a) $2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
b) $n_{H_2SO_4} = 0,05.1 = 0,05(mol)$
Theo PTHH : $n_{NaOH} = 2n_{H_2SO_4} = 0,1(mol)$
$m_{NaOH} = 0,1.40 = 4(gam)$
c) $m_{dd\ NaOH} = \dfrac{4}{30\%} = \dfrac{40}{3}(gam)$
$V_{dd\ NaOH} = \dfrac{ \dfrac{40}{3}}{2,13} = 6,3(ml)$
$V_{dd\ sau\ pư} = 50 + 6,3 = 56,3(ml) = 0,0563(lít)$
$C_{M_{Na_2SO_4}} = \dfrac{0,05}{0,0563} =0,89M$