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.
-$[H_3O^+] = [ClCH_2{CO_2}^-] = x$
-$[ClCH_2CO_2H] = [ClCH_2CO_2H]_{initial} - x$
2. Calculate $[H_3O^+]$
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 1.95}$
$[H_3O^+] = 1.122 \times 10^{- 2}$
$x = 1.122 \times 10^{- 2}$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][ClCH_2{CO_2}^-]}{ [ClCH_2CO_2H]}$
$Ka = \frac{x^2}{[Initial Acid] - x}$
$Ka = \frac{( 0.01122)^2}{ 0.1- 0.01122}$
$Ka = \frac{ 1.259\times 10^{- 4}}{ 0.08878}$
$Ka = 1.418\times 10^{- 3}$
