\(a.n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\\ \left[Na^+\right]=\left[OH^-\right]=\left[NaOH\right]=\dfrac{0,5}{0,5}=1\left(M\right)\\ b.HCl+NaOH\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaOH}=0,5\left(mol\right)\\ V_{ddHCl}=\dfrac{0,5}{2}=0,25\left(l\right)\)