a/ Kết tủa là BaSO4
\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)
\(x-----x----x---2x\)
\(n_{Ba\left(OH\right)_2}=0,09.0,1=0,009\left(mol\right);n_{H_2SO_4}=0,02.0,4=0,008\left(mol\right)\)
⇒Tính theo số mol của H2SO4
\(\Rightarrow m_{BaSO_4}=0,008.233=1,864\left(g\right)\)
b/ \(n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,018\left(mol\right);n_{H^+}=2n_{H_2SO_4}=0,016\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
\(\Rightarrow n_{OH^-\left(du\right)}=0,018-0,016=0,002\left(mol\right)\)
\(\Rightarrow C_{MOH^-\left(du\right)}=\frac{0,002}{0,1+0,4}=4.10^{-3}\left(M\right)\)
\(\Rightarrow pOH^-=2,4;pOH+pH=14\Rightarrow pH=11,6\)
