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