Select True or False: The following net ionic equation correctly shows the result of this experiment: A piece of lead metal was added to an aqueous solution of copper (II) nitrate, and within a few minutes it was observed that the lead turned black and crumbled. The characteristic blue solution of copper (II) ions had faded. (NOTE: Lead forms a 2 + ion when it reacts.) Pb (s) + Cu2 + (aq) → Pb2 + (aq) + Cu (s)

True, Pb (s) + Cu²⁺ (aq) → Pb²⁺ + Cu (s) is the net ionic equation.

Explanation:

A net ionic equation shows all the atoms whose oxidation state change, which is referred to as the species that participate of the reaction.

To write the net ionic equation you follow some basic steps:

Write the molecular chemical equation. Identify the phases of the substances: which remain in solution, which are gases, and which are solids.

Use "aq" for aqueous solutions, "l" for liquids, "g" for gases, and "s"

for solids.

Use the solubility rules or data of solubility of the compounds to

identify which compounds will precipitate (form solid product).

Separate the species in solution into its ions.

Write the total ionic equation

Eliminate the spectator ions (those whose oxidation state to not change)

1) Molecular equation:

Pb (s) + Cu (NO₃) ₂ (aq) → Pb (NO₃) ₂ (aq) + Cu (s)

You know that Cu (NO₃) ₂ is in solution because it is said in the stament that the solution had the characteristic blue color of copper (II) ions.

In the properties of Pb (NO₃) ₂ yo can find that it is soluble, so it is also in solution.

2) Ionization of the compounds.

Cu (NO₃) ₂ (aq) → Cu²⁺ + 2 NO₃⁻

Pb (NO₃) ₂ (aq) → Pb²⁺ + 2NO₃⁻

3) Total ionic equation

Pb (s) + Cu²⁺ (aq) + 2 NO₃⁻ (aq) → Pb²⁺ + 2 NO₃⁻ (aq) + Cu (s)

4) Net ionic equation

The spectator ions are 2 NO₃⁻ (aq) (they did not change meaning that they did not participate in the reaction).

Pb (s) + Cu²⁺ (aq) → Pb²⁺ + Cu (s) ← answer