Question

consider the following reaction:
2ki(aq) + pb(no₃)₂(aq) → pbi₂(s) + 2kno₃(aq)
which of the following net ionic equations correctly represents the precipitation of solid lead iodide from a solution of iodide ions and lead(ii) ions?
○ pb²⁺(s) + 2i⁻(s) → pbi₂(aq)
○ pb²⁺(aq) + 2i⁻(aq) → pbi₂(aq)
○ pb²⁺(s) + 2i⁻(aq) → pbi₂(aq)
○ pb²⁺(aq) + 2i⁻(aq) → pbi₂(s)

Explanation:

In a net - ionic equation for a precipitation reaction, we start with the complete ionic equation. For the reaction \(2KI(aq)+Pb(NO_3)_2(aq)\to PbI_2(s) + 2KNO_3(aq)\), the complete ionic equation is \(2K^{+}(aq)+2I^{-}(aq)+Pb^{2 +}(aq)+2NO_3^{-}(aq)\to PbI_2(s)+2K^{+}(aq)+2NO_3^{-}(aq)\). Spectator ions (\(K^{+}\) and \(NO_3^{-}\)) are removed. Lead (\(Pb^{2+}\)) and iodide (\(I^{-}\)) ions are in aqueous solution (\(aq\)) before reaction, and the product \(PbI_2\) is a solid (\(s\)).

Answer:

\(\text{Pb}^{2 +}(aq)+2\text{I}^{-}(aq)\to\text{PbI}_2(s)\) (the fourth option).