Answer
Percent ionization = 2.17%.
Work Step by Step
$Ka (HN_3) = 1.9 \times 10^{-5}$
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.
$Ka = \frac{[H_3O^+][A^-]}{ [HA]}$
$Ka = 1.9 \times 10^{- 5}= \frac{x * x}{ 4 \times 10^{- 2}}$
$Ka = 1.9 \times 10^{- 5}= \frac{x^2}{ 4 \times 10^{- 2}}$
$ 7.6 \times 10^{- 7} = x^2$
$x = 8.71 \times 10^{- 4}$
Percent Ionization: $\frac{ 8.71 \times 10^{- 4}}{ 4 \times 10^{- 2}} \times 100\% = 2.17\%$
%ionization < 5% : Right approximation.