Question Detail
Question
Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: Pb²⁺ (aq) + 2I⁻(aq) → PbI₂(s). Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many milliliters of 3.550 M HI(aq) must be added to a solution containing 0.700 mol of Pb²⁺ to completely precipitate the lead? [Pb=207; I=127]
Correct Answer
Option B is the correct answer.
Detailed Explanation
The reaction requires 2 moles of I⁻ for every 1 mole of Pb²⁺. Moles of Pb²⁺ = 0.700 mol. Moles of I⁻ needed = 2 × 0.700 = 1.40 mol. HI provides I⁻ in a 1:1 ratio, so moles of HI needed = 1.40 mol. Volume of HI solution = moles / concentration = 1.40 mol / 3.550 M = 0.394 L = 394 mL.
Hint
Use the 1:2 mole ratio from the balanced equation. Don't forget to convert liters to milliliters.