Đáp án D
$pOH = 14 - pH = 14 - 12 = 2 \Rightarrow n_{NaOH} = 0,01.10^{-pOH} = 10^{-4}(mol)$
$pH = 11 \Rightarrow pOH = 14 - 11 = 3 \Rightarrow C_{M_{NaOH\ sau\ khi\ pha}} = 10^{-3}$
$\Rightarrow V_{dd\ sau\ khi\ pha} = \dfrac{10^{-4}}{10^{-3}} = 0,1(lít) = 100(ml)$
$\Rightarrow V_{H_2O} = 100 - 10 = 90(ml)$