Question #5c07c

The acid dissociation constant for this acid is equal to

So, you're dealing with a 0.085-M phenylacetic acid, #C_8H_8O_2#, solution, which has a pH of 2.68.

Use the pH of the solution to determine what the concentration of the hydronium ions, #H_3O^(+)#, is

#[H_3O^(+)] = 10^(-pH_"sol") = 10^(-2.68) = 2.1 * 10^(-3)"M"#

Since the acid produces this much hydronium ions at equilibrium, you can work backwards to determine what the acid dissociation constant, #K_a#, must be.

Use an ICE table for the equilibrium reaction that gets established when phenylacetic acid is placed in aqueous solution to help you determine the value of #K_a#

#" "C_8H_8O_(2(aq)) + H_2O_((l)) rightleftharpoons C_8H_7O_(2(aq))^(-) + H_3O_((aq))^(+)#
I......0.085...........................................0.......................0
C......(-x)..............................................(+x)....................(+x)
E.....0.085-x........................................x........................x

You know that #K_a# is equal to

#K_a = ([H_3O^(+)] * [C_8H_7O_2""^(-)])/([C_8H_8O_2]) = (x * x)/(0.085 - x) = x^2/(0.085 - x)#

You also know that #x# is equal to the concentration of hydronium ions, which you've calculated earlier, so you get

#K_a = (2.1 * 10^(-3))^2/(0.085 - 2.1 * 10^(-3)) = color(green)(5.3 * 10^(-5)#