Question

Write the balanced chemical equations describing the dissolving of the following solids in water. Write the expression for \(K_{\mathrm{sp}}\) for each process. a. \(\operatorname{PbBr}_{2}(s)\) b. \(\mathrm{Ag}_{2} \mathrm{S}(\mathrm{s})\) c. \(\mathrm{PbCO}_{3}(s)\) d. \(\operatorname{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\)

Short answer

a. \(\operatorname{PbBr}_{2}(s) \rightarrow \operatorname{Pb}^{2+}\left(aq\right) + 2\operatorname{Br}^-\left(aq\right)\); \(K_{sp} = [\operatorname{Pb}^{2+}][\operatorname{Br}^-]^2\) b. \(\mathrm{Ag}_{2}\mathrm{S(s)} \rightarrow 2 \mathrm{Ag}^+\left(aq\right) + \mathrm{S}^{2-}\left(aq\right)\); \(K_{sp} = [\mathrm{Ag}^+]^2[\mathrm{S}^{2-}]\) c. \(\mathrm{PbCO}_{3}(s) \rightarrow \mathrm{Pb}^{2+}\left(aq\right) + \mathrm{CO}_{3}^{2-}\left(aq\right)\); \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{CO}_{3}^{2-}]\) d. \(\operatorname{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightarrow 3 \operatorname{Sr}^{2+}\left(aq\right) + 2 \mathrm{PO}_{4}^{3-}\left(aq\right)\); \(K_{sp} = [\operatorname{Sr}^{2+}]^3[\mathrm{PO}_{4}^{3-}]^2\)

Step-by-step solution

Write the Dissolution Equation

The dissolution of \(\operatorname{PbBr}_{2}(s)\) in water can be written as: \[\operatorname{PbBr}_{2}(s) \rightarrow \operatorname{Pb}^{2+}\left(aq\right) + 2\operatorname{Br}^-\left(aq\right)\]

Write the Expression for \(K_{sp}\)

The solubility product constant \(K_{sp}\) for the reaction can be written as: \[K_{sp} = [\operatorname{Pb}^{2+}][\operatorname{Br}^-]^2\] b. Dissolving and Balancing \(\mathrm{Ag}_{2} \mathrm{S}(\mathrm{s})\)

Write the Dissolution Equation

The dissolution of \(\mathrm{Ag}_{2} \mathrm{S}(\mathrm{s})\) in water can be written as: \[\mathrm{Ag}_{2}\mathrm{S(s)} \rightarrow 2 \mathrm{Ag}^+\left(aq\right) + \mathrm{S}^{2-}\left(aq\right)\]

Write the Expression for \(K_{sp}\)

The solubility product constant \(K_{sp}\) for the reaction can be written as: \[K_{sp} = [\mathrm{Ag}^+]^2[\mathrm{S}^{2-}]\] c. Dissolving and Balancing \(\mathrm{PbCO}_{3}(s)\)

Write the Dissolution Equation

The dissolution of \(\mathrm{PbCO}_{3}(s)\) in water can be written as: \[\mathrm{PbCO}_{3}(s) \rightarrow \mathrm{Pb}^{2+}\left(aq\right) + \mathrm{CO}_{3}^{2-}\left(aq\right)\]

Write the Expression for \(K_{sp}\)

The solubility product constant \(K_{sp}\) for the reaction can be written as: \[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{CO}_{3}^{2-}]\] d. Dissolving and Balancing \(\operatorname{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\)

Write the Dissolution Equation

The dissolution of \(\operatorname{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) in water can be written as: \[\operatorname{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightarrow 3 \operatorname{Sr}^{2+}\left(aq\right) + 2 \mathrm{PO}_{4}^{3-}\left(aq\right)\]

Write the Expression for \(K_{sp}\)

The solubility product constant \(K_{sp}\) for the reaction can be written as: \[K_{sp} = [\operatorname{Sr}^{2+}]^3[\mathrm{PO}_{4}^{3-}]^2\]

Key concepts

Chemical Equation Balancing

The process of balancing chemical equations involves providing the correct stoichiometric coefficients to ensure that the number of atoms for each element is the same on both sides of the reaction. This reflects the conservation of mass, a fundamental principle in chemistry.

For example, in the dissolution reaction for lead(II) bromide \(\text{PbBr}_2(s) \rightarrow \text{Pb}^{2+}(aq) + 2\text{Br}^{-}(aq)\), the coefficient '2' in front of \(\text{Br}^{-}\) balances the number of bromide ions, since there are two bromide ions per formula unit of lead(II) bromide.

In simple dissolution reactions like the ones in this exercise, balancing is straightforward as the solid compound dissociates into its constituent ions. However, understanding and applying the principles of chemical equation balancing becomes more complex with multi-step reactions or those involving polyatomic ions, which requires a more in-depth analysis of the reaction stoichiometry.

Solubility Equilibrium

Solubility equilibrium refers to the dynamic balance between the dissolution of a solid and the crystallization (re-formation) of the solid in a saturated solution. At equilibrium, the rate at which the solid dissolves equals the rate at which the solid is formed, resulting in a constant concentration of ions in the solution.

For instance, when solid lead(II) carbonate \(\text{PbCO}_3(s)\) is added to water, it dissolves to form lead(II) ions and carbonate ions until saturation is reached:\(\text{PbCO}_3(s) \rightarrow \text{Pb}^{2+}(aq) + \text{CO}_3^{2-}(aq)\). At this point, any additional solid added to the solution does not change the concentration of dissolved ions because the excess precipitates out, maintaining a dynamic balance.

The solubility product constant \(K_{sp}\) quantitatively expresses this equilibrium. A high \(K_{sp}\) value indicates higher solubility of the compound in water, whereas a low \(K_{sp}\) suggests limited solubility.

Dissolution Reactions

Dissolution reactions describe the process by which a solid dissolves in a solvent to form a homogeneous solution. During dissolution, the ionic or molecular compounds disintegrate into their constituent ions or molecules. In the context of this exercise, the solid is an ionic compound, and the solvent is water, a polar molecule capable of solvating ions.

For example, the dissolution of strontium phosphate \(\text{Sr}_3(\text{PO}_4)_2(s)\) forms strontium ions and phosphate ions in solution:\(\text{Sr}_3(\text{PO}_4)_2(s) \rightarrow 3 \text{Sr}^{2+}(aq) + 2 \text{PO}_4^{3-}(aq)\). Water molecules surround the individual ions, facilitating their separation and stabilizing them in the solution.

The nature and structure of the solid, the polarity of the solvent, and conditions such as temperature and pressure significantly affect these dissolution reactions. Understanding dissolution reactions is crucial for grasping concepts across various scientific disciplines, including chemistry, environmental science, and pharmaceuticals.