Q4P - Organic Chemistry (Mcmurry)

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

$$\begin{gathered} {\mathrm{CI}}_2{\mathrm{CHCO}}_2\mathrm{H}\left(\mathrm{aq}\right) + {\mathrm{H}}_2\mathrm{O}\left(\mathrm{I}\right) \mathrm{֏} {\mathrm{H}}_3{\mathrm{O}}^{+}\left(\mathrm{aq}\right) + {\mathrm{CI}}_2\mathrm{CHC}{{\mathrm{O}}_2}^{-}\left(\mathrm{aq}\right) \\ \mathrm{Initial}\left(\mathrm{M}\right) 0.10 - 0 0 \\ \mathrm{Change}\left(\mathrm{M}\right) -\mathrm{x} - +\mathrm{x} +\mathrm{x} \\ ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ \\ \mathrm{Equilibrium}\left(\mathrm{M}\right) 0.10-\mathrm{x} - \mathrm{x} \mathrm{x} \end{gathered}$$

Step 2: 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 .

Step 3: 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% .