Answer
Percent ionization $\approx$ 0.57%.
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.
-$[OH^-] = [N{H_4}^+] = x$
-$[NH_3] = [NH_3]_{initial} - x = 0.56 - x$
For approximation, we consider: $[NH_3] = 0.56M$
2. Now, use the Kb value and equation to find the 'x' value.
$Ka = \frac{[OH^-][N{H_4}^+]}{ [NH_3]}$
$Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.56}$
$Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.56}$
$ 1.008 \times 10^{- 5} = x^2$
$x = 3.175 \times 10^{- 3}$
Percent ionization: $\frac{ 3.175 \times 10^{- 3}}{ 0.56} \times 100\% = 0.5669\%$
%ionization < 5% : Right approximation.
