Question

Calculate the pH of a 0.334 M aqueous solution of phenol (a weak acid) (C₆H₅OH, K_a = 1.0×10^-10) and the equilibrium concentrations of the weak acid and its conjugate base.

pH	=
[C6H5OH ]equilibrium	=	M
[C6H5O- ]equilibrium	=	M

Answer 1

C₆H₅OH acid dissociates as

C₆H₅OH + H₂O C₆H₅O^- + H₃O⁺

K_a = [C₆H₅O^- ][H₃O⁺] / [C₆H₅OH]

but  [C₆H₅O^- ] = [H₃O⁺] = x

K_a = [x][x] / [C₆H₅OH]

Ka of phenol = 1.0 X10^-10

Substitute the value in equation

1.0 X10^-10 = [x]²/ 0.334

[x]² = 1.0 X10^-10 X  0.334 = 3.34 X 10^-11

[x] = 5.779 X 10^-6

[C₆H₅O^- ] = [H₃O⁺] = x= 5.779 X 10^-6 M

[C₆H₅O^- ] = 5.779 X 10^-6 M

equlibrium concentration of C₆H₅OH = intial concentration of C₆H₅OH - equilibrium cooncentrtion of C₆H₅O^-

[C₆H₅OH] equilibrium = 0.334 - 5.779 X 10^-6 = 0.33399 M

[C₆H₅OH] equilibrium = 0.33399 M

[H₃O⁺] = 5.779 X 10^-6 M

pH = - log[H₃O⁺]

pH = - log (5.779 X 10^-6)

pH = 5.24