Q172CP - Chemistry: Molecular Nature Of Matter And Change

Question

A 1000  m solution of chloroacetic acid $\left({\text{ClCH}}_{\text{2}}\text{COOH}\right)$ freezes at $\mathbf{1}\mathbf{.}\mathbf{93}\mathbf{°}\mathbf{C}$  . Find the Ka of chloroacetic acid. (Assume the molarities equal the molalities.)

Step-by-Step Solution

Answer

${\text{K}}_{\text{a}}\text{ = 1.50}\times {\text{10}}^{\text{ - 3}}.$

Step 1: Find the K a

Important things and equations to note:

$\left[{\text{ClCH}}_{\text{2}}\text{COOH}\right] \text{ = } \text{1.000m} \text{ = } \text{1.000M}$

Freezing point depression: $\Delta {\text{T}}_{\text{f}} \text{ = } {\text{iK}}_{\text{f}}\text{m}$

Degree of dissociation of a weak acid is:

$\alpha \text{ = } \frac{\text{i - 1}}{\text{n - 1}}$, where n is the number of ions.

${\text{K}}_{\text{a}}\text{ = } \frac{\text{[HA]} {\alpha }^{\text{2}}}{\text{1 - }\alpha }$

Now, solve for the i using the freezing point depression formula. Note that for the original freezing point and ${\text{K}}_{\text{f}}$, we will use these values from water since it depends on the solvent.

$\Delta {\text{T}}_{\text{f}}\text{ = } {\text{iK}}_{\text{f}}\text{mi = } \frac{\Delta {\text{T}}_{\text{f}}}{{\text{K}}_{\text{f}}\text{m}} =\frac{\text{0.00°C- }\left(\text{ - 1.93°C}\right)}{\left(\text{1.86°C/m}\right)\text{(1.000m)}}\text{i = } \text{1.038}$

Now, replace the i in the degree of dissociation of a weak acid formula to solve $\alpha$. Note that ${\text{ClCH}}_{\text{2}}\text{COOH}$ will produce 2 ions.

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

$\alpha \text{ = } \frac{\text{i - 1}}{\text{n - 1}} \text{ = } \frac{\text{1.038 - 1}}{\text{2 - 1}} \alpha \text{ = } \text{0.038}$

Lastly, solve for the ${\text{K}}_{\text{a}}$.

$\begin{array}{rcl}& & {\text{K}}_{\text{a}}\text{ = } \frac{\text{[HA]} {\alpha }^{\text{2}}}{\text{1 - }\alpha } \\ & & \text{ = } \frac{\text{(1.000)}\left({\text{0.038}}^{\text{2}}\right)}{\text{1 - 0.038}}\\ & & \text{ = } \text{1.50}\times {\text{10}}^{\text{ - 3}}.\end{array}$

 

 

Hence, ${\text{K}}_{\text{a}}\text{ = } \text{1.50}\times {\text{10}}^{\text{ - 3}}.$