Use Appendix to calculate the percent dissociation of $\mathbf{\text{0.55M}}$ benzoic acid, ${\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}\mathbf{\text{COOH}}$.

In chemistry, dissociation is the breaking up of a chemical into simpler elements that may normally recombine under different conditions. The addition of a solvent or energy in the form of heat causes molecules or crystals of a substance to break up into ions in electrolytic, or ionic, dissociation (electrically charged particles).

For the dissociation of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$, write the reaction equation.

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COOH+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}$

After that, create the ICE table to get the ${\text{K}}_{\text{a}}$equation.

$\begin{array}{l}{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COOH]+H}}_{\text{2}}\text{O}\rightleftharpoons \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{+}\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\right]\\ \begin{array}{cccc}\text{I}& \text{0.55M}& \text{0}& \text{0}\\ \text{C}& \text{-x}& \text{+x}& \text{+x}\\ \text{E}& \text{0.55M-x}& \text{+x}& \text{+x}\end{array}\end{array}$

$\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\right]}{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.55-x}}\end{array}$

As $\text{x=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}\right]$is well-known. As ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ is a weak acid, its ${\text{K}}_{\text{a}}$value is quite low. Assume that the has no effect on the denominator's $\text{0.55M}$.

Then, to solve for $\text{x}$, substitute the ${\text{K}}_{\text{a}}$.

$\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.55}}\\ {\text{x}}^{\text{2}}\text{=}\left({\text{K}}_{\text{a}}\right)\text{(0.55)}\\ \text{x=}\sqrt{\left({\text{K}}_{\text{a}}\right)\text{(0.55)}}\\ \text{=}\sqrt{\left({\text{6.3×10}}^{\text{-5}}\right)\text{(0.55)}}\\ {\text{x=5.89×10}}^{\text{-3}}\text{M}\end{array}$

We already know that $\text{x=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$. The percent dissociation may now be calculated.

$\begin{array}{c}\text{\%dissociation=}\frac{\text{x}}{\left[{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}\right]}\text{×100\%}\\ \text{=}\frac{{\text{5.89×10}}^{\text{-3}}}{\text{0.55}}\text{×100\%}\\ \text{\%dissociation=1.07\%}\end{array}$

Therefore, the percent dissociation is $\text{1.07\%}$.