When aqueous \(\mathrm{KI}\) is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. ( Hint: \(\mathrm{Hg}^{2+}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{HgI}_{4}{ }^{2-} .\) )

Understanding the process of precipitate formation is vital in chemistry, especially when it comes to reactions like the one between potassium iodide (KI) and mercury(II) nitrate. A precipitate is an insoluble solid that emerges from a liquid solution. The color or appearance of the precipitate often gives clues to its composition.

In our exercise, the reaction between aqueous KI and mercury(II) nitrate leads to the formation of mercury(II) iodide, an orange solid. This happens because the iodide ions (I^-) and mercury(II) ions (Hg²⁺) combine to form the mercury(II) iodide (HgI₂) precipitate. The potassium ions (K⁺) and nitrate ions (NO₃^-) remain in solution, forming potassium nitrate.

The formation of a precipitate is an indispensable concept not only in laboratory chemistry but also in industrial processes, where precipitation reactions are used for purification and isolation of specific compounds.

The concept of complex ion chemistry comes into play when exploring the behavior of metal ions in solution. A complex ion consists of a central metal ion surrounded by molecules or anions, termed ligands. These ligands donate electron pairs to the metal ion, forming a coordination compound.

When the exercise mentions the addition of more potassium iodide to the existing solution containing the orange precipitate, the extra I^- ions act as ligands and interact with HgI₂ to form a complex ion: HgI₄^2-. Although the initial precipitate is insoluble, the newly formed complex ion is soluble in water, which results in the dissolution of the orange precipitate. This dissolution is a striking example of complex ion chemistry in action, where the formation of complex ions can alter the solubility of compounds, leading to practical applications in various fields including metallurgy, analytical chemistry, and medicine.

The skill of balancing chemical equations is foundational in chemistry. Every equation must obey the law of conservation of mass, meaning the number of atoms of each element must be the same on both sides of the equation.

Let's review our exercise. The first reaction between KI and Hg(NO₃)₂ is already balanced because there are equal amounts of each element on both sides of the equation. In chemical reactions like this, we must ensure conservation by checking that each element's atom count is consistent from reactants to products.

For the second reaction involving the complex ion, the equation must be balanced to reflect that two additional iodide ions were needed to fully coordinate with the mercury(II) ion and dissolve the precipitate. This highlights that in balancing equations, it is essential to account for all species involved, including the ions that form complexes. Balancing these equations correctly is crucial for accurately describing the stoichiometry of the reaction and for quantitative analysis in chemistry.