Question

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.

Short answer

Yes, the solution above the precipitate will conduct electricity because mobile ions (K⁺(aq) and NO₃⁻(aq)) are still present in the solution after the formation and settling of the PbI₂(s) precipitate. The complete ionic equation for this reaction is: Pb(NO₃)₂(s) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq).

Step-by-step solution

Write the balanced molecular equation

First, we need to write the balanced molecular equation for the reaction between lead nitrate, Pb(NO₃)₂ and potassium iodide, KI: Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃ The balanced molecular equation is: Pb(NO₃)₂(s) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

Write the complete ionic equation

In a complete ionic equation, we split all of the aqueous compounds into their respective ions: Pb(NO₃)₂(s) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)

Write the net ionic equation

Now, let's identify and cancel the spectator ions. In this case, the spectator ions are K⁺(aq) and NO₃⁻(aq) because they appear on both sides of the equation without undergoing any change: Net ionic equation:\( \) Pb(NO₃)₂(s) + 2I⁻(aq) → PbI₂(s) + 2NO₃⁻(aq)

Analyze the conductivity of the solution

To conduct electricity, a solution must have mobile ions. In this case, after the formation and settling of the precipitate (PbI₂) at the bottom of the solution, the remaining species in the solution are K⁺(aq) and NO₃⁻(aq) ions. As these ions are still mobile and present in the solution, they are capable of conducting electricity. Therefore, the solution above the precipitate will conduct electricity, as there are still mobile ions present in the form of K⁺(aq) and NO₃⁻(aq).