#### Answer

(a) 0.00533
(b) 0.533%

#### Work Step by Step

1. Convert the pKa into Ka:
$Ka = 10^{-pKa} = 10^{-4.89} = 1.29 \times 10^{-5}$
2. Drawing the ICE table we get:
-$[H_3O^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x$
For approximation, we consider: $[HA] = [HA]_{initial}$
3. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][A^-]}{ [HA]}$
$Ka = 1.29 \times 10^{- 5}= \frac{x * x}{ 4.5 \times 10^{- 1}}$
$Ka = 1.29 \times 10^{- 5}= \frac{x^2}{ 4.5 \times 10^{- 1}}$
$ 5.8 \times 10^{- 6} = x^2$
$x = 2.4 \times 10^{- 3}$
4. Find the degree of ionization:
Degree of ionization: $\frac{ 2.4 \times 10^{- 3}}{ 4.5 \times 10^{- 1}} = 0.00533$
5. Calculate the percent ionization.
0.00533 * 100% = 0.533%