C6H5OHÂ Â Â ----------------------------> C6H5O-
+ H+
0.11Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0 ----------------> initial
0.11-x                                              Â
x              Â
x -------------------> equilibrium
Ka = [C6H5O-][H+]/[C6H5OH]
Ka = x^2 / 0.11-x
1.3 x 10^-10 = x^2 / 0.11-x
x^2 + 1.3 x 10^-10 x -1.43 x 10^-11 = 0
by solving this
x = 3.78 x 10^-6
[H+] = x = 3.78 x 10^-6 M
pH = -log [H+] = -log (3.78 x 10^-6)
pH = 5.42 ----------------------->
answer
