Answer
$[OBr^-] \approx [H_3O^+ \approx 3.354 \times 10^{- 5} M$
$[HOBr] \approx 0.45M$
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.
-$[H_3O^+] = [OBr^-] = x$
-$[HOBr] = [HOBr]_{initial} - x = 0.45 - x$
For approximation, we consider: $[HOBr] = 0.45M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][OBr^-]}{ [HOBr]}$
$Ka = 2.5 \times 10^{- 9}= \frac{x * x}{ 0.45}$
$Ka = 2.5 \times 10^{- 9}= \frac{x^2}{ 0.45}$
$ 1.125 \times 10^{- 9} = x^2$
$x = 3.354 \times 10^{- 5}$
Percent ionization: $\frac{ 3.354 \times 10^{- 5}}{ 0.45} \times 100\% = 0.007454\%$
%ionization < 5% : Right approximation.
Therefore: $[OBr^-] = [H_3O^+] = x = 3.354 \times 10^{- 5} $
$[HOBr] \approx 0.45M$
