The exact volumes of 1 M $\mathrm{NaOH}$ solution required to neutralize 50 mL of 1 M ${\mathrm{H}}_3{\mathrm{PO}}_3$ solution and 100 mL of 2 M ${\mathrm{H}}_3{\mathrm{PO}}_2$ solution, respectively, are:

The correct option is D

100 mL and 200 mL

Explanation of the correct option:

The correct answer is option D:

The reaction of $\mathrm{NaOH}$ with ${\mathrm{H}}_3{\mathrm{PO}}_3$ is,

$$\begin{gathered} 2\mathrm{NaOH}+{\mathrm{H}}_3{\mathrm{PO}}_3\to {\mathrm{Na}}_2{\mathrm{HPO}}_3+2{\mathrm{H}}_2\mathrm{O} \\ 100\mathrm{mmole}50\mathrm{mmole} \\ 100\mathrm{mmole}=M\times V_{\mathrm{ml}} \\ 100\mathrm{mmole}=1\times V_{\mathrm{ml}} \\ \mathit{\Rightarrow }{\mathrm{V}}_{\mathrm{ml}}=100\mathrm{ml} \end{gathered}$$

The reaction of $\mathrm{NaOH}$ with ${\mathrm{H}}_3{\mathrm{PO}}_2$ is,

$$\begin{gathered} 2\mathrm{NaOH}+{\mathrm{H}}_3{\mathrm{PO}}_2\to {\mathrm{Na}}_2{\mathrm{H}}_2{\mathrm{PO}}_2+{\mathrm{H}}_2\mathrm{O} \\ 200\mathrm{mmole}200\mathrm{mmole} \\ 200\mathrm{mmole}=M\times V_{\mathrm{ml}} \\ 200\mathrm{mmole}=1\times V_{\mathrm{ml}} \\ \mathit{\Rightarrow }{\mathrm{V}}_{\mathrm{ml}}=200\mathrm{ml} \end{gathered}$$

Hence, the exact volumes of 1 M $\mathrm{NaOH}$ solution required to neutralize 50 mL of 1 M ${\mathrm{H}}_3{\mathrm{PO}}_3$ solution and 100 mL of 2 M ${\mathrm{H}}_3{\mathrm{PO}}_2$ solution, respectively, are 100 mL and 200 mL.