Lead(II) nitrate solution reacts with potassium iodide solution to produce lead(II) iodide and potassium nitrate.

The answer in the book is:
Pb(NO3)2(aq) + 2 KI(aq) ---> 2KNO3(aq) + PbI2(s)

I understand how to write and balance the equation. However, I am confused about how to determine the states of matter. Why are some aqueous and one solid? Does it have something to do with the solubility rules? Thanks for your help.

yes.

PbI2 is not soluble in water; therefore, it is written as PbI2(s).
KNO3 is soluble in water; therefore, in solution it is written as KNO3(aq).
Pb(NO3)2(aq) means you have a solution of Pb(NO3)2 and not solid Pb(NO3)2.

Determining the states of matter in a chemical equation involves considering the solubility rules. Solubility is the ability of a substance to dissolve in a solvent, typically water in this context. Let's break down the equation and explain why each compound is assigned the specific state of matter.

1. Pb(NO3)2(aq): This compound is lead(II) nitrate, and it is indicated as aqueous (aq) because nitrates (NO3-) are generally soluble in water. However, it's important to note that lead(II) compounds are an exception to this rule. Lead(II) compounds, including lead(II) nitrate, are generally slightly soluble and tend to form a precipitate.

2. KI(aq): This compound is potassium iodide, and it is also indicated as aqueous (aq). Potassium compounds, such as potassium iodide, are typically soluble in water. Iodides (I-) are generally soluble except when paired with certain elements like lead, silver, or mercury.

3. 2KNO3(aq): This is potassium nitrate, which is indicated as aqueous (aq). Potassium compounds, as mentioned earlier, are usually soluble in water. Nitrates (NO3-) are generally soluble, so potassium nitrate readily dissolves in water.

4. PbI2(s): This compound is lead(II) iodide, and it is indicated as solid (s). Iodides typically form insoluble compounds when paired with certain metals, including lead. Therefore, the lead(II) iodide precipitates out of solution as a solid.

So, in summary, the solubility rules help us determine the states of matter in this particular reaction. The lead(II) nitrate and potassium nitrate compounds remain dissolved in water since their respective ions are generally soluble. However, the lead(II) iodide forms a solid because lead(II) compounds tend to be slightly soluble and iodides form insoluble compounds with certain metals, including lead.

Yes, the states of matter in a chemical equation can be determined based on the solubility rules. In this case, lead(II) nitrate (Pb(NO3)2) and potassium iodide (KI) are both soluble in water, so they are represented as aqueous (aq).

On the other hand, lead(II) iodide (PbI2) is insoluble in water, so it is represented as a solid (s). This means that it will precipitate out of the solution as a solid during the reaction.

Potassium nitrate (KNO3) is also soluble in water, so it remains in the solution and is represented as aqueous (aq).

The balanced chemical equation you provided is correct:

Pb(NO3)2(aq) + 2 KI(aq) ---> 2KNO3(aq) + PbI2(s)