Question

A balanced net ionic equation for the reaction of Pb(NO₃)₂(aq) with NaI(aq) is:

A) Pb²⁺(aq) + 2 NO₃⁻(aq) + 2 Na⁺(aq) + 2 I⁻(aq) → Pb²⁺(aq) + 2I⁻(aq) + 2 Na⁺(aq) + 2 NO₃⁻(aq)

B) Pb(NO₃)₂(aq) + 2 NaI(aq) → PbI₂(s) = 2 NaNO₃(aq)

C) Pb²⁺(aq) + 2 I⁻(aq) → PbI₂(s)

D) Pb²⁺(aq) + 2 NO₃⁻(aq) + 2 Na⁺(aq) + 2 I⁻(aq) → PbI₂(s) + 2 Na⁺(aq) + 2 NO₃⁻(aq)

Answer 1

The balanced net ionic equation for the reaction between Pb(NO₃)₂(aq) and NaI(aq) is Pb²⁺(aq) + 2 I⁻(aq) → PbI₂(s), which shows the formation of a precipitate from soluble salts.

Step-by-step explanation:

The balanced net ionic equation for the reaction of Pb(NO₃)₂(aq) with NaI(aq) involves removing the spectator ions to focus on the ions that participate in the formation of the precipitate. Spectator ions are ions that do not change oxidation state or composition during the course of a reaction. The balanced net ionic equation is:

Pb²⁺(aq) + 2 I⁻(aq) → PbI₂(s)

This reaction indicates that lead(II) cations (²⁺) react with iodide anions (⁻) to form solid lead(II) iodide (PbI₂). This process is a demonstration of a precipitation reaction, where an insoluble solid (precipitate) is formed from the mixing of two solutions containing soluble salts.