\(pH=12\Rightarrow pOH=2\\ \Rightarrow\left[OH^-\right]_{sau}=10^{-2}\left(M\right)\)
Ta có : \(n_{OH^-\left(củaBa\left(OH\right)_2\right)}=0,002.2.0,1=4.10^{-4}\left(mol\right)\)
=> \(V_{sau}=\dfrac{4.10^{-4}}{10^{-2}}=0,04\left(lít\right)=40\left(ml\right)\)
Mà ta có : \(V_{sau}=V_{dd}+V_{H_2O}=2+V_{H_2O}=40\left(ml\right)\\ \Rightarrow V_{H_2O}=38\left(ml\right)\)