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 millilitres of 3.550 mol L^-1 HI(aq) must be added to a solution containing 0.700 mol of Pb(NO₃) ₂(aq) completely precipitate the lead?

Question

Quizplus · Accepted Answer

The balanced equation is: Pb²⁺(aq)+ 2I^-(aq)→ PbI₂(s)From the equation, 1 mol of Pb²⁺ requires 2 mol of I^-.Therefore, 0.700 mol of Pb²⁺ requires 2 × 0.700 = 1.40 mol of I^-.The volume of 3.550 mol L^-1 HI(aq) required to provide 1.40 mol of I^- is:Volume = moles of I^- / concentration of HI(aq)= 1.40 mol / 3.550 mol L^-1= 0.394 L= 394 mLTherefore, the correct answer is B) 394 mL.

Lead Ions Can Be Precipitated from Aqueous Solutions by the Addition