Ba(OH)2 + 2HCl \(\rightarrow\)BaCl2 + 2H2O (1)
Ba(OH)2 + H2SO4 \(\rightarrow\)BaSO4 + 2H2O (2)
nHCl=0,2.0,3=0,06(mol)
nH2SO4=0,2.0,1=0,02(mol)
Theo PTHH 1 và 2 ta có:
\(\dfrac{1}{2}\)nHCl=nBa(OH)2=0,03(mol)
nBa(OH)2=nH2SO4=0,02(mol)
\(\sum\)nBa(OH)2=0,03+0,02=0,05(mol)
V dd Ba(OH)2=\(\dfrac{0,05}{0,2}=0,25\left(lít\right)\)