Question

What is the molecular equation for the reaction between lead(II) acetate and potassium iodide. Which product forms a precipitate and which are spectator ions?

Answer 1

What is the chemical formula of lead(II) acetate and potassium iodide? Do you know how to write a chemical formula based on the chemical name of the compound?

Add lead(II) acetate and potassium iodide together and exchange their anions.
For example:
\(AB+CD \rightarrow AD+BC\)

Label their physical state and see which compound AD or BC forms a solid. Break the aqueous solutions into ions but leave the solid compound by itself.

Cancel ions that appear in the both sides of the equation (ions that cancel out is your spectator ions).

Write the net equation by writing whatever ions and compound(s) you're left with.

Answer 2

Let us say the reaction formed AD as a solid. (Look at the solubility rules if you want to know which compound formed is a solid)

\(AB(aq)+CD(aq) \rightarrow AD(s)+BC(aq)\)
\(A^+(aq)+\color{red}{B^-(aq)}+\color{red}{C^+(aq)}+D^-(aq) \rightarrow AD(s)+\color{red}{C^+(aq)}+\color{red}{B^-(aq)}\)

Letters in red are the spectator ions, hey appear in both sides of the equation. They cancel out. Your net equation is the equation after canceling out the spectator ions.

\(A^+(aq)+D^-(aq) \rightarrow AD(s)\)

Answer 3

Thank you so very much!