Problem 27 - Chemistry The Molecular Nature of Matter

Question

Write the \(K_{\mathrm{sp}}\) expressions for each of the following com- pounds: (a) \(\mathrm{CaF}_{2},\) (b) \(\mathrm{Ag}_{2} \mathrm{CO}_{3},\) (c) \(\mathrm{PbSO}_{4},\) (d) \(\mathrm{Fe}(\mathrm{OH})_{3}\), (e) \(\mathrm{PbF}_{2}\), (f) \(\mathrm{Cu}(\mathrm{OH})_{2}\)

Step-by-Step Solution

Answer

\(\mathrm{CaF}_{2}: K_{\mathrm{sp}} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2\), \(\mathrm{Ag}_{2} \mathrm{CO}_{3}: K_{\mathrm{sp}} = [\mathrm{Ag}^{+}]^2[\mathrm{CO}_{3}^{2-}]\), \(\mathrm{PbSO}_{4}: K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{SO}_{4}^{2-}]\), \(\mathrm{Fe}(\mathrm{OH})_{3}: K_{\mathrm{sp}} = [\mathrm{Fe}^{3+}][\mathrm{OH}^{-}]^3\), \(\mathrm{PbF}_{2}: K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{F}^{-}]^2\), \(\mathrm{Cu}(\mathrm{OH})_{2}: K_{\mathrm{sp}} = [\mathrm{Cu}^{2+}][\mathrm{OH}^{-}]^2\)

Step 1: Identify the Ions in Each Compound

For each compound listed, identify the cations (positively charged ions) and anions (negatively charged ions) formed when the compound dissolves in water.

Step 2: Write the Dissociation Equations

Write an equation for each compound showing how it dissociates into its respective cations and anions in water.

Step 3: Construct the K_{\mathrm{sp}} Expressions

Use the dissociation equations to construct the solubility product constant (\(K_{\mathrm{sp}}\)) expressions for each ionic compound, based on the stoichiometry of the dissolution reaction.

Step 4: List the K_{\mathrm{sp}} Expressions for Each Compound

(a) For \(\mathrm{CaF}_{2}\), which dissociates into one \(\mathrm{Ca}^{2+}\) ion and two \(\mathrm{F}^{-}\) ions:\[K_{\mathrm{sp}} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^{2}\](b) For \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\), which dissociates into two \(\mathrm{Ag}^{+}\) ions and one \(\mathrm{CO}_{3}^{2-}\) ion:\[K_{\mathrm{sp}} = [\mathrm{Ag}^{+}]^{2}[\mathrm{CO}_{3}^{2-}]\](c) For \(\mathrm{PbSO}_{4}\), which dissociates into one \(\mathrm{Pb}^{2+}\) ion and one \(\mathrm{SO}_{4}^{2-}\) ion:\[K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{SO}_{4}^{2-}]\](d) For \(\mathrm{Fe}(\mathrm{OH})_{3}\), which dissociates into one \(\mathrm{Fe}^{3+}\) ion and three \(\mathrm{OH}^{-}\) ions:\[K_{\mathrm{sp}} = [\mathrm{Fe}^{3+}][\mathrm{OH}^{-}]^{3}\](e) For \(\mathrm{PbF}_{2}\), which dissociates into one \(\mathrm{Pb}^{2+}\) ion and two \(\mathrm{F}^{-}\) ions:\[K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{F}^{-}]^{2}\](f) For \(\mathrm{Cu}(\mathrm{OH})_{2}\), which dissociates into one \(\mathrm{Cu}^{2+}\) ion and two \(\mathrm{OH}^{-}\) ions:\[K_{\mathrm{sp}} = [\mathrm{Cu}^{2+}][\mathrm{OH}^{-}]^{2}\]

Key Concepts

Dissociation Equations

Dissociation equations illustrate the process by which ionic compounds break apart into their constituent ions when dissolved in water. This process is essential for understanding the formation of electrolytes in solution, which are crucial in many chemical reactions and biological processes.

To write a dissociation equation, one must consider the chemical formula of the ionic compound and determine into which ions the compound separates. For example, \(\mathrm{CaF}_{2}\) dissociates into one calcium ion (\(\mathrm{Ca}^{2+}\)) and two fluoride ions (\(\mathrm{F}^{-}\)). The equation reflects this by representing the solid compound yielding its ions: \[\mathrm{CaF}_{2}(s) \rightarrow \mathrm{Ca}^{2+}(aq) + 2\mathrm{F}^{-}(aq)\].

This dissolution into ions facilitates various chemical phenomena such as conductivity and reactions in solution. The ratio of cations to anions in these equations is determined by the ionic charges and is essential for the stoichiometry calculations that follow.

Ionic Compounds

Ionic compounds are chemical compounds composed of ions held together by electrostatic forces termed ionic bonding. These compounds typically form when a metal reacts with a nonmetal, transferring electrons and creating ions with positive (cations) and negative (anions) charges. In solid form, ionic compounds constitute a crystalline lattice structure, contributing to their characteristic high melting and boiling points.

During the dissolution process, the attractive forces in the lattice are overcome by the interaction with the solvent, such as water. Water molecules, being polar, surround the ions and 'pull' them into solution due to the electrostatic interactions. This distinct characteristic is why ionic compounds tend to be soluble in water and other polar solvents. The solubility in water and the ability to conduct electricity when melted or dissolved classify these compounds as electrolytes.

Stoichiometry of Dissolution

Stoichiometry of dissolution refers to the quantitative relationship between the reactants and products in a dissolution process. For ionic compounds, it's crucial to consider the moles of each ion produced when the compound dissolves. This stoichiometry is directly reflected in the solubility product constant (\(K_{\mathrm{sp}}\)) expressions.

Take the dissociation of \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\) as an example. It dissociates into two silver ions (\(\mathrm{Ag}^{+}\)) and one carbonate ion (\(\mathrm{CO}_{3}^{2-}\)). Thus, the stoichiometry of the dissolution will have a 2:1 ratio, represented as: \[\mathrm{Ag}_{2}\mathrm{CO}_{3}(s) \rightarrow 2\mathrm{Ag}^{+}(aq) + \mathrm{CO}_{3}^{2-}(aq)\].

The solubility product expression takes this stoichiometry into account. Therefore, in the expression for \(K_{\mathrm{sp}}\) of \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\), the concentration of \(\mathrm{Ag}^{+}\) will be squared, reflecting its 2:1 stoichiometric coefficient in the dissolution equation. This precise stoichiometric inclusion ensures that the solubility product constant is grounded in the mole ratio of ions in the solution.