Chapter 16, Problem 24PE

(a)

Interpretation Introduction

Interpretation:

Percent ionization and $\text{pH}$ of solution of $\text{1}\text{.0 }M$ benzoic acid $\left({\text{HC}}_7{\text{H}}_5{\text{O}}_2\right)$ 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 benzoic acid, ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is as follows:

  ${\text{HC}}_7{\text{H}}_5{\text{O}}_2\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}\left(aq\right)$

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

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}\right]}{\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]}$        (1)

Through chemical equation it is evident that one ${\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}$ is formed for every ${\text{H}}^{\text{+}}$. Thus concentration of ${\text{H}}^{\text{+}}$ is equal to concentration of ${\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}$. Thus consider $\text{x}$ for ${\text{H}}^{\text{+}}$ and ${\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}$ and value of $\text{x}$ is small so concentration of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ can be considered as $1.0M$.

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

  $\left[{\text{H}}^{\text{+}}\right]=\frac{K_a\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]}{\left[{\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}\right]}$        (2)

Substitute $\text{x}$ for $\left[{\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}\right]$, $\text{x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $6.3\times {10}^{-5}$ for $K_a$ and $1.0$ for $\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]$ in equation (2).

  $\begin{array}{c}\text{x}=\frac{\left(6.3\times {10}^{-5}\right)\left(1.0\right)}{\text{x}}\\ {\text{x}}^2=6.3\times {10}^{-5}\end{array}$

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

  $\text{x}=7.9\times {10}^{-}{}^3$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in of $\text{1}\text{.0 }M$ benzoic acid, ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $7.9\times {10}^{-}{}^{3}M$.

$\text{pH}$ 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 $7.9\times {10}^{-}{}^3$ for ${\text{H}}^{+}$ in equation (3).

  $\begin{array}{c}\text{pH}=-\log \left[7.9\times {10}^{-}{}^3\right]\\ =-\log \text{7}\text{.9}+3\text{log10}\\ \text{=}-0.897+3\\ =2.10\end{array}$

Hence, $\text{pH}$ in $1.0M$ benzoic acid, ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $2.10$.

Formula for percent ionization of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ 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 $7.9\times {10}^{-}{}^{3}M$ for ${\text{H}}^{+}$ and $1.0M$ for $\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]$ in equation (4).

  $\begin{array}{c}\text{Percent ionization}=\left(\frac{7.9\times {10}^{-}{}^{3}M}{1.0M}\right)100\\ =0.79\text{ \%}\end{array}$

Hence, percent ionization of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $0.79\text{ \%}$.

(b)

Interpretation Introduction

Interpretation:

Percent ionization and $\text{pH}$ of solution of $\text{0}\text{.10 }M$ benzoic acid, ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The ionization of benzoic acid, ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is as follows:

  ${\text{HC}}_7{\text{H}}_5{\text{O}}_2\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}\left(aq\right)$

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

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}\right]}{\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]}$        (1)

Through chemical equation it is evident that one ${\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}$ is formed for every ${\text{H}}^{\text{+}}$. Thus concentration of ${\text{H}}^{\text{+}}$ is equal to concentration of ${\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}$. Thus consider $\text{x}$ for ${\text{H}}^{\text{+}}$ and ${\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}$ and value of $\text{x}$ is small so concentration of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ can be considered as $\text{0}\text{.10}M$.

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

  $\left[{\text{H}}^{\text{+}}\right]=\frac{K_a\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]}{\left[{\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}\right]}$        (2)

Substitute $\text{x}$ for $\left[{\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}\right]$, $\text{x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $6.3\times {10}^{-5}$ for $K_a$ and $\text{0}\text{.10}$ for $\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]$ in equation (2).

  $\begin{array}{c}\text{x}=\frac{\left(6.3\times {10}^{-5}\right)\left(0.10\right)}{\text{x}}\\ {\text{x}}^2=6.3\times {10}^{-6}\end{array}$

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

  $\text{x}=2.5\times {10}^{-}{}^3$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in of $\text{0}\text{.10 }M$ benzoic acid, ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $2.5\times {10}^{-}{}^{3}M$.

$\text{pH}$ 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 $2.5\times {10}^{-}{}^3$ for ${\text{H}}^{+}$ in equation (3).

  $\begin{array}{c}\text{pH}=-\log \left[2.5\times {10}^{-}{}^3\right]\\ =-\log 2.5+3\text{log10}\\ \text{=}-0.397+3\\ =2.60\end{array}$

Hence, $\text{pH}$ in $0.10M$ benzoic acid, ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $2.60$.

Formula for percent ionization of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ 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 $2.5\times {10}^{-}{}^{3}M$ for ${\text{H}}^{+}$ and $0.10M$ for $\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]$ in equation (4).

  $\begin{array}{c}\text{Percent ionization}=\left(\frac{2.5\times {10}^{-}{}^{3}M}{0.10M}\right)100\\ =2.5\text{ \%}\end{array}$

Hence, percent ionization of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $2.5\text{ \%}$.

(c)

Interpretation Introduction

Interpretation:

Percent ionization and $\text{pH}$ of solution of $\text{0}\text{.010 }M$ benzoic acid, ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The ionization of benzoic acid, ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is as follows:

  ${\text{HC}}_7{\text{H}}_5{\text{O}}_2\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}\left(aq\right)$

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

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}\right]}{\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]}$        (1)

Through chemical equation it is evident that one ${\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}$ is formed for every ${\text{H}}^{\text{+}}$. Thus concentration of ${\text{H}}^{\text{+}}$ is equal to concentration of ${\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}$. Thus consider $\text{x}$ for ${\text{H}}^{\text{+}}$ and ${\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}$ and value of $\text{x}$ is small so concentration of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ can be considered as $\text{0}\text{.010}M$.

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

  $\left[{\text{H}}^{\text{+}}\right]=\frac{K_a\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]}{\left[{\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}\right]}$        (2)

Substitute $\text{x}$ for $\left[{\text{C}}_7{\text{H}}_5{\text{O}}_2^{-}\right]$, $\text{x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $6.3\times {10}^{-5}$ for $K_a$ and $\text{0}\text{.010}$ for $\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]$ in equation (2).

  $\begin{array}{c}\text{x}=\frac{\left(6.3\times {10}^{-5}\right)\left(0.010\right)}{\text{x}}\\ {\text{x}}^2=6.3\times {10}^{-7}\end{array}$

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

  $\text{x}=7.93\times {10}^{-4}$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in of $\text{1}\text{.0 }M$ benzoic acid, ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $7.93\times {10}^{-4}M$.

$\text{pH}$ 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 $7.93\times {10}^{-4}$ for ${\text{H}}^{+}$ in equation (3).

  $\begin{array}{c}\text{pH}=-\log \left[7.93\times {10}^{-4}\right]\\ =-\log 7.93+4\text{log10}\\ \text{=}-0.899+4\\ =3.10\end{array}$

Hence, $\text{pH}$ in $0.010M$ benzoic acid, ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $3.10$.

Formula for percent ionization of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ 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 $7.93\times {10}^{-4}M$ for ${\text{H}}^{+}$ and $0.010M$ for $\left[{\text{HC}}_7{\text{H}}_5{\text{O}}_2\right]$ in equation (4).

  $\begin{array}{c}\text{Percent ionization}=\left(\frac{7.93\times {10}^{-4}M}{0.010M}\right)100\\ =7.93\text{ \%}\end{array}$

Hence, percent ionization of ${\text{HC}}_7{\text{H}}_5{\text{O}}_2$ is $7.93\text{ \%}$.