1. Write the balanced molecular equation:
2KCl(aq) + Pb(CH₃COO)₂(aq) → PbCl₂(s) + 2KCH₃COO(aq)
2. Identify the soluble and insoluble compounds:
* Soluble: KCl, Pb(CH₃COO)₂, and KCH₃COO (generally, salts containing alkali metals (Group 1) and acetate ions are soluble).
* Insoluble: PbCl₂ (generally, halides of lead are insoluble).
3. Write the complete ionic equation:
2K⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2CH₃COO⁻(aq) → PbCl₂(s) + 2K⁺(aq) + 2CH₃COO⁻(aq)
4. Cancel out the spectator ions:
The spectator ions are K⁺ and CH₃COO⁻ because they appear on both sides of the equation without undergoing any change.
5. The net ionic equation is:
Pb²⁺(aq) + 2Cl⁻(aq) → PbCl₂(s)
Therefore, the net ionic equation for the reaction of potassium chloride and lead(II) acetate is Pb²⁺(aq) + 2Cl⁻(aq) → PbCl₂(s).
