Q33 / 155 · WBJEE 2025 · Chemistry
As per the following equation, 0.217 g of HgO (molecular mass $=217 \mathrm{~g} \mathrm{~mol}^{-1}$ ) reacts with excess iodide. On titration of the resulting solution, how many mL of 0.01 M HCl is required to reach the equivalence point?
$\mathrm{HgO}+4 \mathrm{I}^{-}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HgI}_4^{2-}+2 \mathrm{OH}^{-}$
- A50 mL
- Correct answerB200 mL
- C10 mL
- D5 mL
Solution
$\mathrm{HgO}+4 \mathrm{I}^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{HgI}_4^{-2}+2 \mathrm{OH}^{-}$
(limiting (excess
reagent) reagent)
$\begin{aligned}
& 2 \mathrm{H}^{+}(\mathrm{HC} \ell)+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\ell) \\
& \therefore \mathrm{n}_{\mathrm{HgO}}=\frac{1}{2} \mathrm{n}_{\mathrm{H}^{+}}=\frac{1}{2} \mathrm{n}_{\mathrm{HC} \ell} \\
& \frac{0.217}{217}=\frac{1}{2} \times \mathrm{M} \times \frac{\mathrm{V}(\mathrm{~mL})}{1000}=\frac{1}{2} \times 0.01 \times \frac{\mathrm{V}(\mathrm{~mL})}{1000} \\
& \therefore \mathrm{~V}(\mathrm{~mL})=200
\end{aligned}$
Hence, the correct answer is option: 'B'.