a, PT: \(KOH+HCl\rightarrow KCl+H_2O\)
Ta có: \(n_{KOH}=0,2.2=0,4\left(mol\right)\)
Theo PT: \(n_{HCl}=n_{KOH}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
b, Theo PT: \(n_{KCl}=n_{KOH}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{KCl}}=\dfrac{0,4}{0,2+0,2}=1\left(M\right)\)