If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, how many moles of Pb²⁺ were originally in the solution A) 7.25 $\rarr$ 10^-3 moles of Pb²⁺ B) 7.35 $\rarr$ 10^-3 moles of Pb²⁺ C) 7.45 $\rarr$ 10^-3 moles of Pb²⁺ D) 7.55 $\rarr$ 10^-3 moles of Pb²⁺ E) None of the above

Question

Quizplus · Accepted Answer

First, we need to use the balanced chemical equation for the reaction between lead(II) nitrate and potassium iodide: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) From the equation, we can see that 1 mole of Pb(NO3)2 reacts with 2 moles of KI to form 1 mole of PbI2. We know that 73.5 mL of 0.200 M KI was required to precipitate all of the lead(II) ion, which means that 0.200 moles of KI was used (since Molarity = moles/volume). Since every 2 moles of KI is used to form 1 mole of PbI2, then 0.100 moles of PbI2 was formed. Finally, from the balanced equation, we see that 1 mole of PbI2 contains 1 mole of Pb²⁺. So, the original amount of Pb²⁺ in the solution is also 0.100 moles. Therefore, the answer is 7.35 x 10^−10 moles of Pb²⁺ (Choice B).

If 735 ML of 0 →\rarr→ 10-3 Moles of Pb2+ B) 7

If 735 ML of 0 $\rarr$ 10^-3 Moles of Pb²⁺
B) 7