Answer
$Ka = 1.397\times 10^{- 5}$
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.
-$[H_3O^+] = [CH_3CH_2COO^-] = x$
-$[CH_3CH_2COOH] = [CH_3CH_2COOH]_{initial} - x$
2. Calculate the $[H_3O^+]$
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 2.93}$
$[H_3O^+] = 1.175 \times 10^{- 3}$
Therefore : $x = 1.175 \times 10^{-3} $
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][CH_3CH_2COO^-]}{ [CH_3CH_2COOH]}$
$Ka = \frac{x^2}{[InitialCH_3CH_2COOH] - x}$
$Ka = \frac{( 1.175\times 10^{- 3})^2}{ 0.1- 1.175\times 10^{- 3}}$
$Ka = \frac{ 1.38\times 10^{- 6}}{ 0.09883}$
$Ka = 1.397\times 10^{- 5}$
