## Chemistry 9th Edition

The $K_a$ value for $CCl_3CO_2H$ is equal to $0.16$.
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: $Ka = \frac{[H_3O^+][CCl_3^-]}{ [CCl_3CO_2H]}$ $Ka = \frac{x^2}{[InitialCCl_3CO_2H] - x}$ $Ka = \frac{( 0.040)^2}{ 0.050- 0.040}$ $Ka = \frac{ 1.6\times 10^{- 3}}{ 0.010}$ $Ka = 0.16$