" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
```` o pH = - log [H3O⁺]
[H3O⁺]
pOH = log 1
```` o pOH = - log [OH⁻]
[OH⁻]
PARA EL AGUA: [H3O⁺] ∙[OH⁻] = 10⁻¹⁴
1 1 1
------- ------- = ------
[H3O⁺] ∙ [OH⁻] 10⁻¹⁴
APLICANDO LOGARITMOS:
1 1
log ------- + log ------- = log 10¹⁴
[H3O⁺] [OH⁻]
entonces:
l 1 1
log ------- = pH log ------- = pOH
[H3O⁺] [OH⁻]
SE ESCRIBE ENTONCES: pH +pOH = 14