Prove that Ph+Poh= 14
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pH is negative log of [H+] and pOH is negative log of [OH-] right. so we can say: -log[H+] + (-log[OH-]) = 14
taking negative sign common
-( log[H+] + log[OH-]) =14
using log formula
-(log([H+][OH-])) =14
we know product of [H+] ions and [OH-] ions is equal to the ionization constant Kw which at 25 degrees has been experimentally determined to be 1 x 10^-14.
we put that value in the above equation
-(log(1 x 10^-14)) =14
-(-14)=14
14=14
Mathematically proved.
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