$n_{CO_2}= 0,04(mol) > n_{BaCO_3} = 0,02(mol)$ nên có tạo muối axit
CO2 + Ba(OH)2 → BaCO3 + H2O
0,02........0,02...........0,02.................(mol)
2CO2 + Ba(OH)2 → Ba(HCO3)2
0,02........0,01......................................(mol)
$n_{Ba(OH)_2} = 0,02 + 0,01 = 0,03(mol)$
$a = \dfrac{0,03}{0,2} = 0,15(M)$
\(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)
\(0.02..........0.02............0.02\)
\(Ba\left(OH\right)_2+2CO_2\rightarrow Ba\left(HCO_3\right)_2\)
\(0.01........0.04-0.02\)
\(n_{Ba\left(OH\right)_2}=0.02+0.01=0.03\left(mol\right)\)
\(C_{M_{Ba\left(OH\right)_2}}=\dfrac{0.03}{0.2}=0.15\left(M\right)\)
