Cho từ từ 350 ml dung dịch NaOH 1M vào

$n_{NaOH}=0,35mol;{\text{ n}}_{Al{\left(OH\right)}_3}=0,05mol;{\text{ n}}_{AlC{l}_3}=0,1\text{x\hspace{0.33em}}mol$

Phương trình ion:

 $\begin{array}{l}A{l}^{3+}+3\text{O}{H}^{-}\to Al{\left(OH\right)}_3\\ 0,1\text{x}\to 0,3\text{x }\to x\text{ mol}\end{array}$

$\begin{array}{l}Al{\text{ (OH)}}_3+O{H}^{-}\to Al{O}_2^{-}+2H_{2}O\\ 0,1\text{x}-0,05\to 0,1\text{x}-0,05\text{ mol}\end{array}$

Ta có: $n_{NaOH}=n_{O{H}^{-}}=0,3\text{x}+0,1\text{x}-0,05=0,34\to x=1$