Bài $1$$pH=12\Rightarrow [H^+]=10^{-12}{\Rightarrow }[OH^-]=10^{-2}$
$\Rightarrow n_{OH^-}=\frac{10^{-2}(250+250)}{1000}=5.10^{-3}$
$\Rightarrow 5.10^{-3}=2.\frac{250a}{1000}-2.\frac{250.0,01}{1000}-\frac{250.0,08}{1000}\Rightarrow a=0,06$
$\Rightarrow n_{H_2SO_4}<n_{Ba(OH)_2}\Rightarrow n_{BaSO_4}=n_{H_2SO_4}\Rightarrow m=0,5825 gam$