\(pH=2\Rightarrow-log\left[H^+\right]=2\\ \Leftrightarrow\left[H^+\right]=0,01\left(M\right)\\ \Rightarrow\left[H_2SO_4\right]=\dfrac{0,01}{2}=0,005\left(M\right)\\ pH=12\Rightarrow\left[OH^-\right]=0,01\left(M\right)\\ \Rightarrow\left[Ba\left(OH\right)_2\right]=\dfrac{0,01}{2}=0,005\left(M\right)\\ n_{H^+}=0,2.0,005.2=0,002\left(mol\right);n_{OH^-}=2.0,3.0,005=0,003\left(mol\right)\\ H^++OH^-\rightarrow H_2O\\ n_{OH^-\left(dư\right)}=0,003-0,002=0,001\left(mol\right)\\ pH_{dd.thu.được}=14+log\left[OH^-\left(dư\right)\right]=14+log\left[\dfrac{0,001}{0,2+0,3}\right]\approx11,30103\\ m_{\downarrow}=m_{BaSO_4}=\dfrac{0,002}{2}.233=0,233\left(g\right)\)
