\(n_{H^+}=0,5.0,2=0,1\left(mol\right)\)
\(n_{OH^-}=0,5.0,3=0,15\left(mol\right)\)
\(\Rightarrow n_{OH^-dư}=0,15-0,1=0,05\left(mol\right)\)
\(\Rightarrow\left[OH^-\right]_{\text{sau pư}}=\dfrac{0,05}{0,5}=0,1\)
\(\Rightarrow\left[H^+\right]=10^{-13}\)
\(\Rightarrow pH=13\)