Respuesta :

Answer:

pOH = -log[OH-]

log x =a <=> 10ª = x

so, if pOH is 10.75, we need to aply the anti log of it to get the [OH-], where x is [OH-] and a is pOH

using the calc, we have

-log [OH-] = 10.75 <=>10^(-10.75) = [OH-]

[OH-]= 1.78 x 10^(-11)