Bài 1:
Ta có: \(n_{OH^-}=n_{Na^+}=n_{NaOH}=0,2.0,4=0,08\left(mol\right)\)
\(n_{H^+}=n_{Cl^-}=n_{HCl}=0,4.0,3=0,12\left(mol\right)\)
PT ion: \(OH^-+H^+\rightarrow H_2O\)
_____0,08_____0,12 (mol)
⇒ nOH- (dư) = 0,04 (mol)
\(\Rightarrow\left\{{}\begin{matrix}\left[Na^+\right]=\frac{0,08}{0,6}\approx0,133M\\\left[Cl^-\right]=\frac{0,12}{0,6}=0,2M\\\left[OH^-\right]=\frac{0,04}{0,6}\approx0,066M\end{matrix}\right.\)
Câu 2:
Ta có: \(\Sigma n_{K^+}=n_{KCl}+2n_{K_2SO_4}=0,2.1,5+0,3.2.2=1,5\left(mol\right)\)
\(n_{Cl^-}=n_{KCl}=0,2.1,5=0,3\left(mol\right)\)
\(n_{SO_4^{2-}}=0,3.2=0,6\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\left[K^+\right]=\frac{1,5}{0,5}=3M\\\left[Cl^-\right]=\frac{0,3}{0,5}=0,6M\\\left[SO_4^{2-}\right]=\frac{0,6}{0,5}=1,2M\end{matrix}\right.\)
Bạn tham khảo nhé!