PTHH: \(HCl+KOH\rightarrow KCl+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{HCl}=0,4\cdot0,5=0,2\left(mol\right)\\n_{KOH}=\dfrac{56\cdot10\%}{56}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Axit còn dư
\(\Rightarrow n_{KCl}=0,1\left(mol\right)=n_{HCl\left(dư\right)}\) \(\Rightarrow C_{M_{HCl\left(dư\right)}}=C_{M_{KCl}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)