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)
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.