Ta có: \(n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,006\left(mol\right)\)
\(n_{OH^-}=n_{NaOH}=0,005\left(mol\right)\)
PT ion: \(H^++OH^-\rightarrow H_2O\)
_____0,006__0,005 (mol)
⇒ H+ dư. nH+ (dư) = 0,001 (mol)
\(\Rightarrow\left[H^+\right]_{\left(dư\right)}=\frac{0,001}{1}=0,001M\Rightarrow pH=3\)
Bạn tham khảo nhé!