Answer
$Ka \approx 2\times 10^{- 9}$
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.
-$[H_3O^+] = [CCl_3C{O_2}^-] = x$
-$[CCl_3CO_2H] = [CCl_3CO_2H]_{initial} - x$
2. Using the pH, calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 4.95}$
$[H_3O^+] = 1.122 \times 10^{- 5}$
So: $x = 1.122 \times 10^{-5}$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][OBr^-]}{ [HOBr]}$
$Ka = \frac{x^2}{[InitialHOBr] - x}$
$Ka = \frac{( 1.122\times 10^{- 5})^2}{ 0.063- 1.122\times 10^{- 5}}$
$Ka = \frac{ 1.259\times 10^{- 10}}{ 0.06299}$
$Ka = 1.999\times 10^{- 9}$
