Question

When \(1.0 \mathrm{~mol}\) of solid lead nitrate is added to \(2.0 \mathrm{~mol}\) 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 because it contains free-moving ions, specifically potassium ions (K⁺), nitrate ions (NO₃⁻), and any unreacted aqueous iodide ions (I⁻), in the aqueous environment. The complete ionic equation for this reaction is: \(Pb^{2+}(aq) + 2NO_3^-(aq) + 2K^+(aq) + 2I^-(aq) \rightarrow PbI_2(s) + 2K^+(aq) + 2NO_3^-(aq)\).

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 represented as: \[Pb(NO_3)_2 + 2KI \rightarrow PbI_2 + 2KNO_3\]

Identify the soluble and insoluble ionic compounds and their constituent ions.

According to the solubility rules, nitrates (NO₃⁻) and group 1 ions (such as potassium, K⁺) are always soluble in water. However, lead iodide (PbI₂) is an insoluble compound, and it forms a yellow precipitate in this reaction. Thus, the soluble ionic compounds are lead nitrate (Pb(NO₃)₂) and potassium iodide (KI), whose ions in solution are: - Pb²⁺ (from lead nitrate) - 2NO₃⁻ (from lead nitrate) - 2K⁺ (from potassium iodide) - 2I⁻ (from potassium iodide)

Write the complete ionic equation.

To write the complete ionic equation, we represent all soluble ionic compounds as dissociated ions: \[Pb^{2+}(aq) + 2NO_3^-(aq) + 2K^+(aq) + 2I^-(aq) \rightarrow PbI_2(s) + 2K^+(aq) + 2NO_3^-(aq)\]

Discuss why the solution conducts electricity or not.

The solution above the precipitate consists of potassium ions (K⁺), nitrate ions (NO₃⁻), and any unreacted aqueous iodide ions (I⁻). These ions are free in the aqueous environment, and their movement allows the conduction of electricity. Therefore, the solution above the precipitate does conduct electricity because it contains free-moving ions in the aqueous environment.

Key concepts

Complete Ionic Equations

Understanding complete ionic equations is crucial for students wanting to delve into the intricacies of precipitation reactions. In simple terms, a complete ionic equation breaks down aqueous ionic compounds into their constituent ions, showing exactly what species are present in the reaction mixture. For example, when lead nitrate interacts with potassium iodide in an aqueous solution, the complete ionic equation provides a transparent view of the reaction:

[\[Pb^{2+}(aq) + 2NO_3^-(aq) + 2K^+(aq) + 2I^-(aq) \rightarrow PbI_2(s) + 2K^+(aq) + 2NO_3^-(aq)\]]
The ionic equation splits the soluble compounds into ions and leaves the insoluble ones whole. This helps illuminate the process, setting the stage for yellow lead iodide precipitate to form, while the potassium and nitrate ions remain dissolved. These kinds of equations are invaluable for predicting reaction outcomes and are key building blocks for a solid chemistry foundation.

Solubility Rules

To predict whether a reaction in solution will produce a precipitate, students need to become familiar with solubility rules. These are guidelines that help determine the solubility of various compounds in water. A solid grasp on these rules is essential for identifying the behavior of ions in a reaction. For instance, the solubility rules state that nitrates and alkali metal ions (like potassium) are always soluble, while some other compounds, such as lead iodide, are not.

• Nitrates (NO_3^-) are always soluble
• Alkali metal ions (such as K^+) are always soluble
• Compounds like PbI_2 are insoluble and form precipitates

When lead nitrate reacts with potassium iodide, the solubility rules predict that lead iodide (PbI_2) will precipitate out. Recognizing these patterns empowers students to predict and explain outcomes of reactions confidently.

Electrical Conductivity in Solutions

Electrical conductivity in solutions is a fascinating property that arises from the presence of ions that are free to move. In an aqueous solution, if ions are present, they will carry electric current, making the solution conductive. This forms the basis for understanding why the mixture above the yellow lead iodide precipitate can still conduct electricity.

Even after some ions come together to form the precipitate, other ions such as potassium and nitrate remain dissolved in the solution. These free-moving ions (2K^+(aq) and 2NO_3^-(aq)) are what enable the solution to conduct electricity. This concept is not only significant in chemistry but also in various industrial applications where electrolytes and conductive solutions play critical roles.

Chemical Reaction Equations

Lastly, understanding chemical reaction equations is an integral part of mastering chemistry. A well-balanced chemical equation provides a snapshot of a reaction, showing the reactants transforming into products. It's the chemist's equation for storytelling. Reflecting on the example given:

[\[Pb(NO_3)_2 + 2KI \rightarrow PbI_2 + 2KNO_3\]]
The equation tells us that one molecule of lead nitrate reacts with two molecules of potassium iodide to create one molecule of lead iodide and two molecules of potassium nitrate. The equation balances the atoms on both sides, adhering to the law of conservation of mass, and offers a clear, concise summary of the transformation occurring during the reaction. Whether a student is identifying reactants or predicting products, crafting and interpreting these equations is an invaluable skill in their chemical toolkit.