MCQ
$1\,\,M$ of $10 \,ml$ ${H_2}S{O_4}$ will completely neutralise
- A$10\,\,ml$ of $1\,\,M\,\,NaOH$ solution
- ✓$10\,\,ml$ 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
✓
Answer
Correct option: B.
$10\,\,ml$ of $2\,\,M\,\,NaOH$ solution
(b) ${H_2}S{O_4} + 2{H_2}O$ $ \rightleftharpoons $ $2{H_3}{O^ + } + SO_4^{ - \, - }$
$NaOH$ $ \rightleftharpoons $ $N{a^ + } + O{H^ - }$
$1$ 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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