$n_{NaOH}=0,25.0,2=0,05(mol)$
Đặt $x$ là số mol $Br_2$
$SO_2+Br_2+2H_2O\to 2HBr+H_2SO_4$
$\Rightarrow n_{HBr}=2x (mol); n_{H_2SO_4}=x (mol)$
$HBr+NaOH\to NaBr+H_2O$
$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$
$\Rightarrow 2x+2.x=0,05$
$\Leftrightarrow x=0,0125$
$\to C_{M_{Br_2}}=\dfrac{0,0125}{0,5}=0,025M$