1.
$n_{H^+}=2n_{H_2SO_4}=0,06(mol)$
$n_{OH^-}=2n_{Ba(OH)_2}=0,02(mol)$
$\Rightarrow n_{H^+\text{dư}}=0,06-0,02=0,04(mol)$
$[H^+]=\dfrac{0,04}{0,1+0,1}=0,2M$
$\to pH=-\lg[H^+]=0,7$
2.
$n_{H^+}=n_{HCl}=0,02(mol)$
$n_{OH^-}=n_{NaOH}=0,03(mol)$
$\Rightarrow [OH^-]=\dfrac{0,03-0,02}{0,04+0,06}=0,1M$
$\to pH+14-pOH=14+\lg[OH^-]=13$
3. (Thiếu đề)
4.
$n_{Al}=\dfrac{6,345}{27}=0,235(mol)$
$2Al+6HCl\to 2AlCl_3+3H_2$
$n_{HCl}=0,4.2=0,8(mol)$
$\Rightarrow n_{H^+\text{dư}}=0,8-0,235.3=0,095(mol)$
$\to pH=-\lg\dfrac{0,095}{0,4}=0,62$