The volume in $\mathrm{mL}$ of $0.1 \mathrm{M}$ solution of $\mathrm{NaOH}$ required to completely…

The volume in $\mathrm{mL}$ of $0.1 \mathrm{M}$ solution of $\mathrm{NaOH}$ required to completely neutralise $100 \mathrm{~mL}$ of $0.3 \mathrm{M}$ solution of $\mathrm{H}_3 \mathrm{PO}_3$ is
  1. 60
  2. 600
  3. 300
  4. 30

Solution

$\because$ Phosphorus acid $\left(\mathrm{H}_3 \mathrm{PO}_3\right)$ is a dibasic acid. $\begin{aligned} 0.3 \mathrm{M} \mathrm{H}_3 \mathrm{PO}_3 & =0.6 \mathrm{~N} \mathrm{H}_3 \mathrm{PO}_3 \\ N_1 V_1 & =N_2 V_2 \\ 0.1 \times V_1 & =0.6 \times 100 \\ V_1 & =600 \mathrm{~mL} \end{aligned} $

Asked in: AP EAMCET 2011

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