The exact volumes of 1 M Noah solution required to neutralize 50 mL

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Question:

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 :

  1. $100 \mathrm{~mL}$ and $100 \mathrm{~mL}$

  2. $100 \mathrm{~mL}$ and $50 \mathrm{~mL}$

  3. $100 \mathrm{~mL}$ and $200 \mathrm{~mL}$

  4. $50 \mathrm{~mL}$ and $50 \mathrm{~mL}$


Correct Option: 3,

Solution:

$\mathrm{H}_{3} \mathrm{PO}_{3}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{HPO}_{3}+2 \mathrm{H}_{2} \mathrm{O}$

$\begin{array}{ll}50 \mathrm{ml} & 1 \mathrm{M} \\ 1 \mathrm{M} & \mathrm{V}=?\end{array}$

$\Rightarrow \frac{\mathrm{n}_{\mathrm{NaoH}}}{\mathrm{n}_{\mathrm{H}_{3} \mathrm{PO}_{3}}}=\frac{2}{1}$

$\Rightarrow \frac{1 \times \mathrm{V}}{50 \times 1}=\frac{2}{1} \Rightarrow \mathrm{V}_{\mathrm{NaOH}}=100 \mathrm{ml}$

$\mathrm{H}_{3} \mathrm{PO}_{2}+2 \mathrm{NaOH} \rightarrow \mathrm{NaH}_{2} \mathrm{PO}_{3}+\mathrm{H}_{2} \mathrm{O}$

$\begin{array}{lc}100 \mathrm{ml} & 1 \mathrm{M} \\ 2 \mathrm{M} & \mathrm{V}=?\end{array}$

$\Rightarrow \frac{\mathrm{n}_{\mathrm{NaOH}}}{\mathrm{n}_{\mathrm{H}_{3} \mathrm{PO}_{3}}}=\frac{1}{1} \Rightarrow \frac{1 \times \mathrm{V}}{2 \times 100}=\frac{1}{1} \Rightarrow \mathrm{V}_{\mathrm{NaOH}}=200 \mathrm{ml}$

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