Answer:
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%.
