Answer
$Ka = 1.418\times 10^{- 3}$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
\** The image is in the end of this answer.
-$[H_3O^+] = [ClCH_2COO^-] = x$
-$[ClCH_2COOH] = [ClCH_2COOH]_{initial} - x$
2. Calculate the $[H_3O^+]$
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 1.95}$
$[H_3O^+] = 1.122 \times 10^{- 2}M$
Therefore : $x = 0.01122$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][ClCH_2COO^-]}{ [ClCH_2COOH]}$
$Ka = \frac{x^2}{[InitialClCH_2COOH] - x}$
$Ka = \frac{( 0.01122)^2}{ 0.1- 0.01122}$
$Ka = \frac{ 1.259\times 10^{- 4}}{ 0.08878}$
$Ka = 1.418\times 10^{- 3}$
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