Question

The ${\mathbf{K}}_{\mathbf{a}}$for dichloroacetic acid is$\mathbf{3}\mathbf{.}\mathbf{32}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}$. Approximately what percentage of the acid is dissociated in a 0.10 M aqueous solution?

Short answer

The percent dissociation of dichloroacetic acid in a 0.10 M aqueous solution is 43% .

Step-by-step solution

Setup the ICE table for the given conditions.

The chemical equation for the dissociation of dichloroacetic acid in water is written below:

${\mathrm{CI}}_2{\mathrm{CHCO}}_2\mathrm{H}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right)\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}\left(\mathrm{aq}\right){\mathrm{CI}}_2{{\mathrm{CHCO}}_2}^{-}\left(\mathrm{aq}\right)$

The acid dissociation constant expression fKor the above reaction is expressed as follows:

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{CI}}_2{{\mathrm{CHCO}}_2}^{-}\right]}{\left[{\mathrm{CI}}_2{\mathrm{CHCO}}_2\mathrm{H}\right]}$

Dichloroacetic acid is a weak acid and dissociates partially in water.

Setup the ICE table as shown below: Let x be the change in concentration.

$$\begin{gathered} {\mathrm{CI}}_2{\mathrm{CHCO}}_2\mathrm{H}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right)\mathrm{֏}{\mathrm{H}}_3{\mathrm{O}}^{+}\left(\mathrm{aq}\right)+{\mathrm{CI}}_2\mathrm{CHC}{{\mathrm{O}}_2}^{-}\left(\mathrm{aq}\right) \\ \mathrm{Initial}\left(\mathrm{M}\right)0.10-00 \\ \mathrm{Change}\left(\mathrm{M}\right)-\mathrm{x}-+\mathrm{x}+\mathrm{x} \\ ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ \\ \mathrm{Equilibrium}\left(\mathrm{M}\right)0.10-\mathrm{x}-\mathrm{x}\mathrm{x} \end{gathered}$$

Calculate the change in concentration value.

Substitute these values into equilibrium constant expression as given below:

$3.32\times {10}^{-2}=\frac{\left(\mathrm{x}\right)\left(\mathrm{x}\right)}{(0.10-\mathrm{x})}$

Solve for x.

$\mathrm{x}=0.0434$

Therefore, the change in concentration is 0.0434 M .

Determine the percentage of dichloroacetic acid is dissociated in water.

Use the below expression to find the percent dissociation of dichloroacetic acid in water.

$$\begin{gathered} \mathrm{Percent}\mathrm{dissociation}=\frac{\mathrm{Change}\mathrm{in}\mathrm{concentration}}{\mathrm{Initial}\mathrm{concentration}\mathrm{of}\mathrm{dichoroacetic}\mathrm{acid}}\times 100\% \\ =\frac{0.043}{0.10}\times 100\% \\ =43\% \end{gathered}$$

Hence, the percent dissociation of dichloroacetic acid in a 0.10 M aqueous solution is 43% .