$n_{NaOH}=0,5.2=1\ (mol)$
$n_{H_2SO_4}=0,2.0,1=0,02\ (mol)$
PTHH: $H_2SO_4+2NaOH\to Na_2SO_4+2H_2O$
Ta có: $n_{H_2SO_4}=0,02<\dfrac{n_{NaOH}}{2}=0,5\to H_2SO_4$ hết
Theo PT: $n_{Na_2SO_4}=n_{H_2SO_4}=0,02\ (mol)$
$n_{NaOH\ pứ}=2n_{H_2SO_4}=0,04\ (mol)$
$\to n_{NaOH\ dư}=1-0,04=0,96\ (mol)$
$V_{dd\ spứ}=0,5+0,2=0,7\ (l)$
$C_{M_{ddNa_2SO_4}}=\dfrac{0,02}{0,7}=\dfrac1{35}\ (M)$
$C_{M_{ddNaOH\ dư}}=\dfrac{0,96}{0,7}=\dfrac{48}{35}\ (M)$