91Ó°ÊÓ

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:$\mathrm{Δ}{\text{T}}_{\text{f}}\text{=}{\text{iK}}_{\text{f}}\text{m}$

Degree of dissociation of a weak acid is:

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

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

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.

$\mathrm{Δ}{\text{T}}_{\text{f}}\text{=}{\text{iK}}_{\text{f}}\text{mi=}\frac{\mathrm{Δ}{\text{T}}_{\text{f}}}{{\text{K}}_{\text{f}}\text{m}}=\frac{\text{0.00°°ä-}\left(\text{- 1.93°C}\right)}{\left(\text{1.86°°ä/³¾}\right)\text{(1.000m)}}\text{i =}\text{1.038}$

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

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

$\mathrm{Î\pm }\text{=}\frac{\text{i - 1}}{\text{n - 1}}\text{=}\frac{\text{1.038 - 1}}{\text{2 - 1}}\mathrm{Î\pm }\text{=}\text{0.038}$

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

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

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