Question

Lead ions can be removed from solution by precipitation with sulfate ions. Suppose a solution contains lead(II) nitrate. Write a complete ionic and net ionic equation to show the reaction of aqueous lead(II) nitrate with aqueous potassium sulfate to form solid lead(II) sulfate and aqueous potassium nitrate.

Short answer

Net ionic equation: \(\text{Pb}^{2+}(aq) + \text{SO}_{4}^{2-}(aq) \rightarrow \text{PbSO}_{4}(s)\)

Step-by-step solution

Write the balanced molecular equation

First, write down the balanced molecular equation for the reaction between aqueous lead(II) nitrate and aqueous potassium sulfate to form solid lead(II) sulfate and aqueous potassium nitrate: \[\text{Pb(NO_{3})_{2}(aq)} + \text{K}_{2}\text{SO}_{4}(aq) \rightarrow \text{PbSO}_{4}(s) + 2\text{KNO}_{3}(aq)\]

Write the complete ionic equation

Next, write out all the ions individually for the reactants and products where they exist in aqueous solution: \[\text{Pb}^{2+}(aq) + 2\text{NO}_{3}^{-}(aq) + 2\text{K}^{+}(aq) + \text{SO}_{4}^{2-}(aq) \rightarrow \text{PbSO}_{4}(s) + 2\text{K}^{+}(aq) + 2\text{NO}_{3}^{-}(aq)\]

Cancel the spectator ions

Identify and remove the spectator ions (ions that appear on both sides of the ionic equation) to write the net ionic equation: Spectator ions: \(2\text{K}^{+}(aq)\) and \(2\text{NO}_{3}^{-}(aq)\)\[\text{Pb}^{2+}(aq) + \text{SO}_{4}^{2-}(aq) \rightarrow \text{PbSO}_{4}(s)\]

Key concepts

Precipitation Reactions

Precipitation reactions are chemical processes where ions in solution combine to form an insoluble compound, known as a precipitate. This type of reaction can be used to remove harmful or unwanted ions from a solution, such as the removal of lead ions using sulfate ions.

Let's dive a bit deeper into the mechanics of a precipitation reaction. Consider the following scenario: a solution containing lead(II) nitrate reacts with another containing potassium sulfate. The lead(II) nitrate is in aqueous form, meaning it's dissolved in water and separated into its constituent ions, lead (Pb²⁺) and nitrate (NO₃⁻). Similarly, potassium sulfate separates into potassium (K⁺) and sulfate (SO₄²⁻) ions.

When these two solutions are mixed, the lead ions (Pb²⁺) and the sulfate ions (SO₄²⁻) find each other and form lead(II) sulfate (PbSO₄), which is insoluble in water. This solid precipitates out of the solution, as it's unable to remain dissolved. The formation of a precipitate is often indicated by the appearance of a cloudy or solid substance within the otherwise clear solution.

Balancing Chemical Equations

Balancing chemical equations is a crucial skill in chemistry that ensures the law of conservation of mass is upheld during a chemical reaction. This means the number of atoms for each element must be the same on both sides of the equation. Start by writing down the unbalanced equation and then adjust the coefficients, the numbers that appear before the chemical formulas, to balance the atoms.

For instance, in the reaction between aqueous lead(II) nitrate and potassium sulfate, the balanced molecular equation is:
\[\text{Pb(NO_3)_2(aq)} + \text{K_2SO_4(aq)} \rightarrow \text{PbSO_4(s)} + 2\text{KNO_3(aq)}\]
This equation is balanced because there is one lead atom, four nitrogen atoms, and four oxygen atoms from the nitrate group on both sides of the equation. Additionally, there are two potassium atoms and one sulfate group on each side as well. Mastering this process involves understanding the stoichiometry of the reaction, which refers to the quantitative relationships between the amounts of reactants and products.

Spectator Ions

Spectator ions may seem like participants in a chemical reaction, but they don't undergo any change themselves. These are ions that appear in the exact same form on both the reactant and product sides of a complete ionic equation. Essentially, they are present to balance the electric charge in the solution but don't take part in the formation of the precipitate.

In the complete ionic equation for our reaction, potassium (K⁺) and nitrate (NO₃⁻) ions are spectators:
\[\text{Pb}^{2+}(aq) + 2\text{NO}_3^{-}(aq) + 2\text{K}^{+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) + 2\text{K}^{+}(aq) + 2\text{NO}_3^{-}(aq)\]
These ions are eliminated when writing the net ionic equation, focusing only on the ions that form the precipitate. Removing them simplifies the equation to:
\[\text{Pb}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s)\]
By focusing on the actual chemical change, the net ionic equation provides a clearer picture of the reaction's essence—showing exactly which ions are responsible for the formation of the new compound.