Chemistry, asked by kaundalrahul6pagsip, 6 months ago

Derive the relation pH+pOH=14

Answers

Answered by JENETCHACKO

Answer:

How do we derive pH+pOH=14?

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

and voila

-(-14)=14

14=14

mathematically proved.

Answered by babitasingh6846

Explanation:

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

and voila

-(-14)=14

14=14

mathematically proved.

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