The exact volume of 1M NaOH solution required to neutralize 50 mL 1M $H_{3} {PO}_{3}$ solution and $100 m l 2 M H_{3} P O_{2}$ respectively are

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$100 m l, 200 m l$

$100 m l, 50 m l$

$50 m l, 50 m l$

$100 m l, 100 ml$

answer is C.

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Detailed Solution

$\begin{array}{l} M_{b} = 1 M M_{a} = 1 M H_{3} P O_{3} + 2 N a O H \to {Na}_{2} {HPO}_{3} + 2 H_{2} O \\ \begin{array}{ll} V_{b} = x & V_{a} = 50 mL \\ n_{b} = 2 & n_{a} = 1 \end{array} \frac{M_{a} V_{a}}{n_{a}} = \frac{M_{b} V_{b}}{n_{b}} \frac{1 \times 50}{1} = \frac{1 \times x}{2} \\ x = 100 ml \\ NaOH + H_{3} {PO}_{2} \to {NaH}_{2} {PO}_{2} + H_{2} O \\ M_{b} = 1 M M_{a} = 2 M \frac{μ_{a} V_{a}}{n_{a}} = \frac{μ_{b} V_{b}}{n_{b}} \frac{2 \times 100}{1} = \frac{1 \times x}{1} \\ V_{b} = x V_{a} = 100 \\ n_{b} = 1 n_{a} = 1 \\ x = V_{b} = 200 mL or use V 1 N 1 = V 2 N 2 \end{array}$