Question

In which of the following cases is the solution of \(\mathrm{AgCl}\) unsaturated? (a) \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]K_{s p}\) (c) \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=K_{s p}\) (d) \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \leq K_{s p}\)

Short answer

The solution of AgCl is unsaturated in case (a), where the product of ionic concentrations is less than Ksp.

Step-by-step solution

Understanding Saturated Solutions and Solubility Product

A solution is said to be saturated when it contains the maximum concentration of ions that can be dissolved in it without forming a precipitate. The solubility product constant (Ksp) helps to determine this saturation point. If the product of the ionic concentrations is equal to the Ksp, the solution is at equilibrium and is saturated. If it's less, the solution is unsaturated. If it's more, precipitation occurs, and the solution is supersaturated.

Comparing Ionic Product with Ksp

In option (a), the product of the ions' concentrations is less than Ksp, which indicates that more solute can be dissolved; therefore, it's unsaturated. In option (b), the product is greater than Ksp, implying that the solution is supersaturated and precipitation should occur. Option (c) equates the product of ions' concentration with Ksp, pointing to a saturated solution. Option (d), however, suggests that the ion product is either less than or equal to Ksp, so it may be saturated or unsaturated.

Identifying the Condition for an Unsaturated Solution

An unsaturated solution is characterized by the potential to dissolve additional solute. Therefore, we look for a condition where \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] < K_{\text {sp }}\). Option (a) \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]<K_{\text {sp }}\) clearly indicates that the solution is unsaturated because the ionic product is less than the solubility product constant.

Key concepts

Unsaturated Solution

Understanding the concept of an unsaturated solution is fundamental when studying solubility and solutions in chemistry. An unsaturated solution is one that has the ability to dissolve more solute into it.

In simpler terms, if you have a glass of water and you start stirring sugar into it, there will be a point where no more sugar dissolves and it starts to accumulate at the bottom of the glass. Before reaching that point, while sugar can still dissolve, the solution is considered unsaturated.

Regarding the exercise concerning \(\mathrm{AgCl}\), we identified that option (a) \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]

Ionic Product

The term ionic product refers to the product of the concentrations of the ions in a solution. For ionic compounds like \(\mathrm{AgCl}\), which dissociate into \(\mathrm{Ag}^{+}\) and \(\mathrm{Cl}^{-}\) ions, the ionic product is the product of the molar concentrations of these ions.

The comparison of the ionic product with the solubility product constant (Ksp) is a crucial step in determining whether a solution is unsaturated, saturated, or supersaturated. As per the exercise:

• If the ionic product is less than Ksp, the solution can still dissolve more ions and is unsaturated (option a).
• If the ionic product is equal to Ksp, the solution is saturated and in a state of dynamic equilibrium (option c).
• If it's greater than Ksp, the solution has too many ions and precipitation will occur (option b).

Knowing the ionic product of a solution helps us predict whether a solute will continue to dissolve or start to precipitate.

Chemical Equilibrium

The chemical equilibrium in the context of solubility is a state where the rate of dissolution of the solute equals the rate of precipitation. At this point, the system's concentrations of reactants and products remain constant over time. It is depicted mathematically by the solubility product constant (Ksp).

When the ionic product in a solution equals Ksp, the solution is saturated, meaning it's at equilibrium as option (c) \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=K_{s p}\) suggests.

This equilibrium is dynamic, meaning that individual ions continue to dissolve and precipitate, but the overall amounts stay the same. Understanding this concept is crucial for predicting the behavior of salts in solution and for conducting various analytical chemistry techniques.

Precipitation

The term precipitation refers to the formation of a solid from a solution. This occurs when the concentration of dissolved ions exceeds the solubility limit, often described by Ksp.

In the provided exercise, option (b) \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]>K_{s p}\) indicates that \(\mathrm{AgCl}\) will precipitate out of the solution because the ionic product is greater than the solubility product constant.

The understanding of precipitation is essential for various applications in chemistry, from the purification of compounds to the analysis of water hardness. Moreover, being able to control precipitation reactions is fundamental in industrial processes where certain compounds need to be recovered or removed from a solution.

Ionic Product

The term ionic product refers to the product of the concentrations of the ions in a solution. For ionic compounds like \(\mathrm{AgCl}\), which dissociate into \(\mathrm{Ag}^{+}\) and \(\mathrm{Cl}^{-}\) ions, the ionic product is the product of the molar concentrations of these ions.

The comparison of the ionic product with the solubility product constant (Ksp) is a crucial step in determining whether a solution is unsaturated, saturated, or supersaturated. As per the exercise:

• If the ionic product is less than Ksp, the solution can still dissolve more ions and is unsaturated (option a).
• If the ionic product is equal to Ksp, the solution is saturated and in a state of dynamic equilibrium (option c).
• If it's greater than Ksp, the solution has too many ions and precipitation will occur (option b).

Knowing the ionic product of a solution helps us predict whether a solute will continue to dissolve or start to precipitate.

Chemical Equilibrium

The chemical equilibrium in the context of solubility is a state where the rate of dissolution of the solute equals the rate of precipitation. At this point, the system's concentrations of reactants and products remain constant over time. It is depicted mathematically by the solubility product constant (Ksp).

When the ionic product in a solution equals Ksp, the solution is saturated, meaning it's at equilibrium as option (c) \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=K_{s p}\) suggests.

This equilibrium is dynamic, meaning that individual ions continue to dissolve and precipitate, but the overall amounts stay the same. Understanding this concept is crucial for predicting the behavior of salts in solution and for conducting various analytical chemistry techniques.

Precipitation

The term precipitation refers to the formation of a solid from a solution. This occurs when the concentration of dissolved ions exceeds the solubility limit, often described by Ksp.

In the provided exercise, option (b) \(\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]>K_{s p}\) indicates that \(\mathrm{AgCl}\) will precipitate out of the solution because the ionic product is greater than the solubility product constant.

The understanding of precipitation is essential for various applications in chemistry, from the purification of compounds to the analysis of water hardness. Moreover, being able to control precipitation reactions is fundamental in industrial processes where certain compounds need to be recovered or removed from a solution.