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, as there are still dissolved ions (K⁺ and NO₃⁻) in the solution after the reaction. The complete ionic equation is: Pb²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq).

Step-by-step solution

Write the balanced chemical equation

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)

Write the complete ionic equation

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)

Determine if the solution conducts electricity

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.

Key concepts

Ionic Compounds

Ionic compounds are formed when atoms transfer electrons to achieve a full outer shell, resulting in positively and negatively charged ions. These ions are held together by strong electrostatic forces known as ionic bonds. For instance, in lead nitrate

• Lead Nitrate (Pb(NO₃)₂): Composed of lead ions (Pb²⁺) and nitrate ions (NO₃⁻).
• Potassium Iodide (KI): Formed by the combination of potassium ions (K⁺) and iodide ions (I⁻).

When such compounds dissolve in water, they separate into their respective ions, which enables them to participate in various chemical reactions. In the exercise, lead nitrate and potassium iodide dissolve in water, separating into their constituent ions, setting the stage for a reaction.

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions combine to form an insoluble solid, or precipitate. This happens because the ions in the solution react with each other to form a compound that is not soluble in water. In the given exercise:

• Lead ions (Pb²⁺) and iodide ions (I⁻) react to form lead iodide (PbI₂), which is insoluble in water.
• The formation of this yellow precipitate of lead iodide is the evidence of the reaction.

The occurrence of a precipitate is a clear sign of a reaction taking place, as visible in the yellow solid settling at the bottom of the container. Understanding precipitation reactions can help in predicting the products of ionic reactions.

Electrical Conductivity

Electrical conductivity in solutions is determined by the presence of free ions. These ions act as charge carriers, allowing current to pass through the solution. In our exercise:

• The reaction produces a precipitate, but not all ions are removed from the solution.
• Potassium ions (K⁺) and nitrate ions (NO₃⁻) remain dissolved, meaning the solution still contains free-floating ions.

Because these ions persist in the water, the solution above the precipitate can conduct electricity. This conductivity is a crucial implication of dissolving ionic compounds, as it affects the passage of electrical currents in chemical processes.

Ionic Equations

Ionic equations break down the chemical equation into individual ions that participate in a reaction. This detailed representation helps in visualizing which ions are reacting and which remain unchanged in the solution. The process involves:

• Converting compounds to ions: For aqueous substances, separation into ions is necessary, such as turning Pb(NO₃)₂(aq) into Pb²⁺(aq) and NO₃⁻(aq) ions.
• Representing the precipitate: Solid products remain as compounds, like PbI₂(s), since they do not ionize.
• In our given equation: Pb²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2I⁻(aq) → PbI₂(s) + 2K⁺(aq) + 2NO₃⁻(aq)

This clarity allows us to identify remaining ions – potassium and nitrate ions – indicating continuous conductivity. Writing ionic equations is a powerful tool in understanding chemical processes at a molecular level.