Answer
The $K_a$ value for $CCl_3CO_2H$ is equal to $0.16$.
Work Step by Step
Equal pH means equal $H_3O^+$ concentration:
1. Determine the hydronium ion concentration value for the $0.040M $ $HClO_4$ solution.
- $HClO_4$ is a strong acid, so: $[HClO_4]_{initial} = [H_3O^+] = 0.040M$
Therefore, the $CCl_3CO_2H$ solution will have a $[H_3O^+]$ equal to $0.040M
$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$CCl_3CO_2H(aq) + H_2O(l) \lt -- \gt CCl_3CO_2^-(aq) + H_3O^+(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $CCl_3CO_2^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M
-$[H_3O^+] = [CCl_3CO_2^-] = 0 + x = x = 0.040M$
-$[CCl_3CO_2H] = [CCl_3CO_2H]_{initial} - x$
For approximation, we are going to consider $[CCl_3CO_2H]_{initial} = [CCl_3CO_2H]$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][CCl_3^-]}{ [CCl_3CO_2H]}$
$Ka = \frac{x^2}{[InitialCCl_3CO_2H] - x}$
$Ka = \frac{( 0.040)^2}{ 0.050- 0.040}$
$Ka = \frac{ 1.6\times 10^{- 3}}{ 0.010}$
$Ka = 0.16$
