#### Answer

(a) 0.0225
(b) 2.25%

#### Work Step by Step

* Kb for ethylamine = $4.3 \times 10^{-4}$
1. Drawing the ICE table we get:
-$[OH^-] = [B^+] = x$
-$[Base] = [Base]_{initial} - x$
For approximation, we consider: $[Base] = [Base]_{initial}$
2. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][B^+]}{ [Base]}$
$Kb = 4.3 \times 10^{- 4}= \frac{x * x}{ 8.5 \times 10^{- 1}}$
$Ka = 4.3 \times 10^{- 4}= \frac{x^2}{ 8.5 \times 10^{- 1}}$
$ 3.65 \times 10^{- 4} = x^2$
$x = 1.91 \times 10^{- 2}$
- Degree of ionization: $\frac{ 1.91 \times 10^{- 2}}{ 8.5 \times 10^{- 1}} = 0.0225 $
- Percent ionization: 0.0225 * 100% = 2.25%