Which of the following represents the complete ionic equation for the reaction between Pb(NO3)2(aq) and KOH(aq)?

• A. Pb^2+(aq) + NO3^-(aq) + K^+(aq) + OH^-(aq) → Pb(OH)2(s) + K^+(aq) + NO3^-(aq)
• B. Pb^2+(aq) + 2OH^-(aq) → Pb(OH)2(s)
• C. Pb^2+(aq) + 2NO3^-(aq) + 2K^+(aq) + 2OH^-(aq) → Pb(OH)2(s) + 2K^+(aq) + 2NO3^-(aq)
• D. Pb(NO3)2(aq) + 2KOH(aq) → Pb(OH)2(s) + 2KNO3(aq)

Answer: (C)

Verified step by step guidance

1. Identify the reactants and products in the molecular equation: Pb(NO3)2(aq) + 2KOH(aq) → Pb(OH)2(s) + 2KNO3(aq).
2. Write the dissociation of each aqueous compound into its ions: Pb(NO3)2(aq) dissociates into Pb^2+(aq) and 2NO3^-(aq); KOH(aq) dissociates into K^+(aq) and OH^-(aq).
3. Write the complete ionic equation by replacing each aqueous compound with its dissociated ions: Pb^2+(aq) + 2NO3^-(aq) + 2K^+(aq) + 2OH^-(aq) → Pb(OH)2(s) + 2K^+(aq) + 2NO3^-(aq).
4. Identify the spectator ions, which are ions that appear on both sides of the equation and do not participate in the reaction: K^+(aq) and NO3^-(aq).
5. Write the net ionic equation by removing the spectator ions from the complete ionic equation: Pb^2+(aq) + 2OH^-(aq) → Pb(OH)2(s).