Question

Use the following equilibrium to demonstrate why the \(K_{\mathrm{sp}}\) expression does not include the concentration of \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in the denominator: $$ \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3 \mathrm{Ba}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) $$

Short answer

The concentration of the solid \(\mathrm{Ba}_{3}(\mathrm{PO}_{4})_{2}\) is not included in the \(K_{\mathrm{sp}}\) expression because the concentration of a pure solid is constant and does not affect the equilibrium of the dissolved species in solution.

Step-by-step solution

Understand the Nature of the Solid

Recognize that \(\mathrm{Ba}_{3}(\mathrm{PO}_{4})_{2}\) is a solid and solids are not included in the equilibrium constant expression because their concentration is constant and does not affect the equilibrium position.

Write the Expression for the Solubility Product Constant (\(K_{\mathrm{sp}}\))

Write the equilibrium expression for the solubility product, where the concentrations of the ions in solution are raised to the power of their coefficients in the balanced equilibrium equation.\[K_{\mathrm{sp}} = [\mathrm{Ba}^{2+}]^{3} \cdot [\mathrm{PO}_{4}^{3-}]^{2}\]

Explain Why Solids Are Excluded from \(K_{\mathrm{sp}}\) Expression

Explain that the exclusion of the concentration of the solid \(\mathrm{Ba}_{3}(\mathrm{PO}_{4})_{2}\) from the \(K_{\mathrm{sp}}\) expression is because the concentration of a pure solid does not change as the solid is in equilibrium with its ions, hence it would be redundant to include it in the expression.

Key concepts

Equilibrium Constant Expression

When studying chemical equilibria, it's important to understand how to properly formulate an equilibrium constant expression, often symbolized as K. This expression quantifies the relationship between the concentrations of the reactants and products when a reaction has reached equilibrium — a state where the rates of the forward and reverse reactions are equal.

The equilibrium constant expression for the solubility product, K_sp, is unique to ionic compounds that are slightly soluble in water. The K_sp specifically relates to the maximum concentration of ions that will exist in solution at equilibrium without forming a precipitate.

In our example with barium phosphate (Ba₃(PO₄)₂), the expression is:
\[K_{\mathrm{sp}} = [\mathrm{Ba}^{2+}]^{3} \cdot [\mathrm{PO}_{4}^{3-}]^{2}\]
Notice that the expression includes only the ions that are present in the aqueous phase. Solids and liquids are not included because their concentrations are constants and do not change; they are essentially pure substances with fixed compositions. This exclusion simplifies the equation and focuses on the species that are actually in dynamic equilibrium.

Solute-Solvent Interactions

The behavior of solutes dissolving in solvents is governed by solute-solvent interactions, which are crucial for understanding solubility. When an ionic solid like barium phosphate dissolves, it dissociates into its constituent ions, which become solvated — surrounded by solvent molecules.

The strength of the solute-solvent interactions plays a vital role in the solubility of a compound. In the case of water, a polar solvent, the positive and negative parts of water molecules interact with the corresponding oppositely charged ions. In our example, the Ba²⁺ and PO₄^3- ions would be attracted to the partial negative oxygen and partial positive hydrogen ends of water molecules, respectively.

These interactions help stabilize the ions in solution and are critical in determining the extent to which the solid dissolves. A higher strength of solute-solvent interactions leads to greater solubility. This concept links to the idea that 'like dissolves like', meaning that polar solvents best dissolve polar or ionic solutes, while non-polar solvents favor non-polar solutes.

Ionic Solids in Equilibrium

Ionic solids in equilibrium with their dissolved ions are governed by the principles of dynamic equilibrium. At equilibrium, the rates of dissolution and precipitation are equal, meaning the solid phase and the dissolved ions exist in a constant ratio defined by K_sp.

It is important to understand that equilibrium does not imply that the reactants and products are present in equal quantities. Instead, it means that their concentrations do not change over time. For ionic solids, this implies that as long as the product of the ionic concentrations in solution does not exceed the K_sp, no additional solid will precipitate. Conversely, if this product is less than the K_sp, the solid will continue to dissolve until the K_sp is reached, or the solid is completely dissolved.

Different ionic solids have different K_sp values, which reflects their relative solubilities. A small K_sp value suggests a compound is less soluble and will have a greater tendency to remain in the solid form, whereas a large K_sp value indicates a more soluble substance, where more ions can coexist in solution without forming a precipitate.