Answer
The percent dissociation of that $HClO_2$ solution is equal to 21%.
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$HClO_2(aq) + H_2O(l) \lt -- \gt ClO_2^-(aq) + H_3O^+(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $ClO_2^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M
-$[H_3O^+] = [ClO_2^-] = 0 + x = x$
-$[HClO_2] = [HClO_2]_{initial} - x$
For approximation, we are going to consider $[HClO_2]_{initial} = [HClO_2]$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][ClO_2^-]}{ [HClO_2]}$
$Ka = 1.2 \times 10^{- 2}= \frac{x * x}{ 0.22}$
$Ka = 1.2 \times 10^{- 2}= \frac{x^2}{ 0.22}$
$x^2 = 0.22 \times 0.012 $
$x = \sqrt { 0.22 \times 0.012} = 0.051 $
Percent dissociation: $\frac{ 5.1 \times 10^{- 2}}{ 0.22} \times 100\% = 23\%$
%dissociation > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 0.012= \frac{x^2}{ 0.22- x}$
$ 2.6 \times 10^{- 3} - 0.012x = x^2$
$ 2.6 \times 10^{- 3} - 1.2 \times 10^{- 2}x - x^2 = 0$
Bhaskara:
$\Delta = (- 1.2 \times 10^{- 2})^2 - 4 * (-1) *( 2.6 \times 10^{- 3})$
$\Delta = 1.4 \times 10^{- 4} + 1.1 \times 10^{- 2} = 1.1 \times 10^{- 2}$
$x_1 = \frac{ - (- 1.2 \times 10^{- 2})+ \sqrt { 1.1 \times 10^{- 2}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 1.2 \times 10^{- 2})- \sqrt { 1.1 \times 10^{- 2}}}{2*(-1)}$
$x_1 = - 5.8 \times 10^{- 2} (Negative)$
$x_2 = 4.6 \times 10^{- 2}$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
Percent dissociation: $\frac{ 4.6 \times 10^{- 2}}{ 0.22} \times 100\% = 21\%$
