Answer
The percent ionization for this concentration is 0.698%.
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.
$Ka = \frac{[H^+][A^-]}{ [HA]}$
$Ka = 1.9 \times 10^{- 5}= \frac{x * x}{ 0.4}$
$Ka = 1.9 \times 10^{- 5}= \frac{x^2}{ 0.4}$
$ 7.6 \times 10^{- 6} = x^2$
$x = 2.75 \times 10^{- 3}$
%Ionization : $\frac{ 2.75 \times 10^{- 3}}{0.4} \times 100\% = 0.689\%$
Since ionization < 5%, the acid concentration have a right approximation.
