Chapter 18, Problem 69E

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{a}}$ and the percent ionization of $\text{0}\text{.22}\text{M}{\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}$ is to be stated.

Concept introduction:

A strong acid easily releases protons to a base or when dissolved in water. It readily participates in an acid-base reaction. A weak acid undergoes only slight ionization in solution. It does not release protons easily. The major species present in a solution of weak acid is the unionized molecular species. However, some ions are present as minor species. The extent of the ionization is determined by the ionization constant.

Answer to Problem 69E

The value of ${\text{K}}_{\text{a}}$ of $\text{0}\text{.22}\text{M}{\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}$ is $2.6\times {10}^{-4}$ and the percent ionization of $\text{0}\text{.22}\text{M}{\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}$ is $3.5\%$.

Explanation of Solution

The formula to calculate the concentration of hydrogen ions, $\left[{\text{H}}^{\text{+}}\right]$ is given below.

$\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]$…(1)

The $\text{pH}$ of $\text{0}\text{.22}\text{M}{\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}$ is $2.12$.

Substitute the value of $\text{pH}$ in equation (1).

$\begin{array}{rll}2.12& =& -\log \left[{\text{H}}^{\text{+}}\right]\\ \left[{\text{H}}^{\text{+}}\right]& =& \text{anti}\log \left(-2.12\right)\\ & =& 7.6\times {10}^{-3}\text{M}\end{array}$

The equation for the dissociation ${\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}$ is shown below.

${\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}\text{(aq)}\to {\text{C}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{-}\text{(aq)}+{\text{H}}^{+}\text{(aq)}$

The dissociation constant, ${\text{K}}_{\text{a}}$ for the above reaction is given below.

${\text{K}}_{\text{a}}=\frac{\left[{\text{C}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}\right]}$…(2)

For the reaction given above, the concentration of the cation is equal to that of the anion.

$\left[{\text{C}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{-}\right]=\left[{\text{H}}^{+}\right]$

The value of $\left[{\text{H}}^{+}\right]$ is $7.6\times {10}^{-3}\text{M}$.

Therefore, the value of $\left[{\text{C}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{-}\right]$ is $7.6\times {10}^{-3}\text{M}$.

The concentration of ${\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}$, $\left[{\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}\right]$ is $\text{0}\text{.22}\text{M}$.

Substitute the values of $\left[{\text{H}}^{+}\right]$, $\left[{\text{C}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{-}\right]$, and $\left[{\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}\right]$ in equation (2).

$\begin{array}{rll}{\text{K}}_{\text{a}}& =& \frac{7.6\times {10}^{-3}\times 7.6\times {10}^{-3}}{\text{0}\text{.22}}\\ & =& 2.6\times {10}^{-4}\end{array}$

Therefore, the value of ${\text{K}}_{\text{a}}$ is $2.6\times {10}^{-4}$.

The formula to calculate percent ionization is given below.

$\text{Percent}\text{ionization=}\frac{\text{Concentration}\text{of}\text{ionized}\text{acid}}{\text{Initial}\text{concentration}\text{of}\text{acid}}\times 100\%$…(3)

The initial concentration is $\text{0}\text{.22}\text{M}$.

The concentration of ionized acid, $\left[{\text{C}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}{}^{-}\right]$ is $7.6\times {10}^{-3}\text{M}$.

Substitute the initial concentration and the concentration of the ionized acid in equation (3).

$\begin{array}{rll}\text{Percent}\text{ionization}& =& \frac{7.6\times {10}^{-3}\text{M}}{\text{0}\text{.22}\text{M}}\times 100\%\\ & =& 3.5\%\end{array}$

Therefore, the percent ionization of $\text{0}\text{.22}\text{M}{\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}$ is $3.5\%$.

Conclusion

The value of ${\text{K}}_{\text{a}}$ of $\text{0}\text{.22}\text{M}{\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}$ is $2.6\times {10}^{-4}$ and the percent ionization of $\text{0}\text{.22}\text{M}{\text{HC}}_{\text{4}}{\text{H}}_{\text{5}}{\text{O}}_{\text{3}}$ is $3.5\%$.