\(pH=12\Rightarrow pOH=14-12=2\)
=> \(\left[OH^-\right]=10^{-2}\)M
=> \(n_{OH^-}=10^{-2}.0,1=0,001\left(mol\right)\)
\(n_{H^+}=0,012.0,1=0,0012\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
Lập tỉ lệ : \(\dfrac{0,0012}{1}>\dfrac{0,01}{1}\)
=> Sau phản ứng H+ dư
\(n_{H^+\left(dư\right)}=0,0012-0,01=0,0002\left(mol\right)\)
\(\left[H^+\right]=\dfrac{0,0002}{0,2}=0,001M\)
=> \(pH=3\)