Pt ion: OH- + H+ → H2O
Ta có: \(\left\{{}\begin{matrix}n_{OH^-}=0,08.1.1=0,08\left(mol\right)\\n_{H^+}=0,12.0,25.2=0,06\left(mol\right)\end{matrix}\right.\)
⇒ OH- dư
\(\Rightarrow\left[OH^-\right]=\dfrac{0,08-0,06}{0,08+0,12}=0,1M\)
\(\Rightarrow pH=14+log\left(0,1\right)=13\)