Write the net ionic equation for the reaction of lead(II)nitrate and sodium iodide. A)Pb(NO₃)₂(aq)+ 2 NaI(aq)→ PbI₂(s)+ 2 NaNO₃(aq) B)Pb²⁺(aq)+ 2 NaI(aq)→ PbI₂(s)+ 2 Na C)Pb²⁺(aq)+ 2 I^-(aq)→ PbI₂(s) D)Pb²⁺(aq)+ 2 I^-(aq)→ Pb²⁺(aq)+ 2 I^-(aq) E)no reaction

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Quizplus · Accepted Answer

The reactants in this equation are lead(II) nitrate (Pb(NO3)2) and sodium iodide (NaI). The products are lead(II) iodide (PbI2) and sodium nitrate (NaNO3). The net ionic equation removes any spectator ions (ions that don't participate in the reaction) and shows only the ions that are involved in the reaction. In this case, the net ionic equation is Pb2+(aq) + 2I-(aq) → PbI2(s), which corresponds to choice C.

Write the Net Ionic Equation for the Reaction of Lead(II)nitrate