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)
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