Question

Prove. that p poH 14.​

Answer 1

consider dissociation/ self ionization of water:

H2O + H2O→ H3O+ + OH-

the equilibrium constant of this reaction is 1e-14

i.e., [H3O+] [OH-]/[H2O][H2O] = 1e-14

[H2O]=1 (water in bulk) and [H3O+] = [OH-]

→ [H3O+] =[OH-]= 1e-7

now, pH = -log[H+] = -log[H3O+] = 7

(because H3O+ is H2O+H+, for each H3O+, active species is one H+)

and pOH= -log[OH-] =7

→ pH + pOH =14