PTHH: BaCO3+2HCl---->BaCl2+CO2+H2O(1)
HCl+NaOH---->NaCl+H2O(2)
Ta có: nBaCO3=0,02mol\RightarrownHCl (1)=0,04mol
nHCl ban đầu=0,2.0,5=0,1mol
\RightarrownHCl dư=0,06mol
Từ 2\RightarrownNaOH=nHCl dư=0,06mol
\RightarrowV_dd_NaOH=[TEX]\frac{0,06}{0,2}[/TEX]=0,3 l