A) 0.107 M
B) 0.126 M
C) 0.214 M
D) -0.428 M
Correct Answer: A
Solution :
\[0.164\ M\ NaOH\ \cong \ 0.164\ N\ NaOH\] We know, \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}};\] \[{{N}_{1}}\times 25=0.164\times 32.63\] \[0.214\ N\ {{H}_{2}}S{{O}_{4}}\cong \frac{0.214}{2}M\ {{H}_{2}}S{{O}_{4}}\] (\[\because \] basicity of \[{{H}_{2}}S{{O}_{4}}\ is\ 2\]) \[\cong \ 0.107\ M\ {{H}_{2}}S{{O}_{4}}\]
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