Calculate the molar solubility of \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \quad\left(K_{\mathrm{sp}}=\right.\) \(1.75 \times 10^{-13}\) ) in \(0.13 M \mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\)

The solubility product constant, denoted as \(K_{sp}\), is a crucial concept in chemistry. It helps us understand the solubility of sparingly soluble salts in water. \(K_{sp}\) is a type of equilibrium constant that applies specifically to the dissolution process of salts. When a salt dissolves in water, it separates into its constituent ions. The \(K_{sp}\) value quantifies the maximum amount of the salt that can dissolve in water to form a saturated solution.
For example, the solubility product of mercury(I) oxalate (\(\text{Hg}_{2}\text{C}_{2}\text{O}_{4}\)) is given as \(1.75 \times 10^{-13}\). This means, at equilibrium, the product of the molar concentrations of its ions in solution will equal this constant. Understanding \(K_{sp}\) helps predict whether a precipitate will form in a reaction mixture and how soluble different salts are under various conditions.

The dissociation equation shows how a compound breaks down into its ions when dissolved in water. For mercury(I) oxalate, the dissociation equation is:
\[ \text{Hg}_{2}\text{C}_{2}\text{O}_{4} (s) \rightleftharpoons \text{Hg}_{2}^{2+} (aq) + \text{C}_{2}\text{O}_{4}^{2-} (aq) \]
This equation tells us that one mole of solid mercury(I) oxalate produces one mole of mercury(I) ions (\(\text{Hg}_{2}^{2+}\)) and one mole of oxalate ions (\(\text{C}_{2}\text{O}_{4}^{2-}\)) in solution. By writing out the dissociation equation, we can see the relationship between the solid salt and the ions it forms, which is essential for solving the problem of determining molar solubility.

Molar concentration (molarity) refers to the number of moles of solute per liter of solution. In this exercise, we're dealing with the molar concentration of ions in a solution. For mercury(I) oxalate dissolving in water where there's already \(0.13 \text{M}\) of mercury(I) nitrate, the initial concentration of \(\text{Hg}_{2}^{2+}\) ions is \(0.13 \text{M}\). This is crucial because it simplifies our calculation.
The molar concentration of \(\text{C}_{2}\text{O}_{4}^{2-}\) ions, represented by \(s\), is unknown and what we aim to find. The final task will be to identify this \(s\) value which gives us the molar solubility of \(\text{Hg}_{2}\text{C}_{2}\text{O}_{4}\) in the given solution.

The \(K_{sp}\) expression is an equation derived from the solubility product principle. It relates the molar concentrations of ions to the \(K_{sp}\). For \(\text{Hg}_{2}\text{C}_{2}\text{O}_{4}\), the \(K_{sp}\) expression is:
\[ K_{sp} = [\text{Hg}_{2}^{2+}] [\text{C}_{2}\text{O}_{4}^{2-}] \]
In this scenario, the concentration of \(\text{Hg}_{2}^{2+}\) is \(0.13 \text{M}\) from \(\text{Hg}_{2}(\text{NO}_{3})_{2}\) dissolution, and the concentration of \(\text{C}_{2}\text{O}_{4}^{2-}\) is \(s\). We assume \(0.13 + s \approx 0.13\) because the amount of \(\text{Hg}_{2}\text{C}_{2}\text{O}_{4}\) dissolving is very small compared to \(0.13 \text{M}\). Simplifying the \(K_{sp}\) expression, we get:
\[ K_{sp} \approx 0.13 \times s \]
By substituting the given value of \(K_{sp}\), we can solve for the molar solubility \(s\). This approach provides a straightforward method to find out how much of the compound can dissolve in the solution.