The exact volumes of $1 \mathrm{M} \mathrm{NaOH}$ solution required to neutralise $50 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{3}$ solution and $100 \mathrm{~mL}$ of $2 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{2}$ solution, respectively, are :
Correct Option: 3,
$\mathrm{H}_{3} \mathrm{PO}_{3}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{HPO}_{3}+2 \mathrm{H}_{2} \mathrm{O}$
$\begin{array}{ll}50 \mathrm{ml} & 1 \mathrm{M} \\ 1 \mathrm{M} & \mathrm{V}=?\end{array}$
$\Rightarrow \frac{\mathrm{n}_{\mathrm{NaoH}}}{\mathrm{n}_{\mathrm{H}_{3} \mathrm{PO}_{3}}}=\frac{2}{1}$
$\Rightarrow \frac{1 \times \mathrm{V}}{50 \times 1}=\frac{2}{1} \Rightarrow \mathrm{V}_{\mathrm{NaOH}}=100 \mathrm{ml}$
$\mathrm{H}_{3} \mathrm{PO}_{2}+2 \mathrm{NaOH} \rightarrow \mathrm{NaH}_{2} \mathrm{PO}_{3}+\mathrm{H}_{2} \mathrm{O}$
$\begin{array}{lc}100 \mathrm{ml} & 1 \mathrm{M} \\ 2 \mathrm{M} & \mathrm{V}=?\end{array}$
$\Rightarrow \frac{\mathrm{n}_{\mathrm{NaOH}}}{\mathrm{n}_{\mathrm{H}_{3} \mathrm{PO}_{3}}}=\frac{1}{1} \Rightarrow \frac{1 \times \mathrm{V}}{2 \times 100}=\frac{1}{1} \Rightarrow \mathrm{V}_{\mathrm{NaOH}}=200 \mathrm{ml}$
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