A) \[10\,\,ml\] of \[1\,\,M\,\,NaOH\] solution
B) \[10\,\,ml\] of \[2\,\,M\,\,NaOH\] solution
C) \[5\,\,ml\] of \[2\,\,M\,\,KOH\] solution
D) \[5\,\,ml\] of \[1\,\,M\,\,N{{a}_{2}}C{{O}_{3}}\] solution
Correct Answer: B
Solution :
\[{{H}_{2}}S{{O}_{4}}+2{{H}_{2}}O\] ⇌ \[2{{H}_{3}}{{O}^{+}}+SO_{4}^{-\,-}\] NaOH ⇌ \[N{{a}^{+}}+O{{H}^{-}}\] 1 mole of \[{{H}_{2}}S{{O}_{4}}\]acid gives 2 moles of \[{{H}_{3}}{{O}^{+}}\]ions. So 2 moles of \[O{{H}^{-}}\] are required for complete neutralization.
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