Q20 / 225 · JEE Main 2004 · Chemistry
To neutralize completely $20 \mathrm{~mL}$ of $0.1 \mathrm{M}$ aqueous solution of phosphorous acid $\left(\mathrm{H}_3 \mathrm{PO}_3\right)$, the volume of $0.1 \mathrm{M}$ aqueous $\mathrm{KOH}$ solution required is
- A$10 \mathrm{~mL}$
- B$60 \mathrm{~mL}$
- Correct answerC$40 \mathrm{~mL}$
- D$20 \mathrm{~mL}$
Solution
No. of gm equivalents of phosphorous acid $=\mathrm{No}$. of gm equivalents of $\mathrm{KOH}$
$$
\begin{aligned}
& 20 \times 0.1 \times 2(n=\text { factor })=0.1 \times V \\
& =0.1 \times V
\end{aligned}
$$
$\mathrm{V}=\frac{4}{0.1}=40 \mathrm{ml}$
Hence, the correct answer is option: 'C'.