Cho V lít dung dịch NaOH 1M vào dung dịch chứa 0,14 mol AlCl3

* Hướng dẫn giải

$$\begin{gathered} TN1: Vmol NaOH+0,14 mol AlC{l}_3 \\ \to m gam Al{\left(OH\right)}_3 \\ TN2: 1,5Vmol NaOH+0,14 mol AlC{l}_3 \\ \to 0,75m gam Al{\left(OH\right)}_3 \\ \Rightarrow \left\{\begin{array}{l}TN1: Al{\left(OH\right)}_3 hòa \tan một phần \\ hoặc chưa hòa \tan \\ TN2: Al{\left(OH\right)}_3 hòa \tan một phần \end{array}\right. \\ TH1: \left\{\begin{array}{l}TN1: Al{\left(OH\right)}_3 hòa \tan 1 phần\\ TN2: Al{\left(OH\right)}_3 hòa \tan 1 phần\end{array}\right. \\ \begin{array}{l}TN1:n_{O{H}^{-}}=3n_{A{l}^{3+}}+\left(n_{A{l}^{3+}}-n_{Al{\left(OH\right)}_3}\right)\\ TN2:n_{O{H}^{-}}=3n_{A{l}^{3+}}+\left(n_{A{l}^{3+}}-n_{Al{\left(OH\right)}_3}\right)\end{array} \\ \Rightarrow \left\{\begin{array}{l}V=4.0,14-\frac{m}{78}\\ 1,5V=4.0,14-\frac{0,75.m}{78}\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}V=0,186\\ m=29,12\end{array}\right. \\ \Rightarrow TN1: \left\{\begin{array}{l} n_{O{H}^{-}}=0,168\\ n_{A{l}^{3+}}=0,14\end{array}\right. \\ \Rightarrow Al{\left(OH\right)}_3 chưa bị hòa \tan \\ TH2: \left\{\begin{array}{l}TN1: Al{\left(OH\right)}_3 chưa bị hòa \tan \\ TN2: Al{\left(OH\right)}_3 hòa \tan 1 phần\end{array}\right. \\ \begin{array}{l}TN1: n_{O{H}^{-}}=3n_{Al{\left(OH\right)}_3}\\ TN2: n_{O{H}^{-}}=3n_{A{l}^{3+}}+\left(n_{A{l}^{3+}}-n_{Al{\left(OH\right)}_3}\right)\end{array} \\ \Rightarrow \left\{\begin{array}{l}V=\frac{3m}{78}\\ 1,5V=4.0,14-\frac{0,75.m}{78}\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}V=0,32\\ m=8,32\end{array}\right. \end{gathered}$$