How many milliliters of 1.500 M aqueous KI solution must be added to an aqueous solution containing 0.100 mol of Pb(NO₃)₂ to completely precipitate the lead? A)7.50 × 10^-3 B)66.7 C)0.0667 D)0.133 E)133

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The reaction between KI and Pb(NO3)2 is a precipitation reaction where PbI2 is formed. The balanced chemical equation for this reaction is:$$ \text{Pb(NO}_3\text{)}_2 + 2\text{KI} \rightarrow \text{PbI}_2 + 2\text{KNO}_3 $$From the equation, 1 mole of Pb(NO3)2 reacts with 2 moles of KI to form 1 mole of PbI2. Given that there are 0.100 moles of Pb(NO3)2, it will require 0.200 moles of KI to completely react with all of the Pb(NO3)2.The molarity (M) equation is:$$ M = \frac{\text{moles of solute}}{\text{liters of solution}} $$Given a 1.500 M KI solution, we can rearrange the equation to solve for the volume in liters:$$ \text{Volume (L)} = \frac{\text{moles of KI}}{\text{Molarity of KI solution}} = \frac{0.200}{1.500} = 0.133\ \text{L} $$To convert liters to milliliters (since 1 L = 1000 mL):$$ 0.133\ \text{L} \times 1000\ \text{mL/L} = 133\ \text{mL} $$Therefore, 133 mL of the 1.500 M KI solution is needed to completely precipitate the lead.

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