$$Kb = \frac{[NH4+][OH-]}{[NH3]}$$
where Kb is the base ionization constant for ammonia, [NH4+] is the concentration of ammonium ions, [OH-] is the concentration of hydroxide ions, and [NH3] is the concentration of ammonia.
At a concentration of 0.1 M, the ionization of ammonia is not significant, and the concentration of ammonium ions and hydroxide ions can be considered negligible compared to the concentration of ammonia. Therefore, we can simplify the equilibrium constant expression to:
$$Kb = \frac{[OH-]^2}{[NH3]}$$
Assuming complete dissociation of water, the concentration of hydroxide ions is equal to the square root of the ionic product of water (Kw):
$$[OH-] = \sqrt{Kw} = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \ M$$
Substituting the concentration of hydroxide ions into the simplified equilibrium constant expression, we get:
$$Kb = \frac{(1.0 \times 10^{-7})^2}{[NH3]}$$
$$[NH3] = \frac{1.0 \times 10^{-14}}{Kb} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} \ M$$
The percent ionization of ammonia is calculated as follows:
$$Percent \ ionization = \frac{[NH4+]}{[NH3] + [NH4+]} \times 100$$
Since the concentration of ammonium ions is negligible compared to the concentration of ammonia, we can simplify the expression to:
$$Percent \ ionization = \frac{[NH4+]}{[NH3]} \times 100$$
Substituting the concentration of ammonia that we calculated earlier, we get:
$$Percent \ ionization = \frac{5.56 \times 10^{-10}}{0.1} \times 100 = 5.56 \times 10^{-9} \%$$
Therefore, the percent ionization of ammonia at a concentration of 0.1 M is approximately 5.56 × 10^{-9} %.
