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 Pb(NO) ₃) ₂ (aq) to completely precipitate the lead?

Question

Quizplus · Accepted Answer

First, we need to determine the limiting reagent in the reaction between Pb(NO3)2 and HI. From the balanced equation, we see that one mole of Pb(NO3)2 reacts with two moles of HI to produce one mole of PbI2:

Pb(NO3)2(aq) + 2HI(aq) → PbI2(s) + 2HNO3(aq)

So, the number of moles of HI required to react completely with 0.700 mol of Pb(NO3)2 is:

0.700 mol Pb(NO3)2 × 2 mol HI/1 mol Pb(NO3)2 = 1.40 mol HI

Next, we can use the concentration and number of moles of HI to determine the volume required:

1.40 mol HI × (1 L/3.550 mol) × (1000 mL/1 L) = 394 mL

So, the answer is B) 394 mL of 3.550 M HI(aq) must be added to completely precipitate the lead.

Lead Ions Can Be Precipitated from Aqueous Solutions by the Addition