\(n_{NaOH}=\frac{m}{M}=\frac{20}{40}=0,5\left(mol\right)\)
PTHH : \(NaOH+HCl\rightarrow NaCl+H_2O\)
...............0,5.............0,5........................................
Ta có : \(C_{MHCl}=\frac{n}{V}=0,5=\frac{0,5}{V_{HCl}}\)
=> \(V_{HCl}=1\left(l\right)\)
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