Answer
The percent ionization for 0.02M of propionic acid is 2.54%
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.
$Ka = \frac{[H^+][A^-]}{ [HA]}$
$Ka = 1.3 \times 10^{- 5}= \frac{x * x}{ 2 \times 10^{- 2}}$
$Ka = 1.3 \times 10^{- 5}= \frac{x^2}{ 2 \times 10^{- 2}}$
$ 2.6 \times 10^{- 7} = x^2$
$x = 5.09 \times 10^{- 4}$
Percent ionization: $\frac{ 5.09 \times 10^{- 4}}{ 2 \times 10^{- 2}} \times 100\% = 2.54\%$
%ionization < 5% : Right approximation.