin{array}{c}\text{Percent ionization}=\left(\frac{5.48\times {10}^{-7}M}{0.010M}\right)100\\ =0.0055\text{ \%}\end{array}$

Hence, percent ionization of phenol $\left({\text{HC}}_6{\text{H}}_5\text{O}\right)$ is. $0.0055\text{ \%}$.