# Question 5c07c

May 20, 2015

The acid dissociation constant for this acid is equal to

So, you're dealing with a 0.085-M phenylacetic acid, ${C}_{8} {H}_{8} {O}_{2}$, solution, which has a pH of 2.68.

Use the pH of the solution to determine what the concentration of the hydronium ions, ${H}_{3} {O}^{+}$, 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 ${K}_{a}$

$\text{ } {C}_{8} {H}_{8} {O}_{2 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {C}_{8} {H}_{7} {O}_{2 \left(a q\right)}^{-} + {H}_{3} {O}_{\left(a q\right)}^{+}$
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)#