Problem 106 - General Chemistry: Principles and Modern Applications

Question

What is the net ionic equation for the reaction that occurs when an aqueous solution of \(\mathrm{KI}\) is added to an aqueous solution of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Step-by-Step Solution

Answer

The net ionic equation for the reaction that occurs when an aqueous solution of \(\mathrm{KI}\) is added to an aqueous solution of \(\mathrm{Pb(NO}_3)_2\) is \[2 \, \mathrm{I^-} + Pb^{2+} = \mathrm{PbI}_2(s)\]

Step 1: Write down the molecular equation

First, write down the molecular equation. Potassium iodide reacts with lead nitrate to produce lead iodide(which is a precipitate) and potassium nitrate. This translates to the following equation: \[2 \, \mathrm{KI} + \, \mathrm{Pb(NO}_3)_2 \, \rightarrow \, \mathrm{PbI}_2 + 2 \, \mathrm{KNO}_3\] This equation indicates that for each molecule of lead nitrate in solution, two molecules of potassium iodide are needed to produce one molecule of lead iodide and two molecules of potassium nitrate.

Step 2: Write down the total ionic equation

The next step is to write down the ionic equation, which includes all of the ions present in the reaction: \[2 \, \mathrm{K^+} + 2 \, \mathrm{I^-} + Pb^{2+} + 2 \, \mathrm{NO}_3^- \, \rightarrow \, Pb \, \mathrm{I}_2(s) + 2 \, \mathrm{K^+} + 2 \, \mathrm{NO}_3^-\] The (s) next to \(PbI_2\) indicates that it is a solid (precipitate), while the other ions are all in aqueous solution.

Step 3: Write down the net ionic equation

The net ionic equation is obtained by cancelling out the spectator ions that don’t participate in the reaction. In this case, the potassium ions \( \mathrm{K^+} \) and nitrate ions \( \mathrm{NO}_3^- \) are spectator ions. Removing these gives the net ionic equation: \[2 \, \mathrm{I^-} + Pb^{2+} \, \rightarrow \, \mathrm{PbI}_2(s)\] This equation indicates that in the reaction, lead ions combined with iodide ions to form the precipitate, lead iodide.

Key Concepts

Molecular Equation

The molecular equation represents all reactants and products in their undissociated form as if they were compounds. In the reaction between potassium iodide (KI) and lead(II) nitrate \(\mathrm{Pb(NO}_3)_2\), we start by writing the balanced molecular equation:

• \(2 \, \mathrm{KI} + \mathrm{Pb(NO}_3)_2 \rightarrow \mathrm{PbI}_2 + 2 \, \mathrm{KNO}_3\)

This equation shows that two molecules of potassium iodide react with one molecule of lead(II) nitrate. The result is one molecule of lead(II) iodide and two molecules of potassium nitrate. It is important to understand that while the molecular equation illustrates the compounds as whole entities, it doesn't show how these exist in solution.
To prepare for writing the total ionic equation, keep in mind the states of the compounds: \(\mathrm{PbI}_2\) is a solid precipitate formed in the solution, whereas the others remain dissolved and separate into ions.

Total Ionic Equation

The total ionic equation displays all the ions present in the aqueous solution as they actually exist in solution. In water, soluble ionic compounds dissociate into their constituent ions. Therefore, potassium iodide \((\mathrm{KI})\) and lead(II) nitrate \((\mathrm{Pb(NO}_3)_2)\) separate into ions, reflecting the true nature of the substances involved:

• \(2 \, \mathrm{K^+} + 2 \, \mathrm{I^-} + \mathrm{Pb^{2+}} + 2 \, \mathrm{NO}_3^- \rightarrow \mathrm{PbI}_2(s) + 2 \, \mathrm{K^+} + 2 \, \mathrm{NO}_3^-\)

Here, you observe the distinction between solid and aqueous states. \(\mathrm{PbI}_2(s)\) remains as a solid due to its low solubility in water, forming a precipitate.
The total ionic equation lays the groundwork for identifying and canceling out ions on both sides of the equation that do not change and do not participate directly in the formation of the precipitate.

Spectator Ions

Spectator ions are ions that appear on both sides of the total ionic equation but do not participate in the actual chemical change. They essentially 'watch' as the other ions form a new product. In our equation, calcium ions \((\mathrm{K^+})\) and nitrate ions \((\mathrm{NO}_3^-)\) are present on both sides of the reaction:

• They do not change and can be 'cancelled out' from the total ionic equation.
• They remain in the solution and do not form part of the precipitate.

Once you remove these spectator ions, you arrive at the net ionic equation:

• \(2 \, \mathrm{I^-} + \mathrm{Pb^{2+}} \rightarrow \mathrm{PbI}_2(s)\)

This net ionic equation highlights the actual chemical change taking place in the reaction, where only the reacting ions are shown forming the precipitate lead(II) iodide.