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\(n_{H_2SO_4}=0,05.0,4=0,02\left(mol\right)\\ n_{HCl}=0,35.0,2=0,07\left(mol\right)\\ \left[H_2SO_4\right]=\dfrac{0,02}{0,05+0,35}=0,05\left(M\right)\\ \left[HCl\right]=\dfrac{0,07}{0,05+0,35}=0,175\left(M\right)\\ \Rightarrow\left[H^+\right]=0,05.2+0,175.1=0,275\left(M\right)\\ \left[SO^{2-}_4\right]=0,05\left(M\right)\\ \left[Cl^-\right]=0,175\left(M\right)\)
a) Ta có: \(n_{Al\left(NO_3\right)_3}=\dfrac{4,26}{213}=0,02\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al^+}=0,02\left(mol\right)\\n_{NO_3^-}=0,06\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[Al^+\right]=\dfrac{0,02}{0,1}=0,2\left(M\right)\\\left[NO_3^-\right]=\dfrac{0,06}{0,1}=0,6\left(M\right)\end{matrix}\right.\)
b) Ta có: \(\left[Na^+\right]=0,1+0,02\cdot2+0,3=0,304\left(M\right)\)
c) Bạn xem lại đề !!
\(n_{OH^-}=0,5.0,2+0,2.2.0,3=0,22\left(mol\right)\Rightarrow\left[OH^-\right]=\dfrac{0,22}{0,5}=0,44M\)
\(n_{Na^+}=0,5.0,2=0,1\left(mol\right)\Rightarrow\left[Na^+\right]=\dfrac{0,1}{0,5}=0,2M\)
\(n_{Ba^{2+}}=0,2.0,3=0,06\left(mol\right)\Rightarrow\left[Ba^{2+}\right]=\dfrac{0,06}{0,5}=0,12M\)
\(a.\left[H^+\right]=2.0,1=0,2\left(M\right)\\ \left[SO^{2-}_4\right]=\left[H_2SO_4\right]=0,1\left(M\right)\\ b.\left[Ba^{2+}\right]=\left[BaCl_2\right]=0,2\left(M\right)\\ \left[Cl^-\right]=2.0,2=0,4\left(M\right)\\ c.\left[Ca^{2+}\right]=\left[Ca\left(OH\right)_2\right]=0,1\left(M\right)\\ \left[OH^-\right]=0,1.2=0,2\left(M\right)\)
a) \(\left[Ca^{2+}\right]=0,015\left(M\right)\)
\(\left[OH^-\right]=0,015\cdot2=0,03\left(M\right)\)
b) \(\left[K^+\right]=0,2\left(M\right)\)
\(\left[MnO_4^-\right]=0,2\left(M\right)\)