Bài 2: Axit - Bazơ - Muối

5A

6D

a) \([OH^-]=\left[KOH\right]=1,5M\)

b) Để trung hòa dung dịch A: \(n_{\left[OH^-\right]}=n_{\left[H^+\right]}\)

                                          \(\Rightarrow0,15=0,5\cdot V_{HCl}\Rightarrow V_{HCl}=0,3l=300ml\)

a) Ta có: \(n_{HNO_3}=0,1\cdot0,25=0,025\left(mol\right)=n_{H^+}=n_{NO_3^-}\)

b) Ta có: \(n_{Ba\left(OH\right)_2}=0,2\cdot0,15=0,03\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Ba^{2+}}=0,03\left(mol\right)\\n_{OH^-}=0,06\left(mol\right)\end{matrix}\right.\)

c) Ta có: \(n_{Al_2\left(SO_4\right)_3}=0,2\cdot0,2=0,04\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al^{3+}}=0,08\left(mol\right)\\n_{SO_4^{2-}}=0,12\left(mol\right)\end{matrix}\right.\)

\(n_{H^+}=0,2.0,3=0,06\left(mol\right)\)

\(n_{OH^-}=0,5.2.0,001V=0,001V\left(mol\right)\)

\(\Rightarrow n_{OH^-dư}=0,001V-0,06\left(mol\right)\)

\(\left[OH^-\right]=\dfrac{0,001V-0,06}{0,001V+0,3}=0,28\)

\(\Rightarrow V=200\left(ml\right)\)

\(n_{H^+}=n_{HCl}=0,1.1=0,1\left(mol\right)\\ n_{Na^+}=n_{NaCl}=0,15.1=0,15\left(mol\right)\\ n_{Cl^-}=n_{HCl}+n_{NaCl}=0,1+0,15=0,25\left(mol\right)\\ \left[Na^+\right]=\dfrac{0,15}{0,25}=0,6\left(M\right)\\ \left[H^+\right]=\dfrac{0,1}{0,25}=0,4\left(M\right)\\ n_{Cl^-}=\dfrac{0,25}{0,25}=1\left(M\right)\)