Answer
Percent Ionization = 1.27%.
Work Step by Step
Ka (Propionic acid) = $1.3 \times 10^{-5}$
1. Drawing the ICE table we get:
-$[H^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x$
For approximation, we consider: $[HA] = [HA]_{initial}$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H^+][A^-]}{ [HA]}$
$Ka = 1.3 \times 10^{- 5}= \frac{x * x}{ 8 \times 10^{- 2}}$
$Ka = 1.3 \times 10^{- 5}= \frac{x^2}{ 8 \times 10^{- 2}}$
$ 1.04 \times 10^{- 6} = x^2$
$x = 1.01 \times 10^{- 3}$
Ionization : $\frac{ 1.01 \times 10^{- 3}}{ 8 \times 10^{- 2}} \times 100\% = 1.27\%$
%ionization < 5% : Right approximation.