Ta có: \(n_{H^+}=n_{HCl}=0,04.0,75=0,03\left(mol\right)\)
\(n_{OH^-}=2n_{Ba\left(OH\right)_2}+n_{KOH}=2.0,16.0,08+0,16.0,04=0,032\left(mol\right)\)
PT: \(H^++OH^-\rightarrow H_2O\)
____0,03______0,03 (mol)
\(\Rightarrow n_{OH^-\left(dư\right)}=0,032-0,03=0,002\left(mol\right)\)
\(\Rightarrow\left[OH^-\right]=\dfrac{0,002}{0,04+0,16}=0,01\left(M\right)\)
\(\Rightarrow\left[H^+\right]=\dfrac{10^{-14}}{0,01}=10^{-12}\left(M\right)\)
\(\Rightarrow pH=-log\left[H^+\right]=12\)
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