设计实验探讨向AlCl3溶液中加入过量的NaOH和向NaOH溶液中加入过量的Al

问题实验题

设计实验探讨向AlCl₃溶液中加入过量的NaOH和向NaOH溶液中加入过量的AlCl₃现象是否一样。_________________________________

答案

向AlCl₃溶液中逐滴加入NaOH有白色沉淀生成，当NaOH过量时沉淀溶解，最后生成物为NaAlO₂。反应的离子方程式为：Al³⁺+3OH^-==Al(OH)₃↓，Al(OH)₃+OH^-== +2H₂O

向NaOH溶液中逐滴加入AlCl₃溶液时，观察到的现象是：开始有白色沉淀生成，振荡后立即消失，继续滴加AlCl₃溶液直至过量时，又出现白色沉淀，且不消失。这是因为反应开始时，由于NaOH溶液过量，滴入的AlCl₃和NaOH作用转化为NaAlO₂，当加入过量AlCl₃时，AlCl₃和NaAlO₂在溶液里发生反应，生成沉淀Al(OH)₃。其反应的离子方程式为：Al³⁺+3OH^-====Al(OH)₃↓，Al(OH)₃+OH^-==+2H₂O，Al³⁺+3+6H₂O==4Al(OH)₃↓

综上所述，二者滴加顺序不同，现象不一样。

完形填空
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