Assuming that no equilibrium other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: (a) \(\mathrm{KHC}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\) (b) \(\mathrm{PbI}_{2}\) (c) \(\mathrm{Ag}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), a salt containing the \(\mathrm{Fe}(\mathrm{CN})_{4}^{-}\) ion (d) \(\mathrm{Hg}_{2} \mathrm{I}_{2}\)

The solubility product constant, known as Ksp, is a vital concept in chemistry that represents the equilibrium between a solid and its ions in a saturated solution. It is specific to each solute and depends on temperature. Think of Ksp as an indicator of a compound's solubility: the higher the Ksp value, the more soluble the compound.

In our exercise, to approach each part, one must write the Ksp expression based on the balanced dissolution equation. For instance, for the compound \(KHC_4H_4O_6\), the equilibrium represents one \(K^+\) ion and one \(HC_4H_4O_6^-\) ion for each formula unit that dissolves. The Ksp expression is then \( K_{sp} = [K^+] [HC_4H_4O_6^-] \). The square brackets represent concentrations at equilibrium, and the terms are raised to the power of their coefficients in the balanced equation.

Using the Ksp expressions, you can solve for molar solubility, which tells you how many moles of a solute can be dissolved in a liter of solution to reach saturation. Ksp can be valuable when comparing the solubility of different compounds or predicting the outcome of mixing ionic solutions that could lead to precipitation.

Dissolution equations systematically represent how an ionic compound dissociates into its constituent ions when it dissolves. They are pivotal for understanding solubility and for calculating various parameters like molar solubility and the solubility product constant.

The format of a dissolution equation is straightforward: solid reactant \(\rightarrow\) dissolved ionic products. In this exercise, for compound \( PbI_2 \), the dissolution equation is \( PbI_2(s) \rightarrow Pb^{2+}(aq) + 2I^-(aq) \). It's important to note that the states of matter are indicated (solid 's', aqueous 'aq'), and the stoichiometry from the balanced equation is directly reflected in the Ksp expression.

Students must write accurate dissolution equations to ensure their Ksp calculations are correct. It is essential to account for the stoichiometry in the Ksp expression, as it determines how the concentrations of the ions relate to the molar solubility of the compound.

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products remain constant over time. It's a dynamic state where the compounds are constantly reacting, but no net change is observed.

In the context of solubility, equilibrium is achieved when a saturated solution is formed – the point at which additional solute will not dissolve because the dissolving and precipitating rates are equal. Each compound in our exercise has its own unique point of equilibrium determined by its Ksp.

At equilibrium, the reaction has a specific ratio of products to reactants, established by the equilibrium constant, which for solubility is the Ksp. Understanding equilibrium helps in calculating how varying conditions, like changes in concentration or temperature, affect solubility and can even predict whether a precipitate will form when two solutions are mixed.