#2H_2O rightleftharpoons H_3O^+ + OH^-#
The auto-ionization constant of water is given by:
#K_W = [ H_3O^+] [ OH^-]#
Apply the #log# function to both sides of the equation:
#log(K_W )=log( [ H_3O^+] [ OH^-])#
#log(K_W )= log[ H_3O^+] + log[ OH^-]#
Multiply both sides of the equation by a negative sign
#-{log(K_W )}= -{\ log[ H_3O^+] + log[ OH^-]\ }#
# -log(K_W )= -log[ H_3O^+] - log[ OH^-]#
# -log(1.0xx10^-14)= -log[ H_3O^+] - log[ OH^-]#
# 14 = -log[ H_3O^+] - log[ OH^-]#
#pH = -log[ H_3O^+] " and " pOH= - log[ OH^-] #
#14 = pH + pOH =>#
#pH + pOH = 14#
