Answer
Percent dissociation: $ = 20.79\%$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [Cl{O_2}^-] = x$
-$[HClO_2] = [HClO_2]_{initial} - x = 0.22 - x$
For approximation, we consider: $[HClO_2] = 0.22M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][Cl{O_2}^-]}{ [HClO_2]}$
$Ka = 1.2 \times 10^{- 2}= \frac{x * x}{ 0.22}$
$Ka = 1.2 \times 10^{- 2}= \frac{x^2}{ 0.22}$
$ 2.64 \times 10^{- 3} = x^2$
$x = 5.138 \times 10^{- 2}$
Percent dissociation: $\frac{ 5.138 \times 10^{- 2}}{ 0.22} \times 100\% = 23.35\%$
%ionization < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 0.012= \frac{x^2}{ 0.22- x}$
$ 2.64 \times 10^{- 3} - 0.012x = x^2$
$ 2.64 \times 10^{- 3} - 1.2 \times 10^{- 2}x - x^2 = 0$
$\Delta = (- 1.2 \times 10^{- 2})^2 - 4 * (-1) *( 2.64 \times 10^{- 3})$
$\Delta = 1.44 \times 10^{- 4} + 1.056 \times 10^{- 2} = 1.07 \times 10^{- 2}$
$x_1 = \frac{ - (- 1.2 \times 10^{- 2})+ \sqrt { 1.07 \times 10^{- 2}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 1.2 \times 10^{- 2})- \sqrt { 1.07 \times 10^{- 2}}}{2*(-1)}$
$x_1 = - 5.773 \times 10^{- 2} (Negative)$
$x_2 = 4.573 \times 10^{- 2}$
Percent dissociation: $\frac{ 4.573 \times 10^{- 2}}{ 0.22} \times 100\% = 20.79\%$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
