Answer
The $K_a$ value for $HOBr$ is equal to $1.9 \times 10^{-9}$
Work Step by Step
1. Calculate the $[H_3O^+]$ of that solution.
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 4.95}$
$[H_3O^+] = 1.1 \times 10^{- 5}$
Therefore : $x = 1.1\times 10^{- 5}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$HOBr(aq) + H_2O(l) \lt -- \gt OBr^-(aq) + H_3O^+(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $OBr^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M
-$[H_3O^+] = [OBr^-] = 0 + x = x$
-$[HOBr] = [HOBr]_{initial} - x$
For approximation, we are going to consider $[HOBr]_{initial} = [HOBr]$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][OBr^-]}{ [HOBr]}$
$Ka = \frac{x^2}{[InitialHOBr] - x}$
$Ka = \frac{( 1.1\times 10^{- 5})^2}{ 0.063- 1.1\times 10^{- 5}}$
$Ka = 1.9 \times 10^{- 9}$
