Answer
In equilibrium:
$[H_3O^+] = 7.45 \times 10^{-3}M$
$[Cl{O_2}^-] = 7.45 \times 10^{-3}M$
$[HClO_2] = 5.05 \times 10^{-3}M$
Work Step by Step
1. Drawing the ICE table we get:
-$[H_3O^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][A^-]}{ [HA]}$
$Ka = 1.1 \times 10^{- 2}= \frac{x * x}{ 1.25 \times 10^{- 2}- x}$
$Ka = 1.1 \times 10^{- 2}= \frac{x^2}{ 1.25 \times 10^{- 2}- x}$
$ 1.37 \times 10^{- 4} - 1.1 \times 10^{- 2}x = x^2$
$ 1.37 \times 10^{- 4} - 1.1 \times 10^{- 2}x - x^2 = 0$
Bhaskara:
$\Delta = (- 1.1 \times 10^{- 2})^2 - 4 * (-1) *( 1.37 \times 10^{- 4})$
$\Delta = 1.21 \times 10^{- 4} + 5.5 \times 10^{- 4} = 6.71 \times 10^{- 4}$
$x_1 = \frac{ - (- 1.1 \times 10^{- 2})+ \sqrt { 6.71 \times 10^{- 4}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 1.1 \times 10^{- 2})- \sqrt { 6.71 \times 10^{- 4}}}{2*(-1)}$
$x_1 = - 1.84 \times 10^{- 2} (Negative)$
$x_2 = 7.45 \times 10^{- 3}$
The concentration can't be negative.
So: $[H_3O^+] = [ClO_2^-] = x= 7.45 \times 10^{-3}M$
And $[HClO_2] = 0.0125 - 7.45 \times 10^{-3} = 5.05 \times 10^{-3}M$
