Đáp án: $pH=1$
Giải thích các bước giải:
$V_{\text{dd sau}}=0,2+0,1=0,3l$
$n_{NaOH}=0,2.0,1=0,02(mol)$
$\Rightarrow n_{OH^-}=n_{NaOH}=0,02(mol)$
$n_{H_2SO_4}=0,1.0,25=0,025(mol)$
$\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,025.2=0,05(mol)$
$H^++OH^-\to H_2O$
$n_{H^+}>n_{OH^-}\Rightarrow H^+$ dư, $OH^-$ hết
$n_{H^+\text{dư}}=0,05-0,02=0,03(mol)$
$\Rightarrow [H^+]=\dfrac{0,03}{0,3}=0,1M$
Vậy $pH=-\log[H^+]=1$
