Answer
pH = 1.52
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.
-$[H_3O^+] = [Cl{O_2}^-] = x$
-$[HClO_2] = [HClO_2]_{initial} - x = 0.115 - x$
For approximation, we consider: $[HClO_2] = 0.115M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][Cl{O_2}^-]}{ [HClO_2]}$
$Ka = 1.1 \times 10^{- 2}= \frac{x * x}{ 0.115}$
$Ka = 1.1 \times 10^{- 2}= \frac{x^2}{ 0.115}$
$ 1.265 \times 10^{- 3} = x^2$
$x = 3.557 \times 10^{- 2}$
Percent dissociation: $\frac{ 3.557 \times 10^{- 2}}{ 0.115} \times 100\% = 30.93\%$
%dissociation > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 0.011= \frac{x^2}{ 0.115- x}$
$ 1.265 \times 10^{- 3} - 0.011x = x^2$
$ 1.265 \times 10^{- 3} - 1.1 \times 10^{- 2}x - x^2 = 0$
Bhaskara:
$\Delta = (- 1.1 \times 10^{- 2})^2 - 4 * (-1) *( 1.265 \times 10^{- 3})$
$\Delta = 1.21 \times 10^{- 4} + 5.06 \times 10^{- 3} = 5.181 \times 10^{- 3}$
$x_1 = \frac{ - (- 1.1 \times 10^{- 2})+ \sqrt { 5.181 \times 10^{- 3}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 1.1 \times 10^{- 2})- \sqrt { 5.181 \times 10^{- 3}}}{2*(-1)}$
$x_1 = - 4.1 \times 10^{- 2} (Negative)$
$x_2 = 3.0 \times 10^{- 2}$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.030)$
$pH = 1.52$
