# IMAT 2016 Q42[Ionic Equations]

Lead(II) nitrate solution and potassium iodide solution react to form potassium nitrate in solution and a bright yellow precipitate of lead(II) iodide.

Which one of the following correctly represents the ionic equation for this chemical reaction?

A. Pb^{2+}(aq) + I^–(aq) → PbI(s)

B. Pb^{2+}(aq) + I^{2–}(aq) → PbI(s)

C. Pb^+(aq) + 2I^–(aq) → PbI_2(s)

D. Pb^{2+}(aq) + 2I^–(aq) → PbI_2(s)

E. Pb^+(aq) + I^–(aq) → PbI(s)

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This is a double displacement reaction, where both reactants are initially aqueous solutions which form a solid precipitate.

Pb(NO_3)_2 + 2KI \rightarrow PbI_2 + 2KNO_3

We know that the valency of Iodine is -1, and the valency of Lead is +2. With this information, we can eliminate E, C and B.

Therefore, two Iodine Ions must react with one Lead Ion. So, we can eliminate A.

Our answer becomes D, which is true.