18.77P - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

Answer

The $\text{\% dissociation = 1.07 \%}$.

Define dissociation

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).

Explanation

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 $\text{0.55 M}$.

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 $\text{1.07\%}$.