Chapter 16, Problem 20PE

(a)

Interpretation Introduction

Interpretation:

Concentration ${\text{H}}^{+}$ for solution of $0.025M$ lactic acid $\left({\text{HC}}_3{\text{H}}_5{\text{O}}_2\right)$ has to be calculated.

Concept Introduction:

Consider ionization of 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:

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

$K_a$ is defined as the ratio of concentration of products to concentration of reactants.

Explanation of Solution

The ionization of lactic acid is as follows:

  ${\text{HC}}_3{\text{H}}_5{\text{O}}_2\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}\left(aq\right)$

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

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}\right]}{\left[{\text{HC}}_3{\text{H}}_5{\text{O}}_2\right]}$        (1)

Through chemical equation it is evident that one ${\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}$ is formed for every ${\text{H}}^{\text{+}}$. Thus concentration of ${\text{H}}^{\text{+}}$ is equal to concentration of ${\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}$. Thus consider $\text{x}$ for ${\text{H}}^{\text{+}}$ and ${\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}$. Value of $\text{x}$ is small so concentration of ${\text{HC}}_3{\text{H}}_5{\text{O}}_2$ can be considered as $0.025M$.

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

  $\left[{\text{H}}^{\text{+}}\right]=\frac{K_a\left[{\text{HC}}_3{\text{H}}_5{\text{O}}_2\right]}{\left[{\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}\right]}$        (2)

Substitute $\text{x}$ for $\left[{\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}\right]$, $\text{x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $8.4\times {10}^{-4}$ for $K_a$ and $0.025$ for $\left[{\text{HC}}_3{\text{H}}_5{\text{O}}_2\right]$ in equation (2).

  $\begin{array}{c}\text{x}=\frac{\left(8.4\times {10}^{-4}\right)\left(0.025\right)}{\text{x}}\\ {\text{x}}^2=0.000021\end{array}$

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

  $\text{x}=4.6\times {10}^{-3}$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in $0.025M$ lactic acid, ${\text{HC}}_3{\text{H}}_5{\text{O}}_2$ is $4.6\times {10}^{-3}M$.

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ for solution of $0.025M$ lactic acid $\left({\text{HC}}_3{\text{H}}_5{\text{O}}_2\right)$ has to be calculated.

Concept Introduction:

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

Explanation of Solution

The ionization of lactic acid is as follows:

  ${\text{HC}}_3{\text{H}}_5{\text{O}}_2\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}\left(aq\right)$

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

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}\right]}{\left[{\text{HC}}_3{\text{H}}_5{\text{O}}_2\right]}$        (1)

Through chemical equation it is evident that one ${\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}$ is formed for every ${\text{H}}^{\text{+}}$. Thus concentration of ${\text{H}}^{\text{+}}$ is equal to concentration of ${\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}$. Thus consider $\text{x}$ for ${\text{H}}^{\text{+}}$ and ${\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}$. Value of $\text{x}$ is small so concentration of ${\text{HC}}_3{\text{H}}_5{\text{O}}_2$ can be considered as $0.025M$.

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

  $\left[{\text{H}}^{\text{+}}\right]=\frac{K_a\left[{\text{HC}}_3{\text{H}}_5{\text{O}}_2\right]}{\left[{\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}\right]}$        (2)

Substitute $\text{x}$ for $\left[{\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}\right]$, $\text{x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $8.4\times {10}^{-4}$ for $K_a$ and $0.025$ for $\left[{\text{HC}}_3{\text{H}}_5{\text{O}}_2\right]$ in equation (2).

  $\begin{array}{c}\text{x}=\frac{\left(8.4\times {10}^{-4}\right)\left(0.025\right)}{\text{x}}\\ {\text{x}}^2=0.000021\end{array}$

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

  $\text{x}=4.6\times {10}^{-3}$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in $0.025M$ lactic acid, ${\text{HC}}_3{\text{H}}_5{\text{O}}_2$ is $4.6\times {10}^{-3}M$.

$\text{pH}$ 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 $4.6\times {10}^{-3}M$ for ${\text{H}}^{+}$ in equation (3).

  $\begin{array}{c}\text{pH}=-\log \left[4.6\times {10}^{-3}\right]\\ =-\log \text{4}\text{.6}+3\text{log10}\\ \text{=}-0.66+3\\ =2.34\end{array}$

Hence, $\text{pH}$ in $0.025M$ lactic acid, ${\text{HC}}_3{\text{H}}_5{\text{O}}_2$ is $2.34$.

(c)

Interpretation Introduction

Interpretation:

Percent ionization of solution of $0.025M$ lactic acid $\left({\text{HC}}_3{\text{H}}_5{\text{O}}_2\right)$ has to be calculated.

Concept Introduction:

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

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

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:

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

Explanation of Solution

The ionization of lactic acid is as follows:

  ${\text{HC}}_3{\text{H}}_5{\text{O}}_2\left(aq\right)\rightleftharpoons {\text{H}}^{\text{+}}\left(aq\right)+{\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}\left(aq\right)$

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

  $K_a=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}\right]}{\left[{\text{HC}}_3{\text{H}}_5{\text{O}}_2\right]}$        (1)

Through chemical equation it is evident that one ${\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}$ is formed for every ${\text{H}}^{\text{+}}$. Thus concentration of ${\text{H}}^{\text{+}}$ is equal to concentration of ${\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}$. Thus consider $\text{x}$ for ${\text{H}}^{\text{+}}$ and ${\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}$. Value of $\text{x}$ is small so concentration of ${\text{HC}}_3{\text{H}}_5{\text{O}}_2$ can be considered as $0.025M$.

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

  $\left[{\text{H}}^{\text{+}}\right]=\frac{K_a\left[{\text{HC}}_3{\text{H}}_5{\text{O}}_2\right]}{\left[{\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}\right]}$        (2)

Substitute $\text{x}$ for $\left[{\text{C}}_3{\text{H}}_5{\text{O}}_2^{-}\right]$, $\text{x}$ for $\left[{\text{H}}^{\text{+}}\right]$, $8.4\times {10}^{-4}$ for $K_a$ and $0.025$ for $\left[{\text{HC}}_3{\text{H}}_5{\text{O}}_2\right]$ in equation (2).

  $\begin{array}{c}\text{x}=\frac{\left(8.4\times {10}^{-4}\right)\left(0.025\right)}{\text{x}}\\ {\text{x}}^2=0.000021\end{array}$

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

  $\text{x}=4.6\times {10}^{-3}$

Hence, $\left[{\text{H}}^{\text{+}}\right]$ in $0.025M$ lactic acid, ${\text{HC}}_3{\text{H}}_5{\text{O}}_2$ is $4.6\times {10}^{-3}M$.

The percent ionization of ${\text{HC}}_3{\text{H}}_5{\text{O}}_2$ is as follows:

  $\begin{array}{c}\text{Percent ionization}=\left(\frac{\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{HC}}_3{\text{H}}_5{\text{O}}_2\right]}\right)\cdot 100\\ =\left(\frac{4.6\times {10}^{-3}M}{0.025M}\right)\cdot 100\\ =18.4\text{ \%}\end{array}$

Hence, percent ionization of ${\text{HC}}_3{\text{H}}_5{\text{O}}_2$ is $18.4\text{ \%}$.