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 conducts electricity. This is because there are free ions, K⁺ and NO₃⁻, present in the solution. The complete ionic equation for the reaction is: Pb(NO₃)₂ (s) + 2 K⁺ (aq) + 2 I⁻ (aq) → PbI₂ (s) + 2 K⁺ (aq) + 2 NO₃⁻ (aq).

Step-by-step solution

Write the balanced chemical equation for the reaction

The reaction between lead nitrate (Pb(NO₃)₂) and potassium iodide (KI) can be written as: Pb(NO₃)₂ (s) + 2 KI (aq) → PbI₂ (s) + 2 KNO₃ (aq) The yellow precipitate formed is lead iodide (PbI₂), which settles at the bottom. The balanced equation also informs us that potassium nitrate (KNO₃) is the other product formed, which remains in the aqueous phase.

Write the complete ionic equation

We'll now write the complete ionic equation to identify the ions in the solution. We can represent the ionic species as: Pb(NO₃)₂ (s) + 2 K⁺ (aq) + 2 I⁻ (aq) → PbI₂ (s) + 2 K⁺ (aq) + 2 NO₃⁻ (aq) Notice that the equation now includes the ions present in the solution: K⁺ and NO₃⁻.

Determine if the solution conducts electricity

Since there are free ions, K⁺ and NO₃⁻, in the solution above the precipitate, these ions will help in conducting electricity. Therefore, the solution above the precipitate does conduct electricity.

Key concepts

Chemical Equations

Understanding chemical equations is fundamental to grasping how different substances interact during a chemical reaction. A chemical equation is a symbolic representation where the reactants are shown on the left side, and the products are displayed on the right. They must be balanced, meaning the number of atoms for each element should be the same on both sides of the equation to obey the Law of Conservation of Mass.

In our exercise, the balanced chemical equation for the reaction between solid lead nitrate and aqueous potassium iodide is represented as:$${\text{Pb(NO}}_3{{)}_2(s)+2KI(aq)\to PbI}_2{(s)+2KNO}_3(aq)$$The solid that forms, lead iodide (PbI₂), is the yellow precipitate, which indicates a chemical change. Remaining in solution, potassium nitrate (KNO₃) is the aqueous product. This equation provides a clear picture of the reactants and products, aiding in the visualization of the chemical process.

Ionic Equations

When dissolving ionic compounds in water, they disassociate into their respective ions. This process is aptly depicted using ionic equations, which are more detailed than chemical equations since they display individual ions involved in a reaction.

In the exercise, the ionic equation is used to show the details of the reaction in an aqueous solution:$${\text{Pb(NO}}_3{{)}_2(s)+2K}^{+}{\text{(aq) + 2 I}}^{-}{\text{(aq) rightarrow PbI}}_2{\text{ (s) + 2 K}}^{+}{\text{(aq) + 2 NO}}_3^{-}\text{(aq)}$$This equation highlights the species that change state or separate into ions. Lead nitrate remains intact as it is a solid, while potassium iodide separates into potassium ions (K⁺) and iodide ions (I^-). The formation of the solid, lead iodide, removes the iodide ions from the solution, while potassium and nitrate ions remain dissolved, which is significant when discussing the electrical conductivity of the mixture.

Electrical Conductivity in Solutions

The ability of a solution to conduct electricity is primarily dependent on the presence of charged particles, namely ions. Electrical conductivity in solutions is a measure of how well these ions can move and thus carry an electric current. In a solution where ionic compounds are dissolved, free-moving ions facilitate the flow of electricity.

The exercise posits a situation where a yellow precipitate of lead iodide forms, leaving us to ponder whether the remaining solution can conduct electricity. Based on the ionic equation provided, we see that potassium ions (K⁺) and nitrate ions (NO₃^-) are left in the solution. Because these ions are free and mobile in the aqueous solution, they are able to conduct electricity. Thus, we can infer that even after the formation of the precipitate, the solution above remains electrically conductive due to the presence of these dissolved ions.