${{n}_{NaOH}}={{V}_{NaOH}}.{{C}_{{{M}_{NaOH}}}}=0,35.1=0,35\left( mol \right)$

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

0,35    $\to $  0,175                             (mol)

Theo phương trình ta có:

${{n}_{{{H}_{2}}S{{O}_{4}}}}=\frac{{{n}_{NaOH}}}{2}=\frac{0,35}{2}=0,175\left( mol \right)$

${{V}_{{{H}_{2}}S{{O}_{4}}}}=\frac{{{n}_{{{H}_{2}}S{{O}_{4}}}}}{{{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}}=\frac{0,175}{1}=0,175(l)$ =175 ml

$\Rightarrow $Chọn đáp án D