a, \(n_{OH^-}=n_{NaOH}=\dfrac{0,4}{40}=0,01\left(mol\right)\)
\(\Rightarrow\left[OH^-\right]=\dfrac{0,01}{0,1}=0,1\)
\(\Rightarrow\left[H^+\right]=10^{-13}\)
\(\Rightarrow pH=13\)
b, \(n_{OH^-}=2n_{Ba\left(OH\right)_2}=2.0,0025=0,005\left(mol\right)\)
\(\Rightarrow\left[OH^-\right]=\dfrac{0,005}{0,05}=0,1\)
\(\Rightarrow\left[H^+\right]=10^{-13}\)
\(\Rightarrow pH=13\)