#### Answer

The molarity should be about 0.0102M.

#### Work Step by Step

$Kb (NH_3) = 1.8 \times 10^{-5}$
** 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:
$\frac{4.286 \times 10^{-4}}{[Base]} = 0.042$
$[Base] = \frac{4.286 \times 10^{-4}}{0.042}$
$[Base] = 0.0102M$