Final answer:
The net ionic equation for the reaction of aqueous lead nitrate with aqueous potassium chloride, resulting in the formation of lead (II) chloride precipitate, is Pb^{2+}(aq) + 2Cl^{-}(aq) → PbCl2(s).
Step-by-step explanation:
The question asks for the net ionic equation for the reaction between aqueous solutions of lead nitrate and potassium chloride, resulting in the formation of a solid precipitate of lead (II) chloride and an aqueous solution of potassium nitrate. To write the net ionic equation, we first write the complete ionic equation and then eliminate the spectator ions.
The complete ionic equation for the reaction is:
Pb^{2+}(aq) + 2NO3^{-}(aq) + 2K^{+}(aq) + 2Cl^{-}(aq) → PbCl2(s) + 2K^{+}(aq) + 2NO3^{-}(aq)
Observing the equation, we find that the potassium ions, K+, and the nitrate ions, NO3-, are present in both the reactants and the products, so they are spectator ions and can be canceled out.
The net ionic equation is thus:
Pb^{2+}(aq) + 2Cl^{-}(aq) → PbCl2(s)