Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: (aq) + (aq) → (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 to completely precipitate the lead?

Question

Quizplus · Accepted Answer

The balanced chemical equation is:

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

From the equation, it follows that 1 mole of lead nitrate reacts with 2 moles of hydroiodic acid to produce 1 mole of lead iodide. Therefore, in order to completely precipitate 0.700 mol of lead nitrate, we need 1.400 mol of hydroiodic acid.

We can use the formula:

n = C x V

where n is the number of moles, C is the concentration in mol/L, and V is the volume in L.

Rearranging the formula, we get:

V = n / C

Substituting the values, we get:

V = 1.400 mol / 3.550 mol/L = 0.394 L = 394 mL

Therefore, we need 394 mL of 3.550 mol/L hydroiodic acid to completely precipitate 0.700 mol of lead nitrate. The correct answer is option C.

Lead Ions Can Be Precipitated from Aqueous Solutions by the Addition