Hướng dẫn giải                       

Ta có:

$N{{H}_{4}}S{{O}_{4}}+NaOH\to N{{a}_{2}}S{{O}_{4}}+N{{H}_{3}}\uparrow +{{H}_{2}}O.$

 ${{n}_{KOH}}=V\times {{C}_{M}}=0,2\times 0,1=0,02\left( mol \right)$. 

PTHH:

$2KOH+{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+{{H}_{2}}O$                         $2NaOH+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}O$

$0,02\left( mol \right)\to 0,01\left( mol \right)$                           $0,03\left( mol \right)\to 0,015\left( mol \right)$

Ta có: ${{n}_{{{H}_{2}}S{{O}_{4}}}}=0,01+0,015=0,025\left( mol \right)$

Thể tích ${{H}_{2}}S{{O}_{4}}$ cần dùng là: $V=\frac{n}{{{C}_{M}}}=\frac{0,025}{0,5}=0,05\left( l \right)=5\left( ml \right)$. Chọn đáp án B.