PT ion: \(H^++OH^-\rightarrow H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{H^+}=0,15\cdot0,05=0,0075\left(mol\right)\\n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,001\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\) H+ còn dư 0,0065 mol
\(\Rightarrow\left[H^+\right]=\dfrac{0,0065}{0,2}=0,0325\left(M\right)\) \(\Rightarrow pH=-log\left(0,0325\right)\approx1,5\)