Calculate the percent dissociation of trimethylacetic acid(C6H5CO2H) in a aqueous solution of the stuff.

1.112%.

Explanation:

Step one : write out the correct acid dissociation reaction. This is done below:

C6H5CO2H <> C6H5CO2^- + H^+.

At equilibrium, the concentration of C6H5CO2H = 0.5 M - x and the concentration of C6H5CO2^- = x and the concentration of H^+ = x.

The ka for C6H5CO2H = 6.3 × 10^-5.

Step two: find the value for the concentration of C6H5CO2^- and H^+. This can be done by using the equilibrium dissociation Constant formula below;

Ka = [C6H5CO2^- ] [H^+] / C6H5CO2H.

ka = [x] [x] / [0.5 - x].

6.3 × 10^-5 = (x)^2 / [0.5 - x].

x^2 = 3.15 ×10^-5 - 6.3 × 10^-5 x.

x^-2 + 6.3 × 10^-5 x - 3.15 × 10^-5.

Solving for x we have x= 0.0055632289702004 = 0.00556.

Therefore, at equilibrium the concentration of H^+ = 0.00556 M and the concentration of C6H5CO2H= O.5 - 0.00556 = 0.494 M.

Step three: Calculate the equilibrium pH. This stage can be ignored for this question since we are not asked to calculate for the pH.

The formular for pH = - log [H^+].

pH= - log [0.00556].

pH= 2.255.

Step four: find the percentage dissociation.

Percentage dissociation = ( [H^+] at equilibrium/ [ C6H5CO2H] at Initial concentration ) × 100%.

Percentage dissociation=( 0.00556/ 0.5 ) × 100.

=Percentage dissociation= 1.112%.

the correct answer would be d. hope this !