Ta có: \(n_{H^+}=2n_{H_2SO_4}=2.0,1.0,005=0,001\left(mol\right)\)
\(n_{OH^-}=n_{KOH}=0,1.0,012=0,0012\left(mol\right)\)
PT ion: \(H^++OH^-\rightarrow H_2O\)
_____0,001_0,0012________ (mol)
⇒ OH- dư.
\(\Rightarrow n_{OH^-\left(dư\right)}=0,0002\left(mol\right)\)
\(\Rightarrow\left[OH^-\right]_{\left(dư\right)}=\frac{0,0002}{0,1+0,1}=0,001M\)
\(\Rightarrow\left[H^+\right]=\frac{10^{-14}}{0,001}=10^{-11}M\)
\(\Rightarrow x=pH=11\)
Bạn tham khảo nhé!