Question

If a saturated aqueous solution of the shock-sensitive compound lead(II) azide, \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}\), has \(\left[\mathrm{Pb}^{2+}\right]=8.5 \times 10^{-4} \mathrm{M}\), what is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}\) ?

Short answer

The solubility product constant, \(K_{sp}\), is \(2.46 \times 10^{-9}\).

Step-by-step solution

Understanding the Dissolution Equation

First, consider the dissolution of lead(II) azide in water. The reaction is given by: \[\text{Pb(N}_3\text{)}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{N}_3^-(aq)\] This equation indicates that one mole of lead(II) azide dissociates to produce one mole of \(\text{Pb}^{2+}\) and two moles of \(\text{N}_3^-\).

Determining Ion Concentrations

Since each mole of \(\text{Pb(N}_3\text{)}_2\) produces one mole of \(\text{Pb}^{2+}\) in solution, the concentration of \(\text{Pb}^{2+}\) will be the same as provided: \[\text{[Pb}^{2+}] = 8.5 \times 10^{-4} \text{ M}\] For \(\text{N}_3^-\), since two moles are produced for each mole of \(\text{Pb}^{2+}\), its concentration will be:\[\text{[N}_3^-] = 2 \times 8.5 \times 10^{-4} \text{ M} = 1.7 \times 10^{-3} \text{ M}\]

Calculating the Solubility Product Constant (Ksp)

Using the formula for the solubility product \(K_{sp}\):\[K_{sp} = [\text{Pb}^{2+}][\text{N}_3^-]^2\]Substitute the known concentrations:\[K_{sp} = (8.5 \times 10^{-4})(1.7 \times 10^{-3})^2\]Calculate the value:\[K_{sp} = (8.5 \times 10^{-4}) \times (2.89 \times 10^{-6}) = 2.46 \times 10^{-9}\]

Final Answer

The solubility product constant, \(K_{sp}\), expresses the equilibrium condition for the dissolution process and indicates how much of the compound will dissolve under those conditions. In this case, \(K_{sp}\) for lead(II) azide is \(2.46 \times 10^{-9}\).

Key concepts

Lead(II) Azide

Lead(II) azide is a chemical compound that is commonly used as a detonator in explosives due to its shock-sensitive nature. Its chemical formula is \( \text{Pb(N}_3\text{)}_2 \). This compound contains lead cations \( \text{Pb}^{2+} \) and azide anions \( \text{N}_3^- \). The presence of lead makes it an important substance to understand in contexts involving both chemistry and safety. It is crucial to handle lead(II) azide with care, especially in laboratory conditions, because its reactive properties can be hazardous. Understanding compounds like lead(II) azide in solution contexts helps in predicting their behavior in various environments.

Dissolution Equation

The dissolution equation describes how lead(II) azide breaks down in water. For a solid salt dissolving in water, we can write the equation as: \[ \text{Pb(N}_3\text{)}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{N}_3^-(aq) \] This equation shows that when lead(II) azide dissolves, it separates into one lead ion and two azide ions. This balanced equation is crucial as it provides the stoichiometric relationship of the ions formed when the compound dissolves. Knowing the stoichiometry helps in calculating the concentrations of the ions in solution, which is essential for determining the solubility product constant \(K_{sp}\). The chemical equilibrium indicates that the dissolution and precipitation of lead(II) azide happen simultaneously.

Ion Concentrations

When lead(II) azide dissolves, the concentration of its ions in solution depends on its stoichiometry. Given in the problem, the concentration of \(\text{Pb}^{2+} \) is \(8.5 \times 10^{-4} \text{ M}\). Since each mole of \(\text{Pb(N}_3\text{)}_2\) yields one mole of \(\text{Pb}^{2+} \), the given concentration outrightly represents the amount of lead ions.

• For \(\text{N}_3^-\), the dissolution of each mole of lead(II) azide results in two moles of azide ions.
• Therefore, the concentration of azide ions \([\text{N}_3^-]\) is calculated as twice that of lead ions, which amounts to \(1.7 \times 10^{-3} \text{ M}\).

The balance of these ion concentrations is critical for computing the \(K_{sp}\), as they represent the dissolved state of the solid in equilibrium.

Equilibrium Condition

The equilibrium condition is vital in understanding how much of a compound like lead(II) azide will dissolve at a certain temperature. This concept is encapsulated by the solubility product constant, \(K_{sp}\), which quantifies the maximum extent to which a compound can dissolve in a solution. For the given solution, \(K_{sp}\) is calculated using the equilibrium concentrations of the dissociated ions: \[ K_{sp} = [\text{Pb}^{2+}][\text{N}_3^-]^2 \] In our case, substituting the values: \[ K_{sp} = (8.5 \times 10^{-4})(1.7 \times 10^{-3})^2 = 2.46 \times 10^{-9} \] This shows the precise solubility of lead(II) azide in water. A small \(K_{sp}\) value like this signifies that lead(II) azide is not very soluble at equilibrium, indicating only a minor fraction of it dissolves while most remains as a solid.