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. If you have 50.0 mL of a 0.100 M Pb(NO3)2 solution, how many milliliters of 3.550 M HI(aq) are needed to completely precipitate the lead ions as PbI2?

• A. 11.28 mL
• B. 1.41 mL
• C. 2.82 mL
• D. 5.64 mL

Answer: (C)

Verified step by step guidance

1. Start by identifying the balanced chemical equation for the reaction: Pb^{2+}(aq) + 2I^{-}(aq) → PbI_{2}(s). This indicates that one mole of Pb^{2+} reacts with two moles of I^{-}.
2. Calculate the moles of Pb^{2+} present in the solution. Use the formula: \( \text{moles of Pb}^{2+} = \text{volume (L)} \times \text{molarity (M)} \). Convert 50.0 mL to liters by dividing by 1000.
3. Determine the moles of I^{-} required to react with the moles of Pb^{2+}. Since the stoichiometry of the reaction is 1:2, multiply the moles of Pb^{2+} by 2 to find the moles of I^{-} needed.
4. Calculate the volume of HI solution needed to provide the required moles of I^{-}. Use the formula: \( \text{volume (L)} = \frac{\text{moles of I}^{-}}{\text{molarity of HI (M)}} \).
5. Convert the volume from liters to milliliters by multiplying by 1000 to find the final volume of HI solution required.