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CHM 101 CHM 101 Past Questions Objective Question
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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]
Options
A 2.54 × 10⁻³
B 394
Correct Answer
C 197
D 0.197
Correct Answer

Option B is the correct answer.

Detailed Explanation

Each Pb²⁺ requires 2 I⁻. Moles I⁻ needed = 0.700 × 2 = 1.40 mol. HI provides I⁻ in a 1:1 ratio. Volume of HI = moles/concentration = 1.40 mol / 3.550 M = 0.3943 L = 394.3 mL.

Hint

Determine the mole ratio from the balanced equation, then use M₁V₁ = moles.

Question Info

Course: CHM 101
Topic: CHM 101 Past Questions
Type: Objective
Difficulty: Not specified
Question 1111 of 1872
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