$n_{Na_2CO_3}=0,2(mol)$
- Nếu $X$ có $HCl$ dư
$2HCl+Na_2CO_3\to 2NaCl+CO_2+H_2O$
$n_{HCl\rm pứ}=0,2.2=0,4(mol)$
$HCl+KOH\to KCl+H_2O$
$n_{HCl\rm dư}=n_{KOH}=0,05(mol)$
$\to C_{M_{HCl}}=\dfrac{0,4+0,05}{0,1}=4,5M$
- Nếu $X$ có $NaHCO_3$:
$HCl+Na_2CO_3\to NaCl+NaHCO_3$
$\to n_{NaHCO_3}=n_{HCl}$
$2NaHCO_3+2KOH\to K_2CO_3+Na_2CO_3+2H_2O$
$\to n_{HCl}=0,05(mol)$
$\to n_{Na_2CO_3\rm pứ}=n_{HCl}=0,05(mol)<0,2$ (TM)
$\to C_{M_{HCl}}=\dfrac{0,05}{0,1}=0,5M$