Answer
$$K_a = 3.0 \times 10^{- 6}$$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
$HX(aq) + H_2O(l) \lt -- \gt X^-(aq) + H_3O^+(aq) $
\begin{matrix} & [HX] & [H_3O^+] & X^- \\ Initial & 0.085 & 0 & 0 \\ Change & -x & +x & +x\\ Equil & 0.085 -x & x & x \end{matrix}
2. Write the percent dissociation equation, and find 'x':
%dissociation = $\frac{x}{[Initial Acid]} \times 100$
0.59= $\frac{x}{ 0.085} \times 100$
0.0059= $\frac{x}{ 0.085}$
$ 5.0 \times 10^{- 4}= x$
Therefore: $[X^-] and [H_3O^+] = 5.0 \times 10^{- 4}$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][X^-]}{ [HX]}$
$Ka = \frac{x^2}{0.085 - x}$
$Ka = \frac{( 5.0\times 10^{- 4})^2}{ 0.085- 5\times 10^{- 4}}$
$Ka = 3.0 \times 10^{- 6}$
