What is the net ionic equation for the reaction between Pb(NO3)2(aq) and KI(aq)? Pb+(aq) + I−(aq) → PbI(s) Pb2+(aq) + 2 I−(aq) → Pb(s) + I2(s) Pb2+(aq) + 2 I−(aq) → PbI2(s) K+(aq) + NO3−(aq) → KNO3(s) K+(aq) + NO3−(aq) → K(s) + NO(g) + O2(g)
Question
What is the net ionic equation for the reaction between Pb(NO3)2(aq) and KI(aq)? Pb+(aq) + I−(aq) → PbI(s) Pb2+(aq) + 2 I−(aq) → Pb(s) + I2(s) Pb2+(aq) + 2 I−(aq) → PbI2(s) K+(aq) + NO3−(aq) → KNO3(s) K+(aq) + NO3−(aq) → K(s) + NO(g) + O2(g)
Solution
The reaction between Pb(NO3)2(aq) and KI(aq) can be broken down into the following steps:
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Write the balanced molecular equation: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)
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Write the total ionic equation: Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) → PbI2(s) + 2K+(aq) + 2NO3-(aq)
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Write the net ionic equation by removing the spectator ions (ions that appear on both sides of the equation): Pb2+(aq) + 2I-(aq) → PbI2(s)
So, the net ionic equation for the reaction between Pb(NO3)2(aq) and KI(aq) is Pb2+(aq) + 2I-(aq) → PbI2(s).
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