Resposta :
Considerando houve total dissociação!
[OH] = 0,01 / l
pOH = - log [OH-]
pOH= - log [0,01]
pOH = - log [1.10⁻²]
pOH = 2
pOH + pH = 14
2 + pH = 14
pH = 12
[OH] = 0,01 / l
pOH = - log [OH-]
pOH= - log [0,01]
pOH = - log [1.10⁻²]
pOH = 2
pOH + pH = 14
2 + pH = 14
pH = 12