Question

Potassium iodide reacts with lead(II) nitrate in the following precipitation reaction:

2KI(aq)+Pb(NO₃)₂(aq)→2KNO₃(aq)+PbI₂(s)

What minimum volume of 0.200 M potassium iodide solution is required to completely precipitate all of the lead in 155.0 mL of a 0.112 M lead(II) nitrate solution?

Answer 1

To precipitate all the lead from a 0.112 M Pb(NO3)2 solution, 173.6 mL of 0.200 M KI solution is required, based on stoichiometry of the reaction.

Step-by-step explanation:

To determine the minimum volume of potassium iodide solution needed to precipitate all the lead in a given solution, we utilize the stoichiometry of the precipitation reaction:

2KI(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbI2(s)

According to the stoichiometry of the reaction, 2 moles of potassium iodide react with 1 mole of lead(II) nitrate to produce 1 mole of lead(II) iodide precipitate. Given that we have 155.0 mL of a 0.112 M lead(II) nitrate solution, we first need to calculate the moles of lead(II) nitrate:

Moles of Pb(NO3)2 = Molarity × Volume (in liters)

Moles of Pb(NO3)2 = 0.112 mol/L × 0.1550 L = 0.01736 moles

Since it takes 2 moles of KI to react with 1 mole of Pb(NO3)2, we need:

Moles of KI needed = 2 × 0.01736 moles = 0.03472 moles of KI

Now we can find the volume of 0.200 M potassium iodide solution required to provide 0.03472 moles:

Volume = Moles / Molarity

Volume of KI = 0.03472 moles /​ 0.200 mol/L = 0.1736 liters, which is 173.6 mL of KI solution.