\(n_{OH^-}=0,15.2=0,3\left(mol\right)\)
\(n_{H^+}=0,2\left(mol\right)\)
\(\Rightarrow n_{OH^-dư}=0,1\left(mol\right)\)
\(\Rightarrow\left[OH^-\right]=\dfrac{0,1}{0,35}=\dfrac{2}{7}M\)
\(\Rightarrow\left[H^+\right]=3,5.10^{-14}M\)
\(\Rightarrow pH\approx13,46\)