91Ó°ÊÓ

When two aqueous solutions are mixed, a chemical reaction may occur to form an insoluble solid known as a precipitate. In the case of this exercise, when solutions of silver nitrate \((\mathrm{AgNO}_3)\) and zinc chloride \((\mathrm{ZnCl}_2)\) are combined, a precipitate of silver chloride \((\mathrm{AgCl})\) forms. This happens because silver ions \((\mathrm{Ag}^+)\) and chloride ions \((\mathrm{Cl}^-)\) react together and produce the solid precipitate, \(\mathrm{AgCl}\).
\(\mathrm{Ag}^+\) from \(\mathrm{AgNO}_3\) and \(\mathrm{Cl}^-\) from \(\mathrm{ZnCl}_2\) are present in the solution in lower and higher concentrations respectively. Notably, not all ions participate in forming the precipitate. For example, \(\mathrm{Zn}^{2+}\) and \(\mathrm{NO}_3^-\) remain in solution, swimming freely in the liquid, as they don't contribute to the creation of the precipitate.
This is a typical example of a precipitation reaction, where the less soluble compound settles out of the solution as a solid.

In the context of precipitation reactions like this one, chemical equilibrium involves the balance between the dissolved ions and the precipitate that has formed. Once the \(\mathrm{AgCl}\) solid forms, a small fraction of it may dissolve back into the solution until a state of equilibrium is achieved.
This equilibrium can be expressed using the solubility product constant \(K_{sp}\), which defines the concentrations of the ions in solution that are in equilibrium with the solid precipitate. At equilibrium, the rate at which \(\mathrm{AgCl}\) dissolves equals the rate at which it forms, maintaining a balanced system.
In this solution, once all \(\mathrm{Ag}^+\) ions are used up by precipitating \(\mathrm{AgCl}\), equilibrium is reached and further dissolution of \(\mathrm{AgCl}\) gives rise to a negligible amount of \(\mathrm{Ag}^+\) and \(\mathrm{Cl}^-\) ions, important for calculating their concentrations.

The solubility product constant, \(K_{sp}\), is an indicator of the solubility of a compound under given conditions. It reveals how much of a compound can dissolve in water to form a saturated solution.
For \(\mathrm{AgCl}\), the \(K_{sp}\) is very small \((1.77 \times 10^{-10})\), indicating it is only slightly soluble in water. The formula \([\mathrm{Ag}^+][\mathrm{Cl}^-] = K_{sp}\) describes the relationship at equilibrium in saturated solutions.
In this problem, to find the concentration of dissolved ions at equilibrium, you substitute the \(K_{sp}\) value. Due to its low value, changes in \([\mathrm{Ag}^+]\) from dissolution do not significantly affect \([\mathrm{Cl}^-]\) in our scenario, making simplifications possible when determining the concentrations.

Stoichiometry provides a numerical relationship between reactants and products in a chemical reaction. It helps determine the concentrations of substances involved.
In a reaction between \(\mathrm{Ag}^+\) ions and \(\mathrm{Cl}^-\) ions, there's a 1:1 stoichiometry. This means that one mole of \(\mathrm{Ag}^+\) reacts with one mole of \(\mathrm{Cl}^-\) to form one mole of \(\mathrm{AgCl}\). As a result, the decrease in concentration of \(\mathrm{Ag}^+\) is equal to the decrease in \(\mathrm{Cl}^-\).
Understanding stoichiometry allows us to calculate how the initial concentrations of solutions change during reactions. As seen, \(\mathrm{Ag}^+\) is the limiting reactant since it is present at a lower concentration, dictating the extent of the reaction until equilibrium.