Câu 22:
$n_{H^+}=n_{HCl}+n_{HNO_3}=0,2.0,1+0,1.0,1=0,03(mol)$
$\to n_{OH^-}=n_{H^+}=0,03(mol)$
$\to n_{Ba(OH)_2}=\dfrac{n_{OH^-}}{2}=0,015(mol)$
$\to V_{dd Ba(OH)_2}=\dfrac{0,015}{0,02}=0,75l$
Câu 23:
$V_{HCl}: V_{H_2SO_4}=1:1$
$\to V_{HCl}=V_{H_2SO_4}=\dfrac{0,1}{2}=0,05l$
$\to n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,05.0,2+0,05.0,1.2=0,02(mol)$
$\to n_{OH^-}=n_{H^+}=0,02(mol)$
$\to n_{Ba(OH)_2}=\dfrac{n_{OH^-}}{2}=0,01(mol)$
$\to V_{dd Ba(OH)_2}=\dfrac{0,01}{0,02}=0,5l=500ml$
Câu 24:
Đặt $x$, $y$ là nồng độ $HNO_3, KOH$
• TN1:
$n_{HNO_3}=0,02x(mol)$
$n_{KOH}=0,06y(mol)$
$HNO_3+KOH\to KNO_3+H_2O$
$\to 0,02x=0,06y$
$\to 0,02x-0,06y=0$
• TN2:
$n_{KOH}=0,01y(mol)$
$n_{CuO}=\dfrac{2}{80}=0,025(mol)$
$2HNO_3+CuO\to Cu(NO_3)_2+H_2O$
$HNO_3+KOH\to KNO_3+H_2O$
$\to 0,025.2+0,01y=0,02x$
$\to 0,02x-0,01y=0,05$
Giải hệ: $x=3; y=1$
Vậy $C_{M_{HNO_3}}=3M; C_{M_{KOH}}=1M$
