When \(1.0\) mole of solid lead nitrate is added to \(2.0\) moles of aqueous potassium iodide, a yellow precipitate forms. After the precipitate settles to the bottom, does the solution above the precipitate conduct electricity? Explain. Write the complete ionic equation to help you answer this question.

First, we should write the balanced chemical equation for the reaction between lead nitrate and potassium iodide. To do this, we need to identify the products formed in the reaction. Lead nitrate (Pb(NO₃)₂) and potassium iodide (KI) are both ionic compounds. When ionic compounds react, they undergo a double displacement reaction, exchanging ions to form new compounds. The products of the reaction will be lead iodide (PbI₂) and potassium nitrate (KNO₃). The balanced chemical equation for the reaction is: Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

Now that we have the balanced chemical equation, we can write the complete ionic equation. In this process, we separate the aqueous compounds into their individual ions. Note that solid lead iodide (PbI₂) is a precipitate and thus does not dissociate into ions. The complete ionic equation is: Pb²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)

We can now analyze the complete ionic equation to determine if the solution above the precipitate will conduct electricity. For a solution to conduct electricity, it needs to contain charge-carrying particles (ions). From the complete ionic equation, we can see that after the reaction, there are still potassium ions (K⁺) and nitrate ions (NO₃⁻) present in the solution. As they are not part of the precipitate (PbI₂), they remain dissolved in the solution. Since there are still ions present, the solution above the precipitate will conduct electricity. In conclusion, the solution above the precipitate formed by the reaction between lead nitrate and potassium iodide conducts electricity, as there are still dissolved ions (K⁺ and NO₃⁻) in the solution.