## General Chemistry: Principles and Modern Applications (10th Edition)

$Kb (NH_3) = 1.8 \times 10^{-5}$ ** Considering a pure solution: $[OH^-] = [NH_4^+] = x$ ** And for approximation: $[Base] \approx [Base]_{initial}$ 1. If the percent ionization is 4.2%: The degree of ionization = 0.042, therefore: $\frac{x}{[Base]} = 0.042$ 2. Write the Kb equation: $Kb = \frac{[OH^-][NH_4^+]}{[NH_3]}$ $1.8 \times 10^{-5} = \frac{x^2}{[NH_3]} = x * \frac{x}{[NH_3]}$ $1.8 \times 10^{-5} = x * 0.042$ $x = 4.286 \times 10^{-4}$ 3. Therefore: $\frac{4.286 \times 10^{-4}}{[Base]} = 0.042$ $[Base] = \frac{4.286 \times 10^{-4}}{0.042}$ $[Base] = 0.0102M$