Answer
$Ka$ for trichloroacetic acid: 0.16
Work Step by Step
1. Since they have the same pH, their $[H_3O^+]$ should be the same too.
And, $HClO_4$ is a strong acid, therefore :$[H_3O^+] = [HClO_4] = 0.040M$
- Now, we know that, the hydronium concentration in the trichloroacetic solution is 0.040M too.
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [CCl_3C{O_2}^-] = x$
-$[CCl_3CO_2H] = [CCl_3CO_2H]_{initial} - x$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][CCl_3C{O_2}^-]}{ [CCl_3CO_2H]}$
$Ka = \frac{x^2}{[InitialCCl_3CO_2H] - x}$
$Ka = \frac{( 0.04)^2}{ 0.05- 0.04}$
$Ka = \frac{ 1.6\times 10^{- 3}}{ 0.01}$
$Ka = 0.16$
