Dung dịch X gồm 0,1 mol H+, z mol Al3+, t mol và 0,02 mol SO42-. Cho 120 ml dung dịch Y gồm KOH 1,2M và Ba(OH)2 0,1M vào dung dịch X thu được 3,723 gam kết tủa. Giá trị của z, t lần lượt là:

\(\text{X}\left| \begin{align} & {{H}^{+}}:0,1\,\,mol \\ & A{{l}^{3+}}:z\,\,mol \\ & NO_{3}^{-}:t\,\,mol \\ & SO_{4}^{2-}:\,\,0,02\,\,mol \\ \end{align} \right.\,\,\,\,\,\,\,+\,\,\,\,\) \(Y\left| \begin{align} & O{{H}^{-}}:0,168\,\,\,mol \\ & B{{a}^{2+}}:0,012\,\,\,mol \\ & {{K}^{+}} \\ \end{align} \right.\)

Bảo toàn điện tích trong dung dịch X:

\(\begin{align} & 0,1+3\text{z}=t+0,02.2 \\ & \Leftrightarrow t-3\text{z}=0,06\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) \\ \end{align}\)

Khi X + Y:

\(\begin{align} & B{{a}^{2+}}\,\,\,\,\,+\,\,\,\,\,SO_{4}^{2-}\,\,\,\to \,\,\,\,BaS{{O}_{4}}\downarrow \\ & 0,012\,\,\,\,\,\,\,\,\,\,0,012\,\,\,\,\,\,\,\,\,\,\,\,0,012\,\,\,\,\,\,\,\,\,\,\,\,\,\,(mol) \\ & {{m}_{Bas{{O}_{4}}}}=2,796\,\,\,gam \\ & {{m}_{Al{{(OH)}_{3}}}}=3,732-2,796=0,936 \\ & \to {{n}_{Al{{(OH)}_{3}}}}=0,012\,\,\,mol \\ \end{align}\)

\(\begin{align} & {{H}^{+}}\,\,\,\,+\,\,\,\,\,O{{H}^{-}}\,\,\,\,\to \,\,\,\,{{H}_{2}}O \\ & A{{l}^{3+}}\,\,+\,\,\,3\text{O}{{H}^{-}}\,\,\,\to \,\,\,Al{{(OH)}_{3}}\downarrow \\ & Al{{(OH)}_{3}}\,\,\,+\,\,\,\,O{{H}^{-}}\,\,\,\to \,\,\,AlO_{2}^{-}\,\,\,+\,\,\,2{{H}_{2}}O \\ & {{n}_{O{{H}^{-}}}}=4.{{n}_{A{{l}^{3+}}}}-{{n}_{Al{{(OH)}_{3}}}}+{{n}_{{{H}^{+}}}} \\ & \Leftrightarrow 0,1+4\text{z}-0,012=0,168 \\ & \Leftrightarrow z=0,02\,\,\,\,(2) \\ & (1)\,\to \,t=0,12 \\ \end{align}\)