Question

As per the following equation, 0.217 g of HgO (molecular mass = 217 g 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?

• A. 50 mL
• B. 200 mL
• C. 10 mL
• D. 5 mL

Hint:

Using a simple frame or just bolding for the box
Key Points:
Determine moles of reactant (HgO).
Use stoichiometry of the first reaction to find moles of product relevant to titration (OH$^-$).
Use stoichiometry of the titration reaction (acid-base neutralization, usually 1:1 for strong acid/strong base derived ions).
Calculate the required volume using Molarity = Moles / Volume.
Ensure units are consistent (e.g., moles, L, mL).

Answer 1

The Correct Option is B

The problem involves two steps: first, the reaction of HgO with iodide to produce hydroxide ions (OH^-), and second, the titration of these hydroxide ions with HCl. (A) Calculate moles of HgO:

Moles = Mass / Molar Mass
Moles of HgO = 0.217 g / 217 g mol^-1 = 0.001 mol. (B) Calculate moles of OH^- produced:

From the reaction equation: HgO + 4I^- + H₂O → HgI₄^2- + 2OH^-
1 mole of HgO produces 2 moles of OH^-.
Therefore, 0.001 mole of HgO produces 0.001 × 2 = 0.002 moles of OH^-. (C) Titration Reaction:

The hydroxide ions (base) produced are titrated with HCl (acid):
OH^- + HCl → H₂O + Cl^-
The stoichiometry is 1:1, meaning moles of HCl required equals moles of OH^- at the equivalence point. (D) Calculate moles of HCl required:

Moles of HCl = Moles of OH^- = 0.002 mol. (E) Calculate volume of HCl solution:

Volume = Moles / Molarity
Volume of HCl = 0.002 mol / 0.01 M = 0.002 mol / 0.01 mol L^-1 = 0.2 L. (F) Convert volume to mL:

Volume in mL = 0.2 L × 1000 mL/L = 200 mL. The volume of 0.01 M HCl required is 200 mL. This corresponds to option (B).